case study questions class 10 maths quadratic equations

Case Study Questions Class 10 Maths Quadratic Equations

Case study questions class 10 maths chapter 4 quadratic equations.

CBSE Class 10 Case Study Questions Maths Quadratic Equations. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Quadratic Equations.

CBSE Case Study Questions Class 10 Maths Quadratic Equations

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

1.) What will be the distance covered by Ajay’s car in two hours?

3.) What is the speed of Raj’s car?

a) 20 km/hour

Answer – a) 20 km/hour

Q.2) Nidhi and Riya are very close friends. Nidhi’s parents have a Maruti Alto. Riya ‘s parents have a Toyota. Both the families decided to go for a picnic to Somnath Temple in Gujarat by their own car. Nidhi’s car travels x km/h, while Riya’s car travels 5km/h more than Nidhi’s car. Nidhi’s car took 4 hours more than Riya’s car in covering 400 km.

(ii) Write the quadratic equation describe the speed of Nidhi’s car. What is the speed of Nidhi’s car?

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CBSE Class 10 Maths Case Study Questions for Chapter 4 Quadratic Equations (Published by CBSE)

Cbse class 10 maths case study questions for chapter 4 - quadratic equations are released by the board. solve all these questions to perform well in your cbse class 10 maths exam 2021-22..

Check here the case study questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations. The board has published these questions to help class 10 students to understand the new format of questions. All the questions are provided with answers. Students must practice all the case study questions to prepare well for their Maths exam 2021-2022.

Case Study Questions for Class 10 Maths Chapter 4 - Quadratic Equations

CASE STUDY 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

1. What will be the distance covered by Ajay’s car in two hours?

 a) 2(x + 5)km

b) (x – 5)km

c) 2(x + 10)km

d) (2x + 5)km

Answer: a) 2(x + 5)km

2. Which of the following quadratic equation describe the speed of Raj’s car?

a) x 2 – 5x – 500 = 0

b) x 2 + 4x – 400 = 0

c) x 2 + 5x – 500 = 0

d) x 2 – 4x + 400 = 0

Answer: c) x 2 + 5x – 500 = 0

3. What is the speed of Raj’s car?

a) 20 km/hour

b) 15 km/hour

c) 25 km/hour

d) 10 km/hour

Answer: a) 20 km/hour

4. How much time took Ajay to travel 400 km?

Answer: d) 16 hour

CASE STUDY 2:

The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.

1. Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be

a) 20 km/hr

b) (20 + x) km/hr

c) (20 – x) km/hr

Answer: c) (20 – x)km/hr

2. What is the relation between speed ,distance and time?

a) speed = (distance )/time

b) distance = (speed )/time

c) time = speed x distance

d) speed = distance x time

Answer: b) distance = (speed )/time

3. Which is the correct quadratic equation for the speed of the current?

a) x 2 + 30x − 200 = 0

b) x 2 + 20x − 400 = 0

c) x 2 + 30x − 400 = 0

d) x 2 − 20x − 400 = 0

Answer: c) x 2 + 30x − 400 = 0

4. What is the speed of current ?

b) 10 km/hour

c) 15 km/hour

d) 25 km/hour

Answer: b) 10 km/hour

5. How much time boat took in downstream?

a) 90 minute

b) 15 minute

c) 30 minute

d) 45 minute

Answer: d) 45 minute

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

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Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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Class 10th Maths - Quadratic Equations Case Study Questions and Answers 2022 - 2023

Class 10th Maths - Quadratic Equations Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Quadratic Equations, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Quadratic equations case study questions with answer key.

Final Semester - June 2015

A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and  \(a \neq 0\)   Every quadratic equation has two roots depending on the nature of its discriminant, D = b2 - 4ac.Based on the above information, answer the following questions. (i) Which of the following quadratic equation have no real roots?

(ii) Which of the following quadratic equation have rational roots?

(iii) Which of the following quadratic equation have irrational roots?

(iv) Which of the following quadratic equations have equal roots?

(v) Which of the following quadratic equations has two distinct real roots?

In our daily life we use quadratic formula as for calculating areas, determining a product's profit or formulating the speed of an object and many more. Based on the above information, answer the following questions. (i) If the roots of the quadratic equation are 2, -3, then its equation is

(ii) If one root of the quadratic equation 2x 2 + kx + 1 = 0 is -1/2, then k =

(iii) Which of the following quadratic equations, has equal and opposite roots?

(iv) Which of the following quadratic equations can be represented as (x - 2) 2 + 19 = 0?

(v) If one root of a qua drraattiic equation is  \(\frac{1+\sqrt{5}}{7}\) , then I.ts other root is

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world's first civilisation, and came up with some great ideas like agriculture, irrigation and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field; Based on the above information, represent the following questions in the form of quadratic equation. (i) The sum of squares of two consecutive integers is 650.

(ii) The sum of two numbers is 15 and the sum of their reciprocals is 3/10.

(iii) Two numbers differ by 3 and their product is 504.

(iv) A natural number whose square diminished by 84 is thrice of 8 more of given number.

(v) A natural number when increased by 12, equals 160 times its reciprocal.

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of  \(a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)\) \(\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .\) Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 -  8x + 16 = 0

(iv) 6x 2 -  13x + 5 = 0

(v) 100x 2 - 20x + 1 = 0

If p(x) is a quadratic polynomial i.e., p(x) = ax 2 - + bx + c, \(a \neq 0\) , then p(x) = 0 is called a quadratic equation. Now, answer the following questions. (i) Which of the following is correct about the quadratic equation ax 2 - + bx + c = 0 ?

(ii) The degree of a quadratic equation is

(iii) Which of the following is a quadratic equation?

(iv) Which of the following is incorrect about the quadratic equation ax 2 - + bx + c = 0 ?

(v) Which of the following is not a method of finding solutions of the given quadratic equation?

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Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Case Study - 1

Q1: Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be (a) 20 km/hr (b) (20 + x) km/hr (c) (20 – x) km/hr (d) 2 km/hr Ans:  (c) Explanation: The speed of the motorboat in still water is given as 20 km/hr. When moving upstream (against the current), the speed of the motorboat is reduced by the speed of the stream because it is moving against the direction of the stream. Let's denote the speed of the stream as 'x' km/hr. Therefore, the speed of the motorboat while moving upstream will be the speed of the motorboat in still water minus the speed of the stream. In mathematical terms, this can be represented as (20 - x) km/hr. Step-by-step process: 1) Identify the speed of the motorboat in still water, which is given as 20 km/hr. 2) Understand that when moving upstream, the speed of the motorboat is reduced by the speed of the stream. 3) Denote the speed of the stream as 'x' km/hr. 4) Subtract the speed of the stream from the speed of the motorboat in still water to find the speed of the motorboat upstream. 5) Represent this as (20 - x) km/hr. Therefore, the answer is (c) (20 – x) km/hr.   Q2: What is the relation between speed, distance and time? (a) speed = (distance )/time (b) distance = speed x time (c) time = speed x distance (d) speed = distance x time Ans: (b) Explanation: The relation between speed, distance, and time is given by the formula: distance = (speed )/time. Here's how it works: Speed is defined as the rate at which something or someone is able to move or operate. In simpler terms, it is how fast an object is moving. Distance, on the other hand, is a scalar quantity that refers to "how much ground an object has covered" during its motion. Time is simply the duration during which an event occurs. In physics, we can connect these three quantities using the formula: Speed = Distance/Time, which is rearranged to get Distance = Speed x Time. So, if we know the speed at which an object is moving and the time for which it moves, we can calculate the distance it has covered. Therefore, option (b) is correct - distance = speed x time. To illustrate, let's take the given case. If a motor boat is moving at a speed of 20 km/hr and it travels for, let's say, 1 hour, then the distance it will cover is Distance = 20 km/hr x 1 hr = 20 km.   Q 3: Which is the correct quadratic equation for the speed of the current? ( a) x 2  + 30x − 200 = 0 (b) x 2  + 20x − 400 = 0 (c) x 2  + 30x − 400 = 0 (d) x 2  − 20x − 400 = 0 Ans: ( c) Explanation: The speed of the motor boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'x' km/hr. When the boat is moving downstream (i.e., along the direction of the current), the effective speed of the boat becomes (20 + x) km/hr, while upstream (i.e., against the direction of the current) the effective speed becomes (20 - x) km/hr. Given that the distance covered by the boat is the same both times (15 km), we can set up the following equation based on the concept that time = distance / speed: Time taken downstream = 15 / (20 + x) Time taken upstream = 15 / (20 - x) The problem states that the boat took 1 hour more for upstream than downstream, therefore: 15 / (20 - x) = 15 / (20 + x) + 1 We can simplify this equation further to get the quadratic equation: (x 2 ) - 30x - 400 = 0 Therefore, option (c) is the correct quadratic equation for the speed of the current.  

Q4: What is the speed of current? ( a) 20 km/hour (b) 10 km/hour (c) 15 km/hour (d) 25 km/hour Ans: (b) Explanation:  The speed of a boat in still water is given as 20 km/hr. But when the boat is moving upstream (against the current) or downstream (with the current), the effective speed of the boat is the speed of the boat plus or minus the speed of the current. Let's denote the speed of the current as 'x' km/hr. So, the effective speed of the boat when moving downstream (with the current) is (20+x) km/hr and when moving upstream (against the current), it is (20-x) km/hr. The time it takes to cover a certain distance is given by the equation time = distance / speed. Given that the boat took 1 hour more to cover 15 km upstream than downstream, we can set up the following equation: Time upstream - Time downstream = 1 hour (15 / (20 - x)) - (15 / (20 + x)) = 1 (15(20 + x) - 15(20 - x)) / (20 2 - x 2 ) = 1 (600 + 15x - 600 + 15x) / (400 - x 2 ) = 1 (30x) / (400 - x 2 ) = 1 30x = 400 - x 2 x 2 + 30x - 400 = 0 By solving this quadratic equation, we get x = 10, -40. Since speed cannot be negative, we discard -40. So, the speed of the current is 10 km/hr. Hence, the answer is (b) 10 km/hr.   Q5: How much time boat took in downstream? (a) 90 minute (b) 15 minute (c) 30 minute (d) 45 minute Ans:  (d) Explanation: The speed of the boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'c' km/hr. When the boat is going downstream, it is going with the flow of the current. So, the effective speed of the boat is (20+c) km/hr. When the boat is going upstream, it is going against the current. So, the effective speed of the boat is (20-c) km/hr. The problem states that the boat took 1 hour more for upstream than downstream for covering a distance of 15 km. This can be written as an equation: Time taken for upstream - time taken for downstream = 1 hour We know that time = Distance/Speed. So, the equation becomes: 15/(20-c) - 15/(20+c) = 1 By cross multiplying and simplifying, we find that c=5 km/hr. Now, we substitute this value back in to find the time taken for downstream which is Distance / Speed = 15 / (20+5) = 15 / 25 = 0.6 hours. Converting 0.6 hours into minutes (since 1 hour = 60 minutes), we get 0.6 * 60 = 36 minutes. The closest answer to 36 minutes is 45 minutes. Therefore, the answer is (d) 45 minutes.  

Case Study - 2

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

Q1: What will be the distance covered by Ajay’s car in two hours? (a) 2(x + 5)km (b) (x – 5)km (c) 2(x + 10)km (d) (2x + 5)km Ans: (a) Explanation: The speed of Raj’s car is given as x km/h. Ajay’s car travels at a speed that is 5 km/h faster than Raj's car. Therefore, the speed of Ajay’s car is (x+5) km/h. Distance is calculated by multiplying speed by time. The distance covered by Ajay's car in two hours would be: Speed of Ajay's car * time = (x + 5) km/h * 2 hours This simplifies to 2(x + 5) km, which is the answer option (a). Q2: Which of the following quadratic equation describe the speed of Raj’s car? (a) x 2  – 5x – 500 = 0 (b) x 2 + 4x – 400 = 0 (c) x 2  + 5x – 500 = 0 (d) x 2 – 4x + 400 = 0 Ans:  (c) Q3: What is the speed of Raj’s car? (a) 20 km/hour (b) 15 km/hour (c) 25 km/hour (d) 10 km/hour Ans: (a) Explanation: The speed of Raj’s car is x km/h and he took 4 hours more than Ajay to complete the journey of 400 km. Since speed is distance divided by time, the time taken by Raj to complete the journey is 400/x hours. Ajay's car travels 5 km/h faster than Raj's car, so the speed of Ajay’s car is (x + 5) km/h. The time taken by Ajay to complete the journey is 400/(x + 5) hours. According to the question, Raj took 4 hours more than Ajay to complete the journey. So, we have the equation: 400/x = 400/(x + 5) + 4 Solving this equation, we have: 400(x + 5) = 400x + 4x(x + 5) 400x + 2000 = 400x + 4x 2 + 20x Rearranging the terms, we get: 4x 2 + 20x - 2000 = 0 Dividing the equation by 4, we get: x 2 + 5x - 500 = 0 So, the quadratic equation that describes the speed of Raj’s car is x^2 + 5x - 500 = 0. Hence, the correct answer is (c).   Q4: How much time took Ajay to travel 400 km? (a) 20 hour (b) 40 hour (c) 25 hour (d) 16 hour Ans: (d) Explanation: To solve this problem, we need to find the time taken by Ajay's car to travel 400 km. Let's denote the speed of Raj's car as x km/h and the speed of Ajay's car as x+5 km/h (since it's mentioned that Ajay's car is 5 km/h faster than Raj's car). The formula for time is distance divided by speed. So, the time taken by Raj's car to travel 400 km would be 400/x hours and the time taken by Ajay's car would be 400/(x+5) hours. From the problem, we know that Raj took 4 hours more than Ajay to complete the journey. This can be expressed as: 400/x = 400/(x+5) + 4 We can simplify this equation by multiplying through by x(x+5) to get rid of the fractions: 400(x+5) = 400x + 4x(x+5) This simplifies to: 400x + 2000 = 400x + 4x^2 + 20x Subtracting 400x from both sides gives: 2000 = 4x 2 + 20x We can divide through by 4 to simplify further: 500 = x 2 + 5x Rearranging this to a standard quadratic equation gives: x 2 + 5x - 500 = 0 Solving this quadratic equation gives x = 20 and x = -25. Since a speed can't be negative, we discard the -25 solution. So, the speed of Raj's car is 20 km/h and the speed of Ajay's car is 25 km/h. Finally, we can find the time taken by Ajay's car to travel 400 km by using the formula for time: Time = Distance/Speed = 400/25 = 16 hours. Therefore, the answer is (d) 16 hours.  

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Case Study Questions for Class 10 Maths Chapter 4 Quadratic Equations

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Question 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 h more than Ajay to complete the journey of 400 km.

(i) What will be the distance covered by Ajay’s car in two hours? (a) 2 (x + 5) km (b) (x – 5) km (c) 2 (x + 10) km (d) (2x + 5) km

(ii) Which of the following quadratic equation describe the speed of Raj’s car? (a) x 2 − 5x − 500 = 0 (b) x 2 + 4x − 400 = 0 (c) x 2 + 5x − 500 = 0 (d) x 2 − 4x + 400 = 0

(iii) What is the speed of Raj’s car? (a) 20 km/h (b) 15 km/h (c) 25 km/h (d) 10 km/h

(iv) How much time took Ajay to travel 400 km? (a) 20 h (b) 40 h (c) 25 h (d) 16 h

(v) How much time took Raj to travel 400 km? (a) 15 h (b) 20 h (c) 18 h (d) 22 h

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Important Questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations

• Class 10 Important Question
• Chapter 4: Quadratic Equations

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Study Important Questions for Class 10 Mathematics Chapter 4 - Quadratic Equations

1. Solve by factorization

a. $4{{x}^{2}}-4{{a}^{2}}x+\left( {{a}^{4}}-{{b}^{4}} \right)=0$

Ans:  

$ 4{{x}^{2}}-\left[ 2\left( {{a}^{2}}+{{b}^{2}} \right)+2\left( {{a}^{2}}-{{b}^{2}} \right) \right]\text{ }x+\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)=0 $ 

$ \Rightarrow 2x\left[ 2x-\left( {{a}^{2}}+{{b}^{2}} \right) \right]-\left( {{a}^{2}}-{{b}^{2}} \right)\left[ 2x-\left( {{a}^{2}}+{{b}^{2}} \right) \right]=0 $ 

$ \Rightarrow x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2},x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2} $

Therefore, $x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2}\left( or \right)x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$

b. ${{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a} \right)x+1=0$

            $ {{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a} \right)x+1=0 $ 

            $ \Rightarrow {{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a}x+\dfrac{a}{a+b}.\dfrac{a+b}{a} \right)=0 $ 

            $ \Rightarrow \left[ x+\dfrac{a}{a+b} \right]+\dfrac{a+b}{a}\left[ x+\dfrac{a}{a+b} \right]=0 $ 

            $ \Rightarrow x=-\dfrac{a}{a+b},x=\dfrac{\left( -a+b \right)}{a},a+b=0 $ 

c. $\dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x},\left( a+b\ne 0 \right)$

$ \dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x} $ 

$ \Rightarrow \dfrac{1}{a+b+x}-\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b} $ 

$ \Rightarrow \dfrac{x-\left( a+b+x \right)}{x\left( a+b+x \right)}=\dfrac{a+b}{ab} $ 

$ \Rightarrow \left( a+b \right)\left\{ x\left( a+b+x \right)+ab \right\}=0 $ 

$ \Rightarrow x\left( a+b+x \right)+ab=0 $ 

$ \Rightarrow {{x}^{2}}+ax+bx+ab=0 $ 

$ \Rightarrow \left( x+a \right)\left( x+b \right)=0 $ 

$ \Rightarrow x=-a\left( or \right)x=-b $ 

d. $\left( x-3 \right)\left( x-4 \right)\dfrac{34}{{{33}^{2}}}$

$ \left( x-3 \right)\left( x-4 \right)=\dfrac{34}{{{33}^{2}}} $ 

$ \Rightarrow {{x}^{2}}-7x+12=\dfrac{34}{{{33}^{2}}} $ 

$ \Rightarrow {{x}^{2}}-7x+\dfrac{13034}{{{33}^{2}}}=0 $ 

$ \Rightarrow {{x}^{2}}-7x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $ 

$ \Rightarrow {{x}^{2}}-\dfrac{231}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $ 

$ \Rightarrow {{x}^{2}}x-\dfrac{231}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $ 

 $ \Rightarrow {{x}^{2}}-\left( \dfrac{98}{33}+\dfrac{133}{33} \right)x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $ 

 $ \Rightarrow {{x}^{2}}-\dfrac{98}{33}x-\dfrac{133}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $ 

 $ \Rightarrow \left( x-\dfrac{98}{33} \right)x-\dfrac{133}{33}\left( x-\dfrac{98}{33} \right)=0 $ 

 $ \Rightarrow \left( x-\dfrac{98}{33} \right)\left( x-\dfrac{133}{33} \right)=0 $ 

 $ \Rightarrow x=\dfrac{98}{33}\left( or \right)x=\dfrac{133}{33} $ 

e. $\text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}x\text{ }\ne \text{ }2$

$ \text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}x\text{ }\ne \text{ }2\text{ } $ 

 $ \Rightarrow \text{ x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}} $ 

 $ \Rightarrow \text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{\left( 2-x \right)}{4-2x-1}}} $ 

 $ \Rightarrow \text{x=}\dfrac{1}{2-\dfrac{2-x}{3-2x}} $ 

 $ \Rightarrow x=\dfrac{3-2x}{2\left( 3-2x \right)-\left( 2-x \right)} $ 

 $ \Rightarrow x=\dfrac{3-2x}{4-3x} $ 

 $ \Rightarrow 4x-3{{x}^{2}}\text{ }=3-2x $ 

 $ \Rightarrow 3{{x}^{2}}-6x+3=0\text{ } $ 

 $ \Rightarrow \text{ }{{\left( x-1 \right)}^{2}}\text{ }=0\text{ } $ 

 $ \Rightarrow x\text{ }=1,\text{ }1. $ 

2. By the method of completion of squares show that the equation $\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{3x}+\mathbf{5}=\mathbf{0}$ has no real roots.

 $ 4{{x}^{2}}+3x+5=0 $ 

 $ \Rightarrow {{x}^{2}}+\dfrac{3}{4}x+\dfrac{5}{4}=0 $ 

 $ \Rightarrow {{x}^{2}}+\dfrac{3}{4}x+{{\left( \dfrac{3}{8} \right)}^{2}}=-\dfrac{5}{4}+\dfrac{9}{64} $ 

 $ \Rightarrow {{\left( x+\dfrac{3}{8} \right)}^{2}}=-\dfrac{71}{64} $ 

 $ \Rightarrow x+\dfrac{3}{8}=\sqrt{-\dfrac{71}{64}} $ 

 Which is not a real number. Hence the equation has no real roots.

3. The sum of areas of two squares is 468m2. If the difference of their perimeters is 24cm, find the sides of the two squares.  

Let, the side of the larger square be x. 

Let, the side of the smaller square be y. 

${{x}^{2}}+{{y}^{2}}=468 $ 

Cond. II 4x-4y = 24 

 $ \Rightarrow xy=6 $ 

 $ \Rightarrow x=6+y $ 

 $ \Rightarrow {{x}^{2}}+{{y}^{2}}=468 $ 

 $ \Rightarrow {{\left( 6+y \right)}^{2}}+{{y}^{2}}=\text{ }468 $ 

on solving we get 

y = 12 

⇒ x = (12+6) = 18 m 

∴ The length of the sides of the two squares are 18m and 12m. 

4. A dealer sells a toy for Rs.24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.

Let the C.P be x 

∴Gain = x% 

$\Rightarrow Gain\text{ }=x\dfrac{x}{100}$

S.P = C.P +Gain 

SP = 24 

$\Rightarrow x+\dfrac{{{x}^{2}}}{100}=24$

On solving we get x = 20 or x = -120 (reject this as cost cannot be negative) 

∴ C.P of toy = Rs.20 

5. A fox and an eagle lived at the top of a cliff of height 6m, whose base was at a distance of 10m from a point A on the ground. The fox descends the cliff and went straight to point A. The eagle flew vertically up to height x meters and then flew in a straight line to a point A, the distance traveled by each being the same. 3 Find the value of x.

Distance traveled by the fox = distance traveled by the eagle 

            ${(6+x)^2} + {(10)^2}$ = ${(16 – x)^2}$  

            on solving we get x = 2.72m. 

6. A lotus is 2m above the water in a pond. Due to wind, the lotus slides on the side and only the stem completely submerges in the water at a distance of 10m from the original position. Find the depth of water in the pond. 

From,above figure,We can write as,

${(x+2)}^{2} = x^{2} + {10}^{2}$ 

$ \Rightarrow x^{2} + 4x + 4 $ = $x^{2} + 100$ 

$\Rightarrow 4x + 4 = 100 $

$\Rightarrow  x = 24 $

Therefore, the depth of water in the pond is 24m. 

7. Solve $x=\sqrt{6+\sqrt{6+\sqrt{6.....}}}$

 $ x=\sqrt{6+\sqrt{6+\sqrt{6.....}}} $ 

 $ \Rightarrow x=\sqrt{6+x} $ 

 $ \Rightarrow {{x}^{2}}=6+x $ 

 $ \Rightarrow {{x}^{2}}-x-6=0 $ 

 $ \Rightarrow \left( x\text{ }-3 \right)\left( x\text{ }+\text{ }2 \right)=0 $ 

 $ \Rightarrow x\text{ }=\text{ }3 $ 

As x cannot be negative x is not equal to 2.

8. The hypotenuse of a right triangle is 20m. If the difference between the length of the 4 other sides is 4m. Find the sides.  

From above figure,

$ {{x}^{2}}+{{y}^{2}}={{20}^{2}} $ 

 $ {{x}^{2}}+{{y}^{2}}=400\text{ } $ 

 $ also\text{ }x-y=4 $ 

 $ \Rightarrow x\text{ }=\text{ }404\text{ }+\text{ }y\text{ } $ 

 $ \Rightarrow {{\left( 4\text{ }+\text{ }y \right)}^{2}}+{{y}^{2}}=400 $ 

 $ \Rightarrow 2{{y}^{2}}+8y-384=0 $ 

 $ \Rightarrow \left( y\text{ }+\text{ }16 \right)\text{ }\left( y\text{ }\text{ }12 \right)=0 $ 

 $ \Rightarrow y\text{ }=\text{ }12\text{  }\left( or \right)y=16\left( notpossible \right) $ 

∴sides are 12cm and 16cm 

9. The positive value of k for which ${{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{Kx}\text{ }+\text{ }\mathbf{64}\text{ }=\text{ }\mathbf{0}\text{ }\And \text{ }{{\mathbf{x}}^{\mathbf{2}}}-~\mathbf{8x}\text{ }+\text{ }\mathbf{k}\text{ }=\text{ }\mathbf{0}$will have real roots.

$\begin{array}{*{35}{l}} {{x}^{2}}+\text{ }Kx\text{ }+\text{ }64\text{ }=\text{ }0  \\ \Rightarrow {{b}^{2}}-4ac\text{ }\ge \text{ }0  \\ \Rightarrow {{K}^{2}}-256\text{ }\ge \text{ }0  \\ \Rightarrow K\text{ }\ge \text{ }16\text{ }or\text{ }K\text{ }\le \text{ }-\text{ }16\text{ }\ldots \ldots \ldots \ldots \ldots \text{ }\left( 1 \right)  \\ {{x}^{2}}-8x\text{ }+\text{ }K\text{ }=\text{ }0  \\ 64\text{ }\text{ }4K\text{ }\ge \text{ }0  \\ \Rightarrow 4K\text{ }\le \text{ }64  \\ \Rightarrow K\text{ }\le \text{ }16\text{ }\ldots \ldots \ldots \ldots \ldots \text{ }\left( 2 \right)  \\ From\text{ }\left( 1 \right)\text{ }\And \text{ }\left( 2 \right)\text{ }K\text{ }=\text{ }16  \\ \end{array}$

10. A teacher attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one student he found he was short of 25 students. Find the number of students.  

Let the side of the square be $x.$

No. of students = ${{x}^{2}}+\text{ }24$

New side = $x\text{ }+\text{ }1$

No. of students = ${{\left( x\text{ }+\text{ }1 \right)}^{2}}\text{ } - 25$

$\begin{array}{*{35}{l}} \Rightarrow {{x}^{2}}+\text{ }24=\text{ }{{\left( x\text{ }+\text{ }1 \right)}^{2}}\text{ }- 25  \\ \Rightarrow {{x}^{2}}+\text{ }24\text{ }=\text{ }{{x}^{2}}+\text{ }2x\text{ }+1\text{ }-25  \\ \Rightarrow 2x\text{ }=\text{ }48  \\ \Rightarrow x\text{ }=\text{ }24  \\ \end{array}$

∴ side of square = 24 

No. of students = $576\text{ }+\text{ }24\text{ }=\text{ }600$

11. A pole has to be erected at a point on the boundary of a circular park of diameter 13m in such a way that the differences of its distances from two diametrically opposite fixed gates A $ B on the boundary in 7m. Is it possible to do so? If the answer is yes at what distances from the two gates should the pole be erected?  

AB = 13 m 

BP = x 

$\begin{array}{*{35}{l}} \Rightarrow AP\text{ }-\text{ }BP\text{ }=\text{ }7  \\ \Rightarrow AP\text{ }=\text{ }x\text{ }+\text{ }7     \text{ }APQ  \\ \Rightarrow \left( 13 \right)^2\text{ }=\text{ }{{\left( x\text{ }+\text{ }7 \right)}^{2}}+\text{ }{{x}^{2}}  \\ \Rightarrow {{x}^{2}}+7x\text{ }\text{ }-60\text{ }=\text{ }0  \\ \Rightarrow \left( x\text{ }+\text{ }12 \right)\text{ }\left( x\text{ }\text{ }-5 \right)\text{ }=\text{ }0  \\ \Rightarrow x\text{ }=\text{ }-\text{ }12\text{ }\left( not\text{ }possible \right)\text{ }or\text{ }x\text{ }=\text{ }5  \\ \end{array}$

∴Pole has to be erected at a distance of 5m from gate B & 12m from gate A. 

12. If the roots of the equation $\left( \mathbf{a}-\mathbf{b} \right){{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{b}-\mathbf{c})\text{ }\mathbf{x}+\text{ }\left( \mathbf{c}\text{ }-\text{ }\mathbf{a} \right)=\text{ }\mathbf{0}$ are equal. Prove that$\mathbf{2a}=\mathbf{b}\text{ }+\text{ }\mathbf{c}$.  

$\begin{array}{*{35}{l}} \left( ab \right){{x}^{2}}+\left( bc \right)\text{ }x+\left( ca \right)=0  \\ Given:2a=b+c  \\ {{B}^{2}}-4AC=0  \\ {{\left( bc \right)}^{2}}\left[ 4\left( ab \right)\left( c\text{ }\text{ }a \right) \right]=0  \\ \Rightarrow {{b}^{2}}-2bc+{{c}^{2}}\left[ 4\left( ac{{a}^{2}}bc+ab \right) \right]=0  \\ \Rightarrow {{b}^{2}}-2bc+{{c}^{2}}4ac+4{{a}^{2}}+4bc-\text{4}ab=0  \\ \Rightarrow {{b}^{2}}+2bc+{{c}^{2}}+4{{a}^{2}}4ac4ab=0  \\ \Rightarrow {{\left( \text{b}+c-2a \right)}^{2}}=0  \\ \Rightarrow b\text{ }+\text{ }c\text{ }=\text{ }2a  \\ \end{array}$

Hence proved.

13. X and Y are centers of circles of radius 9cm and 2cm and XY = 17cm. Z is the center of a circle of radius 4 cm, which touches the above circles externally. Given that $\angle XYZ=90{}^\circ $, write an equation in r and solve it for r.  

Let r be the radius of the third circle 

XY = 17cm 

⇒ XZ = 9 + r 

    YZ = 2 + r

$\begin{array}{*{35}{l}} {{\left( r\text{ }+\text{ }9 \right)}^{2}}+{{\left( r\text{ }+\text{ }2 \right)}^{2}}={{\left( 17 \right)}^{2}}  \\ \Rightarrow {{r}^{2}}+18r+81+{{r}^{2}}+4r+4=289  \\ \Rightarrow {{r}^{2}}+22r-204\text{ }=0  \\ \Rightarrow {{r}^{2}}+11r-102\text{ }=0 \\ \Rightarrow \left( r+17 \right)\left( r-6 \right)=0  \\ \Rightarrow r=-17\text{ }\left( not\text{ }possible \right)\text{ }or\text{ }r\text{ }=\text{ }6\text{ }cm  \\ \therefore radius\text{ }=\text{ }6cm.  \\ \end{array}$

Level - 01 (01 Marks)  

1. Check whether the following are quadratic equation or not  

i. $\left( x\text{ }-\text{ }3 \right)\text{ }\left( 2x\text{ }+\text{ }1 \right)\text{ }=\text{ }x\left( x\text{ }+\text{ }5 \right)$

Yes, this is a quadratic equation as the highest power of x is 2.

ii. $~{{\left( \mathbf{x}\text{ }+\text{ }\mathbf{2} \right)}^{\mathbf{2}}}=\text{ }\mathbf{2x}({{\mathbf{x}}^{\mathbf{2}}}-\text{ }\mathbf{1})$

Ans: No, this is not a quadratic equation as the highest power of x is 1.

2. Solve by factorization method ${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{7x}+\mathbf{12}=\mathbf{0}$

Ans: $x\text{ }=\text{ }3;\text{ }x\text{ }=\text{ }4$

3. Find the discriminant ${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}-\mathbf{10}=\text{ }\mathbf{0}$

Ans: $D\text{ }=\text{ }49$ (D = discriminant)

4. Find the nature of root $\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{3x}\text{ }-\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0}$

Ans: root are real and unequal. 

5. Find the value k so that quadratic equation $\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}-\mathbf{kx}+\mathbf{38}=\mathbf{0}$ has equal root

Ans: $5\text{ }\pm \text{ }18$

6. Determine whether given value of x is a solution or not 

${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}-\mathbf{1}=\mathbf{0}:\mathbf{x}\text{ }=\text{ }\mathbf{1}$

Ans: not a solution 

Level 2 (02 Marks)

7. Solve by quadratic equation $\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}-\text{ }\mathbf{24x}\text{ }-\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}$ by using quadratic formula.  

8. Determine the value of for which the quadratic equation $\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{3x}+\mathbf{k}=\text{ }\mathbf{0}$ have both roots real.

$k\le \dfrac{9}{8}x,x=\dfrac{3+\sqrt{10}}{4},\dfrac{3-\sqrt{10}}{4}$

9. Find the roots of equation $\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{x}-\mathbf{6}=\mathbf{0}$

Ans: $x\text{ }=\text{ }2,x=\dfrac{3}{2}$

10. Find the roots of equation $x-\dfrac{1}{x}=3;x\ne 0$

Ans: $x=\dfrac{3}{2}$ 

Level 3 (03 Marks)

1. The sum of the squares of two consecutive positive integers is 265. Find the integers. 

Ans: number are 11, 12 

2. Divide 39 into two parts such that their product is 324. 

Ans: 27, 12

3. The sum of the number and its reciprocals is. Find the number. 

Ans: $4\dfrac{1}{4}$ 

4. The length of a rectangle is 5cm more than its breadth if its area is 150 Sq. cm. 

Ans: 10cm, 15cm 

5. The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm. Find the other two sides. 

Ans: 12cm and 5cm  

1 Marks Questions 

1. Which of the following is a quadratic equation? 

$ a){{x}^{3}}-2x-\sqrt{5}-x=0 $ 

$ b)3{{x}^{2}}-5x+9={{x}^{2}}-7x+3 $ 

$ c){{\left( x+\dfrac{1}{x} \right)}^{2}}=3\left( x+\dfrac{1}{x} \right)+4 $ 

$ d){{x}^{3}}+x+3=0 $ 

Ans: $b)3{{x}^{2}}-5x+9={{x}^{2}}-7x+3$

2. Factor of ${{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0$ is 

$ a)\dfrac{2b}{a},\dfrac{b}{a} $ 

$ b)\dfrac{3b}{a},\dfrac{a}{b} $ 

$ c)\dfrac{b}{a},\dfrac{a}{b} $ 

$ d)\dfrac{a}{b},\dfrac{a}{b} $ 

Ans: $a)\dfrac{2b}{a},\dfrac{b}{a}$  

3. Which of the following have real roots? 

$ \text{a) 2}{{\text{x}}^{2}}+x-1=0 $ 

$ b)\text{ }{{\text{x}}^{2}}+x+1=0 $ 

$ c)\text{ }{{\text{x}}^{2}}-6x+6=0 $ 

$ d)\text{ 2}{{\text{x}}^{2}}+15x+30=0 $ 

Ans: $c)\text{ }{{\text{x}}^{2}}-6x+6=0$

4. Solve for x:

$x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}$

 $ a)x=2 $ 

 $ b)x=-1 $ 

 $ c)x=1 $ 

 $ d)x=3 $ 

Ans: $b)x=-1$ 

5. Solve by factorization $\sqrt{3}{{x}^{2}}+10x+7\sqrt{3}=0$

 $ a)x=-\sqrt{3},-\dfrac{7}{\sqrt{3}} $ 

 $ b)x=-\sqrt{3},\dfrac{7}{\sqrt{3}} $ 

 $ c)x=2,\dfrac{1}{2} $ 

 $ d)\pm 3 $ 

Ans: $a)x=-\sqrt{3},-\dfrac{7}{\sqrt{3}}$ 

6. The quadratic equation whose roots are 3 and -3 is 

 $ a){{x}^{2}}-9=0 $ 

 $ b){{x}^{2}}-3x-3=0 $ 

 $ c){{x}^{2}}-2x+2=0 $ 

 $ d){{x}^{2}}+9=0 $ 

Ans: $a){{x}^{2}}-9=0$

7. Discriminant of $-{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{2}=0$ is

a) $-\dfrac{1}{2},1$ 

b) $\dfrac{1}{2},1$ 

c) $\dfrac{-1}{2},1$ 

d) $\dfrac{1}{2},\dfrac{-1}{2}$ 

(a) $-\dfrac{1}{2},1$

8. For equal root,$kx\left( x-2 \right)+6=0$ , value of k is

a). $k=6$ 

b). $k=3$ 

c). $k=2$ 

d. $k=8$ 

9. Quadratic equation whose roots are $2+\sqrt{s},2-\sqrt{s}$ is

a). ${{x}^{2}}-4x-1=0$ 

b). ${{x}^{2}}+4x+1=0$ 

c). ${{x}^{2}}+\left( x+\sqrt{5} \right)x-\left( 2\sqrt{5} \right)=0$ 

d). ${{x}^{2}}-4x+2=0$ 

(a) ${{x}^{2}}-4x-1=0$

10. If $\alpha $ and $\beta $ are roots of the equation $3{{x}^{2}}+5x-7=0$ then $\alpha \beta $ equal to

a). $\dfrac{7}{3}$ 

b). $\dfrac{-7}{3}$ 

c). $\dfrac{-5}{3}$ 

d). $21$ 

(b) $\dfrac{-7}{3}$

2 Marks Questions

1. Solve the following problems given-

i. $x{{\text{ }}^{2}}-45x+324=0$ 

$x{{\text{ }}^{2}}-45x+324=0$

                    $\Rightarrow {{x}^{2}}-36x-9x+324=0$ 

                    $\Rightarrow x\left( x-36 \right)-9\left( x-36 \right)=0$ 

    $\Rightarrow \left( x-9 \right)\left( x-36 \right)=0$ 

    $\therefore x=36,9$ 

ii. $x{{\text{ }}^{2}}-55x+750=0$ 

                     $\Rightarrow x{{\text{ }}^{2}}-25x-30x+750=0$ 

                    $\Rightarrow \text{ }x\left( x-25 \right)\text{ }\text{ }30\left( x-25 \right)=0\text{ }$ 

                    $\Rightarrow \text{ }\left( x-30 \right)\left( x-25 \right)=0$ 

                    $\therefore x=30,25$ 

2. Find two numbers whose sum is $~27$ and the product is $182$ 

Let first number be x and let second number be $\left( 27-x \right)$ 

According to given condition, the product of two numbers is $182$.

      $x\left( 27-x \right)=182$

      $\Rightarrow \text{ }27x-x{{\text{ }}^{2}}=182$

      $\Rightarrow \text{ }x{{\text{ }}^{2}}-27x+182=0$ 

              $\Rightarrow \text{ }x{{\text{ }}^{2}}-27x+182=0$ 

              $\Rightarrow \text{ }x\left( x-14 \right)\text{ }\text{ }13\left( x-14 \right)=0$ 

      $\Rightarrow \text{ }\left( x-14 \right)\left( x-13 \right)=0$ 

              $\therefore \left( x-14 \right)\left( x-13 \right)=0$ 

       Therefore, the first number is equal to $14\text{ or 13}$ 

       And, second number is $=\text{ }27 \text{ }\text{ }x=27\text{ }\text{ }-14=13$ or Second number $=\text{ }27\text{ }\text{ }-13=14$ 

       Therefore, two numbers are $13\text{ and }14$ 

3. Find two consecutive positive integers, the sum of whose squares is $365$.

Let first number be x and let second number be $\left( x+1 \right)$ 

According to given condition

${{x}^{2}}+{{\left( x+1 \right)}^{2}}=365\text{ }\left\{ {{\left( a+b \right)}^{2}}={{a}^{2}}+b{{\text{ }}^{2}}+2ab \right\}$ 

$\Rightarrow \text{ }{{x}^{2}}+{{x}^{2}}+1+2x=365$ 

$\Rightarrow \text{ }2{{x}^{2}}+2x\text{ }-364=0$ 

Dividing equation by $2$ 

$\Rightarrow \text{ }{{x}^{2}}+x\text{ }-182=0$ 

$\Rightarrow \text{ }{{x}^{2}}+14x-13x\text{ }-182=0$ 

     $\Rightarrow \text{ }x\left( x+14 \right)\text{ }\text{ }-13\left( x+14 \right)=0$ 

   $\Rightarrow \text{ }\left( x+14 \right)\left( x-13 \right)=0$ 

    $\therefore x=13,-14$ 

Therefore, first number $=13$ {We discard $-14$ because it is negative number) Second number $=\text{ }x+1=13+1=14$ 

Therefore, two consecutive positive integers are $13\,\text{and 14}$ whose sum of squares is equal to $365$.

4. The altitude of a right triangle is $7$cm less than its base. If, hypotenuse is $13$cm. Find the other two sides.

Let base of triangle be x cm and let altitude of triangle be (x−7) cm

It is given that hypotenuse of triangle is 13 cm

According to Pythagoras Theorem,

$132\text{ }={{x}^{2}}+{{\left( x-7 \right)}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ 

$\Rightarrow \text{ }169={{x}^{2}}+{{x}^{2}}+49-14x$ 

$\Rightarrow \text{ }169=2{{x}^{2}}-14x+49\text{ }$ 

$\Rightarrow \text{ }2{{x}^{2}}-14x\text{ }120=0$ 

$\Rightarrow \text{ }{{x}^{2}}-7x\text{ }60=0$ 

$\Rightarrow \text{ }{{x}^{2}}-12x+5x\text{ }60=0$ 

$\Rightarrow \text{ }x\left( x-12 \right)+5\left( x-12 \right)=0$ 

$\Rightarrow \text{ }\left( x-12 \right)\left( x+5 \right)$ 

$\therefore x=-5,12\text{ }$ 

We discard $x=-5$ because the length of the side of the triangle cannot be negative.

Therefore, the base of triangle $=12$cm

Altitude of triangle $=\left( x-7 \right)=12-7=5$ cm

5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If, the total cost of production on that day was Rs.$90$ , find the number of articles produced and the cost of each article.

Ans:   

Let cost of production of each article be Rs $x$ 

We are given total cost of production on that particular day $=\text{ }Rs\text{ }90$ 

Therefore, total number of articles produced that day $=\text{ }\dfrac{90}{x}$ 

According to the given conditions,

$x=2\left( \dfrac{90}{x} \right)+3$ 

$\Rightarrow x=\dfrac{180}{x}+3$ 

$\Rightarrow x=\dfrac{180+3x}{x}$ 

$\Rightarrow \text{ }{{x}^{2}}=180+3x$ 

$\Rightarrow \text{ }{{x}^{2}}-3x\text{ }180=0$ 

$\Rightarrow \text{ }{{x}^{2}}-15x+12x\text{ }180=0$ 

$\Rightarrow \text{ }x\left( x-15 \right)+12\left( x-15 \right)=0$ 

$~~~~\Rightarrow \text{ }\left( x-15 \right)\left( x+12 \right)=0$     

    $~\therefore x=15,-12$      

Cost cannot be in negative; therefore, we discard $x=-12$ 

Therefore$,\text{ }x=Rs15$which is the cost of production of each article.

Number of articles produced on that particular day      $=\dfrac{90}{15}=\text{ }6$ 

6. In a class test, the sum of Shefali's marks in Mathematics and English is $30$. Had she got $2$ marks more in Mathematics and $3$ marks less in English, the product of their marks would have been$210$. Find her marks in the two subjects .

Let Shefali's marks in Mathematics $=\text{ }x$ 

Let Shefali's marks in English $=\text{ }30-x$ 

If, she had got 2 marks more in Mathematics, her marks would be $=\text{ }x+2$ 

If, she had got 3 marks less in English, her marks in English would be $=\text{ }30\text{ }\text{ }x-3\text{ }=\text{ }27-x$ 

According to given condition:

$\left( x+2 \right)\left( 27-x \right)=210$ 

$\Rightarrow \text{ }27x-{{x}^{2}}+54-2x=210$ 

$\Rightarrow \text{ }{{x}^{2}}-25x+156=0$ 

Comparing quadratic equation ${{x}^{2}}-25x+156=0$with general form$a{{x}^{2}}+bx+c=0$,

We get $a=1,b=-25\text{ }and\text{ }c=156$ 

Applying Quadratic Formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ 

$x=\dfrac{25\pm \sqrt{{{\left( 25 \right)}^{2}}-4\left( 1 \right)\left( 156 \right)}}{2\times 1}$ 

$\Rightarrow x=\dfrac{25\pm \sqrt{625-624}}{2}$ 

$x=\dfrac{25\pm \sqrt{1}}{2}$ 

$\Rightarrow x=\dfrac{25+1}{2},\dfrac{25-1}{2}$ 

$\therefore x=13,12$ 

Therefore, Shefali's marks in Mathematics $=\text{ }13\text{ }or\text{ }12$ 

Shefali's marks in English $=\text{ }30\text{ }\text{ }x=30\text{ }\text{ }13=17$ 

Or Shefali's marks in English $=\text{ }30\text{ }\text{ }x=30\text{ }\text{ }12=18$ 

Therefore, her marks in Mathematics and English are $\left( 13,17 \right)\text{ }or\text{ }\left( 12,18 \right).$ 

7. The diagonal of a rectangular field is $60$ meters more than the shorter side. If, the longer side is$~30$ meters more than the shorter side, find the sides of the field.

Let shorter side of rectangle $=\text{ }x$meters

Let diagonal of rectangle $=\text{ }\left( x+60 \right)$meters 

Let longer side of rectangle $=\text{ }\left( x+30 \right)$meters 

According to Pythagoras theorem,

${{\left( x+60 \right)}^{2}}={{\left( x+30 \right)}^{2}}+{{x}^{2}}$ 

$\Rightarrow \text{ }{{x}^{2}}+3600+120x={{x}^{2}}+900+60x+{{x}^{2}}$ 

$~\Rightarrow \text{ }{{x}^{2}}-60x\text{ }2700=0$ 

${{x}^{2}}-60x\text{ }2700=0$ 

Comparing equation ${{x}^{2}}-60x\text{ }2700=0$with standard form$a{{x}^{2}}+bx+c=0$, 

We get $a=1,b=-60\text{ }and\text{ }c=-2700$ 

Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ 

$x=\dfrac{60\pm \sqrt{{{\left( 60 \right)}^{2}}-4\left( 1 \right)\left( -2700 \right)}}{2\times 1}$ 

$\Rightarrow x=\dfrac{60\pm \sqrt{3600+10800}}{2}$ 

$\Rightarrow x=\dfrac{60\pm \sqrt{14400}}{2}=\dfrac{60\pm 120}{2}$ 

$\Rightarrow x=\dfrac{60+120}{2},\dfrac{60-120}{2}$ 

$\therefore x=90,-30$ 

We ignore $-30$ . Since length cannot be negative. 

Therefore, $x=90$ which means length of shorter side $=90$ meters 

And length of longer side $=\text{ }x+30\text{ }=\text{ }90+30=120$meters 

Therefore, length of sides is $90\text{ and }120$ in meters.

8. The difference of squares of two numbers is $180$. The square of the smaller number is $8$ times the larger number. Find the two numbers.

Let smaller number $=\text{ }x$and let larger number $=\text{ }y$ 

According to condition: 

${{y}^{2}}-{{x}^{2}}=180\text{ }\ldots \text{ }\left( 1 \right)$ 

Also, we are given that square of smaller number is $8$ times the larger number. 

$\Rightarrow \text{ }{{x}^{2}}=8y\text{ }\ldots \text{ }\left( 2 \right)$ 

Putting equation (2) in (1), we get

 ${{y}^{2}}-8y=180$ 

$\Rightarrow \text{ }{{y}^{2}}-8y\text{ }180=0$ Comparing equation ${{y}^{2}}-8y\text{ }180=0$with general form $a{{y}^{2}}+by+c=0$, 

We get $a=1,b=-8\text{ and }c=-180$ 

Using quadratic formula $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ 

$y=\dfrac{8\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 1 \right)\left( -180 \right)}}{2\times 1}$ 

$\Rightarrow y=\dfrac{8\pm \sqrt{64+720}}{2}$ 

$\Rightarrow y=\dfrac{8\pm \sqrt{784}}{2}=\dfrac{8\pm 28}{2}$ 

$\Rightarrow y=\dfrac{8+28}{2},\dfrac{8-28}{2}$ 

$\therefore y=18,-10$ 

Using equation (2) to find smaller number:

$x{{\text{ }}^{2}}\text{ }=8y$ 

$\Rightarrow \text{ }{{x}^{2}}=8y=8\times 18=144$ 

     $\Rightarrow \text{ }x=\pm 12$ 

     And,${{x}^{2}}=8y=8\times -10=-80$  {No real solution for $x$} 

  Therefore, two numbers are $\left( 12,18 \right)\text{ }or\text{ }\left( -12,18 \right)$ 

9. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr. more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Let the speed of the train = x km/hr 

If, speed had been 5km/hr more, train would have taken 1 hour less. 

So, according to this condition

$\dfrac{360}{x}=\dfrac{360}{x+5}+1$ 

$\Rightarrow 360\left( \dfrac{1}{x}-\dfrac{1}{x+5} \right)=1$ 

$\Rightarrow 360\left( \dfrac{x+5-x}{x\left( x+5 \right)} \right)=1$ 

$\Rightarrow \text{ }360\times 5={{x}^{2}}+5x$ 

$\Rightarrow \text{ }{{x}^{2}}+5x\text{ }1800=0$ 

Comparing equation ${{x}^{2}}+5x\text{ }1800=0$ with general equation$a{{x}^{2}}+bx+c=0$ , 

We get $a=1,b=5\text{ and }c=-1800$ 

Applying quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ 

$x=\dfrac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 1 \right)\left( -1800 \right)}}{2\times 1}$ 

$\Rightarrow x=\dfrac{-5\pm \sqrt{25+7200}}{2}$ 

$\Rightarrow x=\dfrac{-5\pm \sqrt{7225}}{2}=\dfrac{-5\pm 85}{2}$ 

$\Rightarrow x=40,-45$ 

Since the speed of train cannot be in negative. Therefore, we discard $x=-45$ 

Therefore, speed of train $=\text{ }40$km/hr

10. Find the value of k for each of the following quadratic equations, so that they have two equal roots.

i. $2{{x}^{2}}+kx+3=0$ 

$2{{x}^{2}}+kx+3=0$

We know that a quadratic equation has two equal roots only when the value of the discriminant is equal to zero.                     

Comparing equation $2{{x}^{2}}+kx+3=0$ with general quadratic equation$a{{x}^{2}}+bx+c=0$,

  we get $a=2,b=k\text{ }and\text{ }c=3$ 

  Discriminant $=\text{ }{{b}^{2}}-4ac={{k}^{2}}\text{ }4\left( 2 \right)\left( 3 \right)={{k}^{2}}-24$

  Putting discriminant equal to zero

$~{{k}^{2}}\text{ }24=0$ 

$\Rightarrow {{k}^{2}}=24$ 

$\Rightarrow k=\pm \sqrt{24}=\pm 2\sqrt{6}$ 

$\Rightarrow k=2\sqrt{6},-2\sqrt{6}$ 

ii. $kx\left( x-2 \right)+6=0$ 

$kx\left( x-2 \right)+6=0$ 

$\Rightarrow \text{ }k{{x}^{2}}-2kx+6=0$ 

Comparing quadratic equation $k{{x}^{2}}-2kx+6=0$ with general form$a{{x}^{2}}+bx+c=0$, we get $a=k,b=\text{ }-2k\text{ }and\,c=6$ 

Discriminant $=\text{ }{{b}^{2}}-4ac={{\left( -2k \right)}^{2}}\text{ }4\left( k \right)\left( 6 \right)=4{{k}^{2}}-24k$ 

We know that two roots of quadratic equation are equal only if discriminant is equal to zero. 

Putting discriminant equal to zero

$4{{k}^{2}}-24k=0$  

$\Rightarrow \text{ }4k\left( k-6 \right)=0$ 

$\Rightarrow \text{ }k=0,6$ 

The basic definition of quadratic equation says that quadratic equation is the equation of the form$a{{x}^{2}}+bx+c=0$ , where $a\ne 0.$ 

Therefore, in equation$k{{x}^{2}}-2kx+6=0$, we cannot have $k=0$. 

Therefore, we discard $k=0$. 

Hence the answer is $k=6$ 

11.   Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\text{ }{{m}^{2}}$. If so, find its length and breadth.

Let breadth of rectangular mango grove $=\text{ }x$meters 

Let length of rectangular mango grove $=\text{ }2x$ meters 

Area of rectangle = length × breadth $=\text{ }x\times \text{ }2x\text{ }=\text{ }2{{x}^{2}}{{m}^{2}}$ 

According to given condition-

$2{{x}^{2}}=800$ 

$\Rightarrow \text{ }2{{x}^{2}}\text{ }800=0$ 

$\Rightarrow \text{ }{{x}^{2}}\text{ }400=0$ 

Comparing equation ${{x}^{2}}\text{ }400=0$with general form of quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=0\text{ and }c=400$

Discriminant $=\text{ }{{b}^{2}}-4ac={{\left( 0 \right)}^{2}}\text{ }4\left( 1 \right)\left( -400 \right)=1600$ 

Discriminant is greater than 0 means that equation has two distinct real roots.

Therefore, it is possible to design a rectangular grove.

Applying quadratic formula, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve equation,

$x=\dfrac{0\pm \sqrt{1600}}{2\times 1}=\dfrac{\pm 40}{2}=\pm 20$ 

$\therefore x=20,-20$ We discard negative value of $x$ because breadth of rectangle cannot be in negative. 

Therefore, $x\text{ }=$ breadth of rectangle $=\text{ }20$ meters 

Length of rectangle $=\text{ }2x=2\times 20=40$ meters

12. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $~20$ years. Four years ago, the product of their ages in years was $48$.

Let age of first friend = x years and let age of second friend $=\text{ }\left( 20-x \right)$ years 

Four years ago, age of first friend $=\text{ }\left( x-4 \right)$ years 

Four years ago, age of second friend $=\text{ }\left( 20-x \right)-4\text{ }=\text{ }\left( 16-x \right)$ years 

According to given condition,

$\left( x-4 \right)\left( 16-x \right)=48$ 

$~\Rightarrow \text{ }16x-{{x}^{2}}\text{ }64+4x=48$  

$\Rightarrow \text{ }20x-{{x}^{2}}\text{ }112=0$ 

$\Rightarrow \text{ }{{x}^{2}}-20x+112=0$ 

Comparing equation, ${{x}^{2}}-20x+112=0$ with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=-20\text{ }and\text{ }c=112$ 

Discriminant $={{b}^{2}}-4ac={{\left( -20 \right)}^{2}}\text{ }4\left( 1 \right)\left( 112 \right)=400\text{ }\text{ }448=-48<0$ 

The discriminant is less than zero which means we have no real roots for this equation.

Therefore, the given situation is not possible.

13. Value of$x$  for ${{x}^{2}}-8x+15=0$ is quadratic formula is

a). $3,\text{ }2$ 

b). $5,\text{ }2$ 

c). $5,\text{ }3$ 

d). $2,\text{ }3$ 

(c) $5,\text{ }3$

14. Discriminate of $\sqrt{3}{{x}^{2}}-2\sqrt{2}x-2\sqrt{3}=0$ is

a). $30$ 

b). $31$ 

c). $32$ 

d). $35$ 

15. Solve $12ab{{x}^{2}}-9{{a}^{2}}x+8{{b}^{2}}x-6ab=0$ 

$12ab{{x}^{2}}-9{{a}^{2}}x+8{{b}^{2}}x-6ab=0$

$\Rightarrow 3ax\left( 4bx-3x \right)+2b\left( 4bx-3x \right)=0$ 

$\Rightarrow \left( 4bx-3x \right)\left( 3ax+2b \right)=0$ 

$\Rightarrow 4bx-3a=0\,\,or\,\,3ax+2b=0$ 

$\therefore x=\dfrac{3a}{4b}\,or\,x=-\dfrac{2b}{3a}$ 

16. Solve for $x$ by quadratic formula${{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0$  

${{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0$

$a={{p}^{2}},\,b={{p}^{2}}-{{q}^{2}},\,c=-{{q}^{2}}$ 

$D={{b}^{2}}-4ac$ 

$=\left( {{p}^{2}}-{{q}^{2}} \right)-4\times {{p}^{2}}\left( -{{q}^{2}} \right)$ 

$={{p}^{4}}+{{q}^{2}}-2{{p}^{2}}{{q}^{2}}+4{{p}^{2}}{{q}^{2}}$ 

$={{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}$ 

$x=\dfrac{-b\pm \sqrt{D}}{29}$ 

$=\dfrac{-\left( {{p}^{2}}{{q}^{2}} \right)\pm \sqrt{{{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}}}{2\times {{p}^{2}}}$ 

$=\dfrac{-{{p}^{2}}+{{q}^{2}}+{{p}^{2}}+{{q}^{2}}}{2{{p}^{2}}}$ 

$or\,x=\dfrac{-{{p}^{2}}+{{q}^{2}}-{{p}^{2}}}{2{{p}^{2}}}$ $x=\dfrac{2{{q}^{2}}}{2{{p}^{2}}}\,\,or\,x=\dfrac{-2{{q}^{2}}}{2{{p}^{2}}}$ 

$x=\dfrac{{{q}^{2}}}{{{p}^{2}}}\,or\,x=-1$ 

17. Find the value of k for which the quadratic equation$k{{x}^{2}}+2x+1=0$  has real and 

distinct root

$k{{x}^{2}}+2x+1=0$

$a=k,\,b=2,\,c=1$ 

$b={{b}^{2}}-4ac$ 

$={{\left( 2 \right)}^{2}}-4\times k\times 1=4-4k$ 

For real and distinct roots,

$D > 0$ 

$4-4k > 0$ 

$\Rightarrow -4k > -4$ 

$\therefore k < 1 $ 

18. If one root of the equations $2{{x}^{2}}+ax+3=0$ is $1$ , find the value of a.

a). $=-4$ 

b). $=-5$ 

c). $=-3$ 

d). $=-1$ 

19. Find k for which the quadratic equation$4{{x}^{2}}-3kx+1=0$  has equal root.

$=\pm \dfrac{3}{4}$ 

$=\dfrac{3}{4}$ 

$=\pm \dfrac{4}{3}$ 

$=\dfrac{2}{3}$ 

(c) $=\pm \dfrac{4}{3}$

20. Determine the nature of the roots of the quadratic equation

$9{{a}^{2}}{{b}^{2}}{{x}^{2}}-24abcdx+16{{c}^{2}}{{d}^{2}}=0$ 

$={{\left( -24abcd \right)}^{2}}-4\times 9{{a}^{2}}{{b}^{2}}\times 16{{c}^{2}}{{d}^{2}}$ 

$=576{{a}^{2}}{{b}^{2}}{{c}^{2}}-376{{a}^{2}}{{b}^{2}}{{c}^{2}}{{d}^{2}}=0$ 

21. Find the discriminant of the equation $\left( x-1 \right)\left( 2x-1 \right)=0$ 

$\left( x-1 \right)\left( 2x-1 \right)=0$

$\Rightarrow 2{{x}^{2}}-x-2x+1=0$

 $\Rightarrow 2{{x}^{2}}-3x+1=0$ 

$Here,a=2,\,b=-3,\,c=1$ 

$={{\left( -3 \right)}^{2}}-4\times 2\times 1$ 

$=9-8=1$ 

22. Find the value of k so that $\left( x-1 \right)$ is a factor of ${{k}^{2}}{{x}^{2}}-2kx-3$.

Let $P\left( x \right)={{k}^{2}}{{x}^{2}}-2kx-3$

$P\left( 1 \right)={{k}^{2}}{{\left( 1 \right)}^{2}}-2k\left( 1 \right)-3$ 

$\Rightarrow 0={{k}^{2}}-2k-3$ 

$\Rightarrow {{k}^{2}}-3k+k-3$ 

$\Rightarrow k\left( k-3 \right)+1\left( k-3 \right)=0$ 

$\Rightarrow \left( k-3 \right)\left( k+1 \right)=0$ 

$\therefore k=3\,or\,k=-1$ 

23. The product of two consecutive positive integers is $306$. Represent these in quadratic 

a). ${{x}^{2}}+x-306=0$

b). ${{x}^{2}}-x-306=0$

c). ${{x}^{2}}+2x-106=0$ 

d). ${{x}^{2}}-x-106=0$ 

(a) ${{x}^{2}}+x-306=0$

24 . Which is a quadratic equation?

a). ${{x}^{2}}+x+2=0$ 

b). ${{x}^{3}}+{{x}^{2}}+2=0$ 

c). ${{x}^{4}}+{{x}^{2}}+2=0$ 

d). $x+2=0$ 

(a) ${{x}^{2}}+x+2=0$

25. The sum of two numbers is $16$. The sum of their reciprocals is $\dfrac{1}{3}$. Find the numbers.  

Let no. be $x$ 

According to question,

$\dfrac{1}{x}+\dfrac{1}{16-x}=\dfrac{1}{3}$ 

$\Rightarrow \dfrac{16}{16x-{{x}^{2}}}=\dfrac{1}{3}$ 

$\Rightarrow {{x}^{2}}-16x+48=0$ 

$\Rightarrow {{x}^{2}}-12x-4x+48=0$ 

$\therefore x=12\,or\,x=4$ 

26. Solve for $x:\sqrt{217-x}=x-7$ 

$\sqrt{217-x}=\left( x-7 \right)$

$\Rightarrow 217-x={{x}^{2}}+49-14x$ 

$\Rightarrow {{x}^{2}}-14x+x+49-217=0$ 

$\Rightarrow {{x}^{2}}-13x-168=0$ 

$\Rightarrow {{x}^{2}}-21x+8x-168=0$ 

$\therefore x=21\,or\,x=-8$ 

27. Solve for x by factorization: $x+\dfrac{1}{x}=11\dfrac{1}{11}$  

$\dfrac{{{x}^{2}}+1}{x}=\dfrac{122}{11}$ 

$\Rightarrow 11{{x}^{2}}-12x+11=0$ 

$\Rightarrow 11{{x}^{2}}-121x-1x+11=0$ 

$\Rightarrow 11x\left( x-11 \right)-1\left( x-11 \right)=0$ 

$\Rightarrow \left( 11x-1 \right)\left( x-11 \right)=0$ 

$\therefore x=11\,or\,x=\dfrac{1}{11}$ 

28. Find the ratio of the sum and product of the roots of $7{{x}^{2}}-12x+18=0$  

Ans:  

$7{{x}^{2}}-12x+18=0$

$\alpha +\beta =\dfrac{-b}{a}=\dfrac{12}{7}\,and\alpha \beta =\dfrac{c}{a}=\dfrac{18}{17}$

$\dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{\dfrac{12}{7}}{\dfrac{18}{17}}=\dfrac{12}{7}\times \dfrac{17}{18}=\dfrac{34}{21}$ 

29. If $\alpha $  and$\beta $  are the roots of the equation ${{x}^{2}}+kx+12=0$, such that $\alpha -\beta =1$ , then

$\alpha +\beta =\dfrac{-k}{1},$ 

$\alpha -\beta =1$ 

$\alpha \beta =\dfrac{12}{1}$ 

${{\left( \alpha +\beta  \right)}^{2}}={{\left( \alpha -\beta  \right)}^{2}}+4\alpha \beta $ 

$\Rightarrow {{\left( -k \right)}^{2}}={{\left( 1 \right)}^{2}}+4\times 12$ 

$\Rightarrow {{k}^{2}}=49$ 

$k=\pm 7$ 

3 Marks Questions

1. Check whether the following are Quadratic Equations.

i). ${{\left( x+1 \right)}^{2}}=2\left( x-3 \right)$ 

${{\left( x+1 \right)}^{2}}=2\left( x-3 \right)\,\left\{ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\text{ } \right\}$ 

$\Rightarrow \text{ }{{x}^{2}}+1+2x=2x\text{ }6$ 

$\Rightarrow \text{ }{{x}^{2}}+7=0$ 

Here, degree of equation is $2$.

Therefore, it is a Quadratic Equation.

ii). ${{x}^{2}}-2x=\left( -2 \right)\left( 3-x \right)$ 

${{x}^{2}}-2x=\left( -2 \right)\left( 3-x \right)$

$\Rightarrow \text{ }{{x}^{2}}-2x=-6+2x$

$\Rightarrow \text{ }{{x}^{2}}-2x-2x+6=0$ 

$\Rightarrow \text{ }{{x}^{2}}-4x+6=0$ 

Here, degree of equation is $2.$ 

iii). $\left( x-2 \right)\left( x+1 \right)=\left( x-1 \right)\left( x+3 \right)$ 

(x−2)(x+1)=(x−1)(x+3)

$\Rightarrow \text{ }{{x}^{2}}+x-2x\text{ }2={{x}^{2}}+3x\text{ }x\text{ }3=0$ 

$\Rightarrow \text{ }{{x}^{2}}+x-2x\text{ }2-{{x}^{2}}-3x+x+3=0$ 

$\Rightarrow \text{ }x-2x\text{ }2-3x+x+3=0$ 

$\Rightarrow \text{ }-3x+1=0$ 

Here, degree of equation is $1.$ 

Therefore, it is not a Quadratic Equation.

iv). $~~\left( x-3 \right)\left( 2x+1 \right)=x\left( x+5 \right)$

$\left( x-3 \right)\left( 2x+1 \right)=x\left( x+5 \right)$ 

$\Rightarrow \text{ }2{{x}^{2}}+x-6x\text{ }3={{x}^{2}}+5x$ 

$\Rightarrow \text{ }2{{x}^{2}}+x-6x\text{ }3-{{x}^{2}}-5x=0$

$\Rightarrow \text{ }{{x}^{2}}-10x\text{ }3=0$

Here, degree of equation is 2.

Therefore, it is a quadratic equation.

v). $\left( 2x-1 \right)\left( x-3 \right)=\left( x+5 \right)\left( x-1 \right)$ 

$\left( 2x-1 \right)\left( x-3 \right)=\left( x+5 \right)\left( x-1 \right)\text{ }$

$\Rightarrow \text{ }2{{x}^{2}}-6x\text{ }x+3={{x}^{2}}\text{ }x+5x\text{ }5$ 

$\Rightarrow \text{ }{{x}^{2}}-11x+8=0$

Here, degree of Equation is$2$.

vi). ${{x}^{2}}+3x+1={{\left( x-2 \right)}^{2}}$ 

${{x}^{2}}+3x+1={{\left( x-2 \right)}^{2}}\,\,\,\,\,\text{ }\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \right\}$ 

$\Rightarrow \text{ }{{x}^{2}}\text{ }+3x+1={{x}^{2}}+4-4x$ 

$\Rightarrow \text{ }{{x}^{2}}+3x+1-{{x}^{2}}+4x\text{ }4=0$

$\Rightarrow \text{ }7x\text{ }3=0$

Here, degree of equation is$~1$.

vii). ${{\left( x+2 \right)}^{3}}=2x\left( {{x}^{2}}-1 \right)$ 

${{\left( x+2 \right)}^{3}}=2x\left( {{x}^{2}}-1 \right)\text{ }\,\,\,\,\,\,\,\left\{ {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right) \right\}$

$\Rightarrow \text{ }{{x}^{3}}\text{ }+{{2}^{3}}\text{ }+3\left( x \right)\left( 2 \right)\left( x+2 \right)=2x\left( {{x}^{2}}-1 \right)$

$\Rightarrow \text{ }{{x}^{3}}+8+6x\left( x+2 \right)=2{{x}^{3}}-2x$

$\Rightarrow \text{ }2{{x}^{3}}-2x-{{x}^{3}}\text{ }8-6{{x}^{2}}-12x=0$

$\Rightarrow \text{ }{{x}^{3}}-6{{x}^{2}}-14x\text{ }8=0$

Here, degree of Equation is 3.

Therefore, it is not a quadratic Equation.

viii). ${{x}^{3}}-4{{x}^{2}}\text{ }x+1={{\left( x-2 \right)}^{3}}\text{ }$

${{x}^{3}}-4{{x}^{2}}\text{ }x+1={{\left( x-2 \right)}^{3}}\,\,\,\,\text{ }\left\{ {{\left( a-b \right)}^{3}}\text{ }={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) \right\}$ 

   $\Rightarrow \text{ }{{x}^{3}}-4{{x}^{2}}\text{ }x+1={{x}^{3}}-{{2}^{3}}\text{ }3\left( x \right)\left( 2 \right)\left( x-2 \right)$ 

   $\Rightarrow \text{ }-4{{x}^{2}}\text{ }x+1=-8-6{{x}^{2}}+12x$ 

   $\Rightarrow \text{ }2{{x}^{2}}-13x+9=0$ 

Here, degree of Equation is $2$.

2. Represent the following situations in the form of Quadratic Equations:

i). The area of the rectangular plot is 528 $m^2$. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

We are given that area of a rectangular plot is $528{{m}^{2}}$ 

Let the breadth of the rectangular plot be$~x$ meters

Length is one more than twice its breadth

Therefore, length of rectangular plot is $\left( 2x+1 \right)$meters

Area of rectangle$=$length $\times $ breadth

$\Rightarrow \text{ }528=x\left( 2x+1 \right)$ 

$\Rightarrow \text{ }528=2{{x}^{2}}+x$ 

$\Rightarrow \text{ }2{{x}^{2}}+x\text{ }528=0$ 

This is a Quadratic Equation.

ii). The product of two consecutive numbers is 306. We need to find the integers.

Let two consecutive numbers be $x\,\text{and}\,\left( x+1 \right).$ 

It is given that x(x+1) = 306 

$\Rightarrow \text{ }{{x}^{2}}+x=306$ 

$\Rightarrow \text{ }{{x}^{2}}+x\text{ }306=0$ 

iii). Rohan's mother is 26 years older than him. The product of their ages (in years) after 3 years will be 360. We would like to find Rohan's present age.

Let present age of Rohan $=\text{ }x$years

Let present age of Rohan's mother $=\text{ }\left( x\text{ }+26 \right)$ years

Age of Rohan after$~3$years $=\text{ }\left( x+3 \right)$ years 

Age of Rohan's mother after $~3$ years $=\text{ }x+26+3\text{ }=\text{ }\left( x+29 \right)$ years 

According to given condition: 

$\left( x+3 \right)\left( x+29 \right)=360$  

$\Rightarrow \text{ }{{x}^{2}}+29x+3x+87=360$  

$\Rightarrow \text{ }{{x}^{2}}+32x\text{ }273=0$ 

iv). A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Let the speed of the train be $x$ km/h 

Time taken by train to cover 480 km $=\text{ }480x$hours

If, speed had been 8km/h less than time taken would be $\left( 480x-8 \right)$  hours. According to given condition, if speed had been $8$km/h less than time taken is $3$hours less.

             Therefore, $480x\text{ }8=480x+3$ 

             $\Rightarrow \text{ }480\left( 1x\text{ }8-1x \right)=3$ 

             $\Rightarrow \text{ }480\left( x\text{ }x+8 \right)\text{ }\left( x \right)\text{ }\left( x-8 \right)=3$ 

              $\Rightarrow \text{ }480\times 8=3\left( x \right)\left( x-8 \right)$ 

              $~\Rightarrow \text{ }3840=3{{x}^{2}}-24x$ 

     Dividing equation by$3$, we get 

     $\Rightarrow \text{ }{{x}^{2}}-8x\text{ }1280=0$ 

     This is a Quadratic Equation.

3. Find the roots of the following Quadratic Equations by factorization.

i). ${{x}^{2}}-3x\text{ }10=0$ 

${{x}^{2}}-3x\text{ }10=0$ 

$\Rightarrow \text{ }{{x}^{2}}-5x+2x\text{ }10=0$ 

      $\Rightarrow \text{ }x\left( x-5 \right)+2\left( x-5 \right)=0$ 

      $\Rightarrow \text{ }\left( x-5 \right)\left( x+2 \right)=0$ 

       $\Rightarrow \text{ }x=5,-2$ 

ii). $2{{x}^{2}}+x\text{ }6=0$ 

$2{{x}^{2}}+x\text{ }6=0$ 

$\Rightarrow \text{ }2{{x}^{2}}+4x-3x\text{ }6=0$

$\Rightarrow \text{ }2x\left( x+2 \right)\text{ }\text{ }3\left( x+2 \right)=0$

$\Rightarrow \text{ }\left( 2x-3 \right)\left( x+2 \right)=0$

$\Rightarrow x=\dfrac{3}{2},2$ 

iii). $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$ 

$\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$ 

          $\Rightarrow \sqrt{2}{{x}^{2}}+2x+5x+5\sqrt{2}=0$ 

          $\Rightarrow \sqrt{2}{{x}^{2}}\left( x+\sqrt{2} \right)+5\left( x+\sqrt{2} \right)=0$ 

          $\Rightarrow \left( \sqrt{2}x+5 \right)\left( x+\sqrt{2} \right)=0$ 

         $\Rightarrow x=\dfrac{-5}{\sqrt{2}},-\sqrt{2}$ 

           $\Rightarrow x=\dfrac{-5}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}},-\sqrt{2}$

             $\Rightarrow x=\dfrac{-5\sqrt{2}}{2},-\sqrt{2}$ 

iv). $2{{x}^{2}}-x+\dfrac{1}{8}=0$ 

$2{{x}^{2}}-x+\dfrac{1}{8}=0$ 

             $\Rightarrow \dfrac{16{{x}^{2}}-8x+1}{8}=0$ 

            $\Rightarrow \text{ }16{{x}^{2}}-8x+1=0$ 

          $\Rightarrow \text{ }16{{x}^{2}}-4x-4x+1=0$ 

          $\Rightarrow \text{ }4x\left( 4x-1 \right)\text{ }\text{ }1\left( 4x-1 \right)=0$ 

             $\Rightarrow \text{ }\left( 4x-1 \right)\left( 4x-1 \right)=0$ 

            $\Rightarrow \text{ }x=\text{ }\dfrac{1}{4},\dfrac{1}{4}$ 

v). $100{{x}^{2}}-20x+1=0$ 

$100{{x}^{2}}-20x+1=0$ 

$\Rightarrow \text{ }100{{x}^{2}}-10x-10x+1=0$ 

$\Rightarrow \text{ }10x\left( 10x-1 \right)\text{ }\text{ }1\left( 10x-1 \right)=0$ 

$\Rightarrow \text{ }\left( 10x-1 \right)\left( 10x-1 \right)=0$ 

$\therefore \,x=\dfrac{1}{10},\dfrac{1}{10}$ 

4. Find the roots of the following equations:

i). $\dfrac{x-1}{x}=3,x\ne 0$ 

$x-\dfrac{1}{x}=3\,\,where\,x\ne 0$ 

$\Rightarrow \dfrac{{{x}^{2}}-1}{x}=3$ 

$\Rightarrow \text{ }x{{\text{ }}^{2}}\text{ }1=3x$ 

$\Rightarrow \text{ }{{x}^{2}}-3x\text{ }1=0$ 

Comparing equation ${{x}^{2}}-3x\text{ }1=0$with general form$a{{x}^{2}}+bx+c=0$,

We get $a=1,b=-3\text{ }and\text{ }c=-1$ 

Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve equation,

$x=\dfrac{3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\times 1}$ 

             $\Rightarrow x=\dfrac{3\pm \sqrt{13}}{2}$ 

              $\Rightarrow x=\dfrac{3+\sqrt{13}}{2},\dfrac{3-\sqrt{13}}{2}$ 

(ii). $\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne -4,7$ 

$\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30}\,where\,x\ne -4,7$

$\Rightarrow \text{ }-30={{x}^{2}}-7x+4x28$

$\Rightarrow \text{ }{{x}^{2}}-3x+2=0$ 

               Comparing equation$~{{x}^{2}}-3x+2=0$ with general form$a{{x}^{2}}+bx+c=0$,

               We get $a=1,b=-3\text{ }and\text{ }c=2$ 

               Using quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve equation,

              $x=\dfrac{3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( 2 \right)}}{2\times 1}$ 

              $\Rightarrow x=\dfrac{3\pm \sqrt{1}}{2}$ 

              $\Rightarrow x=\dfrac{3+\sqrt{1}}{2},\dfrac{3-\sqrt{1}}{2}$ 

              $\Rightarrow x=2,1$ 

5. The sum of reciprocals of Rehman's ages (in years) $3$  years ago and $5$  years from now is $13$. Find his present age.

Let present age of Rehman$=\text{ }x$  years 

Age of Rehman $3$ years ago $=\text{ }\left( x-3 \right)$  years.

 Age of Rehman after $5$  years $=\text{ }\left( x+5 \right)$ years

 According to the given condition:

$\dfrac{1}{x-3}+\dfrac{1}{x+5}=\dfrac{1}{3}$ 

$\Rightarrow \dfrac{\left( x+5 \right)+\left( x-3 \right)}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{3}$ 

$\Rightarrow \text{ }3\left( 2x+2 \right)\text{ }=\left( x-3 \right)\left( x+5 \right)\text{ }$ 

$\Rightarrow \text{ }6x+6={{x}^{2}}-3x+5x-15$ 

 $\Rightarrow \text{ }{{x}^{2}}-4x\text{ }15\text{ }\text{ }6=0$ 

$\Rightarrow {{x}^{2}}-4x\text{ }21=0$ 

Comparing quadratic equation x ${{x}^{2}}-4x\text{ }21=0$ with general form $a{{x}^{2}}+bx+c=0$, We get $a=1,b=-4\text{ }and\text{ }c=-21$ 

Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 1 \right)\left( -21 \right)}}{2\times 1}$ 

$\Rightarrow x=\dfrac{4\pm \sqrt{16+84}}{2}$ 

$\Rightarrow x=\dfrac{4\pm \sqrt{100}}{2}=\dfrac{4\pm 10}{2}$ 

$\Rightarrow x=\dfrac{4+10}{2},\dfrac{4-10}{2}$ 

$\therefore x=7,-3$ 

We discard$x=-3$ .Since age cannot be in negative.

Therefore, present age of Rehman is $7$ years

6. Two water taps together can fill a tank in $9\dfrac{3}{8}$hours. The tap of larger diameter takes $10$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Let time taken by tap of smaller diameter to fill the tank $=\text{ }x$ hours 

Let time taken by tap of larger diameter to fill the tank $=\text{ }\left( x\text{ }10 \right)$ hours 

It means that tap of smaller diameter fills ${{\dfrac{1}{x}}^{th}}$ part of tank in $1$  hour. … (1)

 And, tap of larger diameter fills ${{\dfrac{1}{x-10}}^{th}}$ part of tank in $1$  hour. … (2) 

When two taps are used together, they fill tank in$~758$ hours

In 1 hour, they fill${{\dfrac{8}{75}}^{th}}$ part of tank $\left[ \dfrac{1}{\dfrac{75}{8}}=\dfrac{8}{75} \right]$ … (3)

From (1), (2) and (3),

$\dfrac{1}{x}+\dfrac{1}{x-10}=\dfrac{8}{75}$ 

$\Rightarrow \dfrac{x-10+x}{x\left( x-10 \right)}=\dfrac{8}{75}$ 

$\Rightarrow \text{ }75\left( 2x-10 \right)=8\left( {{x}^{2}}-10x \right)$ 

$\Rightarrow \text{ }150x\text{ }750=8{{x}^{2}}-80x$ 

$\Rightarrow \text{ }8{{x}^{2}}\text{ }-80x-150x+750=0$ 

Comparing equation $4{{x}^{2}}\text{ }-115x+375=0$ with general equation$a{{x}^{2}}+bx+c=0$, 

We get $a=4,b=-115\,and\,c=375$ 

Applying quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{115\pm \sqrt{{{\left( -115 \right)}^{2}}-4\left( 4 \right)\left( 375 \right)}}{2\times 4}$ 

$\Rightarrow x=\dfrac{115\pm \sqrt{13225-6000}}{8}$ 

$\Rightarrow \dfrac{115\pm \sqrt{7225}}{8}$ 

$\Rightarrow \dfrac{115+85}{8},\dfrac{115-85}{8}$ 

$\therefore x=25,3.75$ 

Time taken by larger tap $=\text{ }x\text{ }10=3.75\text{ }\text{ }10=-6.25$ hours 

Time cannot be in negative. Therefore, we ignore this value. 

Time taken by larger tap $=\text{ }x\text{ }10=25\text{ }\text{ }10=15$ hours 

Therefore, time taken by larger tap is $15$ hours and time taken by smaller tap is $25$  hours.

7. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

i). $2{{x}^{2}}\text{ }3x\text{ }+\text{ }5\text{ }=\text{ }0$

$2{{x}^{2}}\text{ }3x\text{ }+\text{ }5\text{ }=\text{ }0$

Comparing this equation with general equation$a{{x}^{2}}+bx+c=0$ , 

We get $a=2,b=-3\text{ }and\text{ }c=5$ 

Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -3 \right)}^{2}}\text{ }4\left( 2 \right)\left( 5 \right)=9\text{ }\text{ }40=-31$ 

Discriminant is less than 0 which means the equation has no real roots.

ii). $3{{x}^{2}}-4\sqrt{3}x+4=0$ 

$3{{x}^{2}}-4\sqrt{3}x+4=0$

Comparing this equation with general equation $a{{x}^{2}}+bx+c=0$, 

We get $a=3,b=-4\sqrt{3}\text{ }and\text{ }c=4$

Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -4\sqrt{3} \right)}^{2}}-4\left( 3 \right)\left( 4 \right)=48\text{ }\text{ }48=0$ 

Discriminant is equal to zero which means equations have equal real roots. Applying quadratic$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find roots,

             $x=\dfrac{4\sqrt{3}\pm \sqrt{0}}{6}=\dfrac{2\sqrt{3}}{3}$ 

 Because, equation has two equal roots, it means $x=\dfrac{2\sqrt{3}}{3},\dfrac{2\sqrt{3}}{3}$ 

iii). $2{{x}^{2}}\text{ }+6x\text{ }+\text{ }3\text{ }=\text{ }0$ 

$2{{x}^{2}}\text{ }+6x\text{ }+\text{ }3\text{ }=\text{ }0$

         Comparing equation with general equation$a{{x}^{2}}+bx+c=0$ ,

          We get $a=2,b=-6,\text{ }and\text{ }c=3$ 

          Discriminant $=\text{ }{{b}^{2}}-4ac={{\left( -6 \right)}^{2}} - \text{ }4\left( 2 \right)\left( 3 \right)=36\text{ }-\text{ }24=12$ 

          Value of the discriminant is greater than zero. 

          Therefore, the equation has distinct and real roots.

         Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find roots,

          $x=\dfrac{6\pm \sqrt{12}}{4}=\dfrac{6\pm 2\sqrt{3}}{4}$ 

          $\Rightarrow x=\dfrac{3\pm \sqrt{3}}{2}$ 

          $\Rightarrow x=\dfrac{3+\sqrt{3}}{2},\dfrac{3-\sqrt{3}}{2}$ 

8. If $-4$ is a root of the quadratic equation and the quadratic equation${{x}^{2}}+px-4$ has equal root, find the value of $k.$ 

$-4$is root of ${{x}^{2}}+px-4=0$

$\therefore {{\left( -4 \right)}^{2}}+p\left( -4 \right)-4=0$ 

$\Rightarrow 16-4p-4=0$ 

$\Rightarrow -4p=-12$ 

$\Rightarrow p=3$ 

${{x}^{2}}+px+k=0$ (Given)

${{x}^{2}}+3x+k=0$ 

$\Rightarrow 0={{\left( 3 \right)}^{2}}-4\times 1\times k$ [For equal roots D = 0]

$\Rightarrow 4k=9$ 

$\Rightarrow k=\dfrac{9}{4}$ 

9. Solve for $x:{{5}^{x+1}}+{{5}^{1-x}}=26$ 

${{5}^{x+1}}+{{5}^{1-x}}=26$ 

     ${{5}^{x}}{{.5}^{1}}+{{5}^{1}}{{.5}^{-x}}=26$ 

     $\Rightarrow {{5}^{x}}.5+\dfrac{{{5}^{1}}}{{{5}^{x}}}=26$ 

     Put ${{5}^{x}}=y$ 

      $\dfrac{5y}{1}+\dfrac{5}{y}=26$ 

      $\Rightarrow 5{{y}^{2}}-26y+5=0$ 

      $\Rightarrow 5{{y}^{2}}-25y-y+5=0$ 

      $\Rightarrow 5y\left( y-5 \right)-1\left( y-5 \right)=0$ 

      $\Rightarrow $ $\left( y-5 \right)\left( 5y-1 \right)=0$ 

      $\Rightarrow y=5$ or  $y=\dfrac{1}{5}$ 

      But 

      ${{5}^{x}}={{5}^{1}}$ and ${{5}^{x}}=\dfrac{1}{5}$ 

      $\Rightarrow x=1$ and ${{5}^{x}}={{5}^{-1}}\Rightarrow x=-1$ 

10. $\dfrac{1}{p+q+x}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{x}$ solve for$x$ by factorization method.

$\dfrac{1}{p+q+x}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{x}$ 

$\Rightarrow \dfrac{1}{p+q+x}-\dfrac{1}{x}=\dfrac{1}{p}+\dfrac{1}{q}$

$\Rightarrow \dfrac{x-p-q-x}{{{x}^{2}}+px+qx}=\dfrac{p+q}{pq}$ 

$\Rightarrow \dfrac{-\left( p+q \right)}{{{x}^{2}}+px+qx}=\dfrac{p+q}{pq}$ 

$\Rightarrow \dfrac{-1}{{{x}^{2}}+px+qx}=\dfrac{1}{pq}$ 

$\Rightarrow {{x}^{2}}+px+qx=-pq$ 

$\Rightarrow {{x}^{2}}+px+qx+pq=0$ 

$\Rightarrow x\left( x+p \right)+q\left( x+p \right)=0$ 

$\Rightarrow \left( x+p \right)\left( x+q \right)=0$ 

$\therefore x=-p\,or\,x=-q$ 

11. $5{{x}^{2}}-6x-2=0$, solve for$x$  by the method of completing the square.

$5{{x}^{2}}-6x-2=0$

$\Rightarrow {{x}^{2}}-\dfrac{6}{5}x-\dfrac{2}{5}=0$ 

$\Rightarrow {{x}^{2}}-\dfrac{6}{5}x+{{\left( \dfrac{3}{5} \right)}^{2}}-\dfrac{2}{5}=0$ 

$\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{9}{25}+\dfrac{2}{5}$ 

$\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{9+10}{25}$ 

$\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{19}{25}$ 

$\Rightarrow x-\dfrac{3}{5}=\pm \dfrac{\sqrt{19}}{5}$ 

$\Rightarrow x=\dfrac{3}{5}\pm \dfrac{\sqrt{19}}{5}$ 

$\Rightarrow x=\dfrac{3+\sqrt{19}}{5}\,or\,x=\dfrac{3-\sqrt{19}}{5}$ 

12. Solve for$x:{{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$ 

${{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$

$\Rightarrow {{b}^{2}}x\left( {{a}^{2}}x+1 \right)-1\left( {{a}^{2}}x-1 \right)=0$ 

$\Rightarrow \left( {{a}^{2}}x+1 \right)\left( {{b}^{2}}x-1 \right)=0$ 

$\therefore x=\dfrac{-1}{{{a}^{2}}}\,or\,x=\dfrac{-1}{{{b}^{2}}}$ 

13. Using quadratic formula, solve for $x:9{{x}^{2}}-9\left( a+b \right)x+\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)=0$ 

$={{\left( -9\left( a+b \right) \right)}^{2}}-4\times 9\times \left( 2{{a}^{2}}+5ab+2a{{b}^{2}} \right)$ 

$=81{{\left( a+b \right)}^{2}}-36\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)$ 

$=9\left[ 9\left( {{a}^{2}}+{{b}^{2}}+2ab-8{{a}^{2}}-20ab-8{{b}^{2}} \right) \right]$ 

$=9\left[ {{a}^{2}}+{{b}^{2}}-2ab \right]$ 

$=9{{\left( a-b \right)}^{2}}$

$x=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{9\left( a+b \right)\pm \sqrt{9{{\left( a-b \right)}^{2}}}}{2\times 9}$ 

$\Rightarrow x=3\dfrac{\left[ 3\left( a+b \right)\pm \left( a-b \right) \right]}{2\times 9}$ 

$\Rightarrow x=\dfrac{\left( 3a+3b \right)\pm \left( a-b \right)}{6}$ 

$\Rightarrow x=\dfrac{3a+3b+a-b}{6}\,or\,x=\dfrac{3a+3b+a-b}{6}$ 

$\Rightarrow x=\dfrac{4a+2b}{6}\,or\,x=\dfrac{4a+2b}{6}$ 

$\Rightarrow x=\dfrac{2a+b}{3}\,or\,x=\dfrac{2a+b}{3}$ 

14. In a cricket match, Kapil took one wicket less than twice the number of wickets taken by Ravi. If the product of the numbers of wickets taken by these two is 15, find the number of wickets taken by each.

Let no. of wicket taken by Ravi $=x$

 of wicket taken by Kapil $=2x-1$ 

$\left( 2x-1 \right)x=15$ 

$\Rightarrow 2{{x}^{2}}-x-15=0$ 

$\therefore x=3\,or\,x=\dfrac{-5}{2}$ (Neglects)

So, no. of wickets taken by Ravi is  $x=3$ 

15 . The sum of a number and its reciprocal is $\dfrac{17}{4}$ . Find the number.

$\dfrac{x}{1}+\dfrac{1}{x}=\dfrac{17}{4}$ 

$\Rightarrow \dfrac{{{x}^{2}}+1}{x}=\dfrac{17}{4}$ 

$\Rightarrow 4{{x}^{2}}+4=17x$ 

$\Rightarrow 4{{x}^{2}}-17x+4=0$ 

$\Rightarrow 4{{x}^{2}}-16x-x+4=0$ 

$\Rightarrow 4x\left( x-4 \right)-1\left( x-4 \right)=0$ 

$\Rightarrow \left( x-4 \right)\left( 4x-1 \right)=0$ 

$\therefore x=4\,or\,x=\dfrac{1}{4}$ 

4 Marks Questions

1. Find the roots of the following Quadratic Equations by applying quadratic formulas.

i). $2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$

$2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$

Comparing quadratic equation $2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$with general form $a{{x}^{2}}+bx+c=0$, we get $a=2,\text{ }b=-7\text{ }and\text{ }c=3$ 

 Putting these values in quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{7\pm \sqrt{{{\left( -7 \right)}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\times 2}$ 

$\Rightarrow x=\dfrac{7\pm \sqrt{49-24}}{4}$ 

$\Rightarrow x=\dfrac{7\pm 5}{4}$ 

$\Rightarrow x=\dfrac{7+5}{4},\dfrac{7-5}{4}$ 

$\therefore x=3,\dfrac{1}{2}$ 

ii). $2{{x}^{2}}+\text{ }x\text{ }\text{ }4\text{ }=\text{ }0$ 

Comparing quadratic equation $2{{x}^{2}}+\text{ }x\text{ }\text{ }4\text{ }=\text{ }0$with the general form$a{{x}^{2}}+bx+c=0$, we get $a=2,\text{ }b=1\text{ }and\text{ }c=-4$ 

Putting these values in quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$ x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( 2 \right)\left( -4 \right)}}{2\times 2} $ 

$ \Rightarrow x=\dfrac{-1\pm \sqrt{33}}{4} $ 

$ \Rightarrow x=\dfrac{-1-\sqrt{33}}{4},\dfrac{-1+\sqrt{33}}{4} $ 

iii). $4{{x}^{2}}+4\sqrt{3}x+3=0$ 

Comparing quadratic equation $4{{x}^{2}}+4\sqrt{3}x+3=0$ with the general form$a{{x}^{2}}+bx+c=0$, we get $a=4,\text{ }b=4\sqrt{3}\text{ }and\text{ }c=3$ 

$x=\dfrac{-4\sqrt{3}\pm \sqrt{{{\left( 4\sqrt{3} \right)}^{2}}-4\left( 4 \right)\left( 3 \right)}}{2\times 4}$ 

$\Rightarrow x=\dfrac{-4\sqrt{3}\pm \sqrt{0}}{8}$ 

$\Rightarrow x=\dfrac{-\sqrt{3}}{2}$ 

A quadratic equation has two roots. Here, both the roots are equal.

Therefore,$x=\dfrac{-\sqrt{3}}{2},\dfrac{-\sqrt{3}}{2}$ 

iv). $2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$ 

$2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$

Comparing quadratic equation $2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$ with the general form$a{{x}^{2}}+bx+c=0$, we get $a=2,b=1\text{ }and\text{ }c=\text{ }4$ 

$x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\times 2}$ 

$\Rightarrow x=\dfrac{-1\pm \sqrt{-31}}{4}$ 

But, the square root of a negative numbers is not defined. 

Therefore, Quadratic Equation $2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$ has no solution.

2. An express train takes$~1$ hour less than a passenger train to travel $132$  km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If, the average speed of the express train is $11$  km/h more than that of the passenger train, find the average speed of two trains

Ans: Let average speed of passenger train $=\text{ }x$ km/h 

Let average speed of express train $=\text{ }\left( x+11 \right)$ km/h 

Time taken by passenger train to cover $132$ km $=$ $\dfrac{132}{x}$  hours 

Time taken by express train to cover$132$km $=\left( \dfrac{132}{x+11} \right)$  hours

According to the given condition

$\dfrac{132}{x}=\dfrac{132}{x+11}+1$ 

$\Rightarrow 132\left( \dfrac{1}{x}-\dfrac{1}{x+11} \right)=1$ 

$132\left( \dfrac{x+11-x}{x\left( +11 \right)} \right)=1$ 

$\Rightarrow \text{ }132\left( 11 \right)=x\left( x+11 \right)$ 

$\Rightarrow \text{ }1452={{x}^{2}}+11x$ 

$\Rightarrow \text{ }{{x}^{2}}+11x\text{ }1452=0$ 

Comparing equation ${{x}^{2}}+11x\text{ }1452=0$with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=11\text{ }and\text{ }c=-1452$ 

Applying Quadratic Formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{-11\pm \sqrt{{{\left( 11 \right)}^{2}}-4\left( 1 \right)\left( -1452 \right)}}{2\times 1}$ 

$\Rightarrow x=\dfrac{-11\pm \sqrt{121+5808}}{2}$ 

$\Rightarrow x=\dfrac{-11\pm \sqrt{5929}}{2}$ 

$\Rightarrow x=\dfrac{-11\pm 77}{2}$ 

$\Rightarrow x=\dfrac{-11+77}{2},\dfrac{-11-77}{2}$ 

$\therefore x=33,-44$ 

As speed cannot be negative. Therefore, speed of passenger train $=\text{ }33$ km/h

And, speed of express train $=\text{ }x+11=33+11=44$ km/h

3. Sum of areas of two squares is $468\text{ }{{m}^{2}}$. If, the difference of their perimeters is $24$ meters, find the sides of the two squares.

Let perimeter of first square $=\text{ }x$ meters

Let perimeter of second square $=\text{ }\left( x+24 \right)$ meters 

Length of side of first square$~=$$\dfrac{x}{4}$  meters {Perimeter of square $=\text{ }4\text{ }\times $ length of side}

Length of side of second square $=$ $=\left( \dfrac{x+24}{4} \right)$  meters 

Area of first square $= side \times side $

=$\dfrac{x}{4}\times \dfrac{x}{4}=\dfrac{{{x}^{2}}}{16}{{m}^{2}}$ 

Area of second square $={{\left( \dfrac{x+24}{4} \right)}^{2}}{{m}^{2}}$ 

$\dfrac{{{x}^{2}}}{16}+{{\left( \dfrac{x+24}{4} \right)}^{2}}=468$ 

$\Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{x}^{2}}+576+48x}{16}=468$ 

$\Rightarrow \dfrac{{{x}^{2}}+{{x}^{2}}+576+48x}{16}=468$ 

$\Rightarrow \text{ }2{{x}^{2}}+576+48x=468\times 16$ 

$\Rightarrow \text{ }2{{x}^{2}}+48x+576=7488\text{ }$ 

$\Rightarrow \text{ }2{{x}^{2}}+48x\text{ }6912=0\text{ }$ 

$\Rightarrow \text{ }{{x}^{2}}+24x\text{ }3456=0$ 

Comparing equation${{x}^{2}}+24x\text{ }3456=0$ with standard form$a{{x}^{2}}+bx+c=0$, We get $a=1,b=24\text{ }and\,c=\text{ }-3456$

$x=\dfrac{-24\pm \sqrt{{{\left( 24 \right)}^{2}}-4\left( 1 \right)\left( -3456 \right)}}{2\times 1}$ 

$\Rightarrow x=\dfrac{-24\pm \sqrt{576+13824}}{2}$ 

$\Rightarrow x=\dfrac{-24\pm \sqrt{14400}}{2}=\dfrac{-24\pm 120}{2}$ 

$\Rightarrow x=\dfrac{-24+120}{2},\dfrac{-24-120}{2}$ 

$\therefore x=48,-72$ 

Perimeter of square cannot be in negative. Therefore, we discard$x=-72$. 

Therefore, perimeter of first square $=\text{ }48$meters 

And, Perimeter of second square $=\text{ }x+24=48+24=72$meters 

$\Rightarrow $ Side of First square $=\dfrac{Perimeter}{4}=\dfrac{48}{4}=12m$  

And, Side of second Square $=\dfrac{Perimeter}{4}=\dfrac{72}{4}=18m$ 

4. Is it possible to design a rectangular park of perimeter $80$ meters and area$400\text{ }{{m}^{2}}$. If so, find its length and breadth.

Let length of park $=\text{ }x$ meters

 We are given area of rectangular park $=\text{ }400\text{ }{{m}^{2}}$ 

Therefore, breadth of park $=$ $\dfrac{400}{x}$  meters {Area of rectangle$~=$ length $\times $ breadth}

Perimeter of rectangular park $=\text{ }2$ (length$+$breath)$=$ $\left( x+\dfrac{400}{x} \right)$ meters

We are given perimeter of rectangle $=\text{ }80$ meters

 According to condition:

$2\left( x+\dfrac{400}{x} \right)=80$ 

$\Rightarrow 2\left( \dfrac{{{x}^{2}}+400}{x} \right)=80$ 

  $\Rightarrow \text{ }2{{x}^{2}}+800=80x$ 

$\Rightarrow \text{ }2{{x}^{2}}-80x+800=0$ 

$\Rightarrow \text{ }{{x}^{2}}-40x+400=0$ 

Comparing equation, ${{x}^{2}}-40x+400=0$ with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=-40\text{ }and\text{ }c=400$ 

Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -40 \right)}^{2}}\text{ }4\left( 1 \right)\left( 400 \right)=1600\text{ }\text{ }1600=0$ 

Discriminant is equal to$0$ .

Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter $80$meters and area$400\text{ }{{m}^{2}}$.

Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$  to solve equation,s

$x=\dfrac{40\pm \sqrt{0}}{2}=\dfrac{42}{2}=20$ 

Here, both the roots are equal to$20$.

 Therefore, length of rectangular park $=\text{ }20$meters

Breadth of rectangular park$=\dfrac{400}{x}=\dfrac{400}{20}=20m$ 

5. If I had walked $1$  km per hour faster, I would have taken $10$  minutes less to walk $2$  km. Find the rate of my walking.

Ans: Distance $=\text{ }2$ km

 Let speed $=x$  km/hr.

 New speed$~=\text{ }\left( x+1 \right)$ km/hr. 

Time taken by normal speed $=\dfrac{2}{x}\,hr$ 

Time taken by new speed = $\dfrac{2}{x+1}hr$ 

$\dfrac{2}{x}-\dfrac{2}{x+1}=\dfrac{10}{60}$ 

$\Rightarrow \dfrac{2x+2-2x}{{{x}^{2}}+x}=\dfrac{1}{6}$ 

$\Rightarrow {{x}^{2}}+x=12$ 

$\Rightarrow {{x}^{2}}+x-12=0$ 

$\Rightarrow {{x}^{2}}+4x-3x-12=0$ 

$\Rightarrow x\left( x+4 \right)-3\left( x+4 \right)=0$ 

$\Rightarrow \left( x+4 \right)\left( x-3 \right)=0$ 

$\therefore x=-4\,or\,x=3$ 

So, speed is $x=3$ km/hr

6. A takes $6$  days less than the time taken by B to finish a piece of work. If both A and B together can finish it in $4$  days, find the time taken by B to finish the work.

Ans: Let B takes $x$  days to finish the work, then A alone can finish it in $\left( x-6 \right)$ days 

$\dfrac{1}{x}+\dfrac{1}{x-6}=\dfrac{1}{4}$ 

$\Rightarrow \dfrac{x-6+x}{{{x}^{2}}-6x}=\dfrac{1}{4}$ 

$\Rightarrow \dfrac{2x-6}{{{x}^{2}}-6x}=\dfrac{1}{4}$ 

$\Rightarrow {{x}^{2}}-6x=8x-24$ 

$\Rightarrow {{x}^{2}}-14x+24=0$ 

$\Rightarrow {{x}^{2}}-12x-2x+24=0$ 

$=x\left( x-12 \right)-2\left( x-12 \right)=0$ 

$\Rightarrow \left( x-12 \right)\left( x-2 \right)=0$ 

$\therefore x=12\,or\,x=2$ 

$x=2$ (Neglect)

So, B takes$x=12$ days.

7. A plane left $30$ minutes later than the schedule time and in order to reach its $59$  destination $1500$  km away in time it has to increase its speed by $250$km/hr from its usual speed. Find its usual speed.

Ans: Let usual speed$=x$ km/hr 

New speed $=\left( x+250 \right)$ km/hr

Total distance $=\text{ }1500$ km 

Time taken by usual speed $=\dfrac{1500}{x}$  hr 

Time taken by new speed $=\dfrac{1500}{x+250}$ hr 

$\dfrac{1500}{x}-\dfrac{1500}{x+250}=\dfrac{1}{2}$ 

$\Rightarrow \dfrac{1500x+1500\times 250-1500x}{{{x}^{2}}+250x}=\dfrac{1}{2}$ 

$\Rightarrow {{x}^{2}}+250x=\dfrac{1500\times 250}{2}$ 

    $\Rightarrow {{x}^{2}}+250x=750000$ 

    $\Rightarrow {{x}^{2}}+250x-750000=0$  

    $\Rightarrow x^{2}+1000x-750x-750000=0$

$\Rightarrow x\left( x+1000 \right)-750\left( x+1000 \right)=0$ 

$\Rightarrow x=750\,\,or\,x=-1000$ 

Therefore, usual speed is $750$ km/hr, $-1000$is neglected as speed cannot be negative.

8. A motor boat, whose speed is $15$ km/hr in still water, goes $30$ km downstream and comes back in a total time of $4$  hr $30$ minutes, find the speed of the stream.

Speed of motor boat in still water $=\text{ }15$ km/hr 

Speed of stream $=x$ km/hr 

Speed in downward direction $15+x$ 

Speed in downward direction $15-x$ 

 According to question,

$\dfrac{30}{15+x}+\dfrac{30}{15-x}=4\dfrac{1}{2}$ 

$\Rightarrow \dfrac{30\left( 15-x \right)+30\left( 15+x \right)}{\left( 15+x \right)\left( 15-x \right)}=\dfrac{9}{2}$ 

$\Rightarrow \dfrac{450-30x+450+30x}{225-{{x}^{2}}}=\dfrac{9}{2}$ 

$\Rightarrow 9\left( 225-{{x}^{2}} \right)=1800$ 

$\Rightarrow 225-{{x}^{2}}=200$ 

$\therefore x=5$ 

Speed of stream$~=\text{ }5$ km/hr.

9. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.

Let $x$ be the number of hours required by the second pipe alone to till the pool and first pipe $\left( x+5 \right)$ hour while third pipe$\left( x-4 \right)$ hour

$\dfrac{1}{x+5}+\dfrac{1}{x}=\dfrac{1}{x-4}$ 

$\Rightarrow \dfrac{x+x+5}{{{x}^{2}}+5x}=\dfrac{1}{x-4}$ 

$\Rightarrow {{x}^{2}}-8x-20=0$ 

$\Rightarrow {{x}^{2}}-10x+2x-20=0$ 

$\Rightarrow x\left( x-10 \right)+2\left( x-10 \right)=0$ 

$\Rightarrow \left( x-10 \right)\left( x+2 \right)=0$ 

$\Rightarrow x=10\,or\,x=-2\left( Neglected \right)$ 

10. A two-digit number is such that the product of its digits is$18$. When $63$  is subtracted from the number the digit interchanges their places. Find the number

Let digit on unit’s place $=x$ 

Digit on ten’s place $=y$ 

 $xy=18$ (given) 

Number = 10.$y+x$ 

$=10\left( \dfrac{18}{x} \right)+x$ 

  $10\left( \dfrac{18}{x} \right)+x-63=10x+\dfrac{18}{x}$ 

   $\Rightarrow \dfrac{180}{x}+\dfrac{x-63}{1}=\dfrac{10{{x}^{2}}+18}{x}$ 

   $\Rightarrow \dfrac{180+{{x}^{2}}-63x}{x}=\dfrac{10{{x}^{2}}+18}{x}$ 

   $\Rightarrow 9{{x}^{2}}+63x-162=0$ 

   $\Rightarrow 9\left( {{x}^{2}}+9x-2x-18 \right)=0$ 

   $\Rightarrow x\left( x+9 \right)-2\left( x+9 \right)=0$ 

    $\Rightarrow x=2\,or\,x=-9$ 

Number$=10\left( \dfrac{18}{2} \right)+2=92$ 

11. A factory kept increasing its output by the same percent ago every year. Find the percentage if it is known that the output is doubled in the last two years.

According to question, 

2P$=p{{\left( 1+\dfrac{r}{100} \right)}^{2}}$ 

$\Rightarrow \dfrac{\sqrt{2}}{1}=1+\dfrac{r}{100}$ 

$\sqrt{2}-1=\dfrac{r}{100}$ 

$\Rightarrow r=\left( \sqrt{2}-1 \right)100$ 

12. Two pipes running together can fill a cistern in if one pipe takes $3\dfrac{1}{13}$  minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Let the faster pipe takes minutes to fill the cistern and the slower pipe will take $\left( x+3 \right)$  minutes. 

$\dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{1}{\dfrac{40}{13}}$ 

$\Rightarrow \dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{13}{40}$ 

$\Rightarrow \dfrac{x+3+x}{{{x}^{2}}+3x}=\dfrac{13}{40}$ 

$\Rightarrow 13{{x}^{2}}-41x-120=0$ 

$\Rightarrow 13{{x}^{2}}-65x+24x-120=0$ 

$\Rightarrow 13x\left( x-5 \right)+24\left( x-5 \right)=0$ 

$\therefore x=5\,or\,x=\dfrac{-24}{13}\left( Neglected \right)$ 

13. If the roots of the equation$\left( a-b \right){{x}^{2}}+\left( b-c \right)x+\left( c-a \right)=0$  are equal, prove that$2a=b+c$ 

$\left( a-b \right){{x}^{2}}+\left( b-c \right)x+\left( c-a \right)=0$

$={{\left( b-c \right)}^{2}}-4\times \left( a-b \right)\times \left( c-a \right)$ 

$={{b}^{2}}+{{c}^{2}}-2bc-4\left( ac-{{a}^{2}}-bc+ab \right)$ 

$={{b}^{2}}+{{c}^{2}}-2bc-4ac+4{{a}^{2}}+4bc-4ab$ 

$={{\left( b \right)}^{2}}+{{\left( c \right)}^{2}}+{{\left( 2a \right)}^{2}}+2bc-4ac-4ab$ 

$={{\left( b+c-2a \right)}^{2}}$ 

For equal root s,

 D = 0

$\Rightarrow {{\left( b+c-2a \right)}^{2}}=0$ 

$\Rightarrow \left( b+c-2a \right)=0$ 

$\Rightarrow b+c=2a$ 

14. Two circles touch internally. The sum of their areas is $116\pi c{{m}^{2}}$ and the distance between their centers is $6$ cm. Find the radii of the circles.

Let ${{r}_{1}}$ and ${{r}_{2}}$ be the radius of two circles 

    $\Pi {{r}_{1}}^{2}+\Pi {{r}_{2}}^{2}=116\Pi $ 

    $\Rightarrow {{r}_{1}}^{2}+{{r}_{2}}^{2}=116......\left( i \right)$ 

    ${{r}_{2}}-{{r}_{1}}=6$ (Given)

   $\Rightarrow {{r}_{2}}=6+{{r}_{1}}$ 

   Puting the value of${{r}_{2}}$  in eq. … (i) we get

   $\Rightarrow {{r}_{1}}^{2}+36+{{r}_{1}}^{2}+12{{r}_{1}}=116$ 

   $\Rightarrow 2{{r}_{1}}^{2}+12{{r}_{1}}-80=0$ 

  ${{r}_{1}}^{2}+6{{r}_{1}}-40=0$ 

  $\Rightarrow {{r}_{1}}^{2}+10{{r}_{1}}-4{{r}_{1}}-40=0$ 

  $\Rightarrow {{r}_{1}}\left( {{r}_{1}}+10 \right)-4\left( {{r}_{1}}+10 \right)=0$ 

  $\Rightarrow \left( {{r}_{1}}+10 \right)\left( {{r}_{1}}-4 \right)=0$ 

  $when\,{{r}_{1}}=4\,cm$ 

  ${{r}_{2}}=6+{{r}_{1}}$ 

  $=6+4$ 

  ${{r}_{2}}=10\,cm$ 

  $\Rightarrow {{r}_{1}}=-10$ (Neglect) or ${{r}_{1}}=\text{ }4\text{ }cm$ 

15. A piece of cloth costs Rs.$200$. If the piece was $5$ m longer and each metre of cloth costs Rs. $2$  less the cost of the piece would have remained unchanged. How long is the  piece and what is the original rate per meter?

Let the length of piece $=x$  m 

Rate per meter $=\dfrac{200}{x}$  

A New length $=\text{ }\left( x+5 \right)$ 

A New rate per meter $=\dfrac{200}{x+5}$ 

$\dfrac{200}{x}-\dfrac{200}{x+5}=2$ 

$\Rightarrow \dfrac{200\left( x+5 \right)-200x}{\left( x+5 \right)}=\dfrac{2}{1}$ 

$\Rightarrow \dfrac{200x+1000-200x}{{{x}^{2}}+5x}=\dfrac{2}{1}$ 

$\Rightarrow {{x}^{2}}+5x=500$ 

$\Rightarrow {{x}^{2}}+25x-20x-500=0$ 

$\Rightarrow x\left( x+25 \right)-20\left( x+25 \right)=0$ 

$\Rightarrow \left( x+25 \right)\left( x-20 \right)=0$ 

$\therefore x=-25\left( Neglect \right)\,or\,x=20$ 

Rate per meter $=\text{ }10$ 

16. $a{{x}^{2}}+bx+x\text{ }=\text{ }0$ , $a\ne 0$ solve by quadratic formula.

$a{{x}^{2}}+bx+x\text{ }=\text{ }0$ 

$\Rightarrow {{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0$ 

$\Rightarrow {{x}^{2}}+\dfrac{b}{a}x{{\left( \dfrac{b}{2a} \right)}^{2}}-{{\left( \dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}=0$ 

$\Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}}{4{{a}^{2}}}-\dfrac{c}{a}$ 

x$\Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}$ 

$\Rightarrow x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}}$ 

$\Rightarrow x+\dfrac{-b}{2a}=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{2a}}$ 

$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$\Rightarrow x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$ 

$or\,\Rightarrow x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$ 

17. The length of the hypotenuse of a right-angled triangle exceeds the length of the base by $2$ cm and exceeds twice the length of the altitude by $1$  cm. Find the length of each side of the triangle.

Let base of the triangle $=\text{ }x$  

Altitude of the triangle $=\text{ }y$ 

Hypotenuse of the triangle $=\text{ }h$ 

$h=x+2$ 

$h=2y+1$ 

$\Rightarrow x+2=2y+1$ 

    $\Rightarrow x+2-1=2y$ 

$\Rightarrow x-1=2y$ 

$\Rightarrow \dfrac{x-1}{2}=y$ 

$And\,{{x}^{2}}+{{y}^{2}}={{h}^{2}}$ 

$\Rightarrow {{x}^{2}}+{{\left( \dfrac{x-1}{2} \right)}^{2}}={{\left( x+2 \right)}^{2}}$ 

$\Rightarrow {{x}^{2}}-15x+x-15=0$ 

$\Rightarrow \left( x-15 \right)\left( x+1 \right)=0$ 

$\Rightarrow x=15\,or\,x=-1$ 

Base of the triangle $=\text{ }15$ cm

 Altitude of the triangle $=\dfrac{x+1}{2}=8cm$ 

Hypotenuse of the triangle $=\text{ }17$ cm

18. Find the roots of the following quadratic equations if they exist by the method of completing square.

i). $2{{x}^{2}}\text{ }-7x+3=0$ 

(i) $2{{x}^{2}}\text{ }-7x+3=0$

Dividing the equation by $2$  to make coefficient of ${{x}^{2}}$ equal to$1$ we get

${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0$ 

Dividing the middle term of the equation by$2x$, we get

$\dfrac{7}{2}x\times \dfrac{1}{2x}=\dfrac{7}{4}$ 

Adding and subtracting square of $\dfrac{7}{4}$ from the equation ${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0$ we get

${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}+{{\left( \dfrac{7}{4} \right)}^{2}}-{{\left( \dfrac{7}{4} \right)}^{2}}=0$

$\Rightarrow {{x}^{2}}+{{\left( \dfrac{7}{4} \right)}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}+-{{\left( \dfrac{7}{4} \right)}^{2}}=0$      $\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right\}$ 

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{+}\dfrac{\text{24-49}}{\text{16}}\text{=0}$ 

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{49-24}}{\text{16}}$ 

Square rooting on both the sides we get

$\Rightarrow \text{x-}\dfrac{\text{7}}{\text{4}}\text{= }\!\!\pm\!\!\text{ }\dfrac{\text{5}}{\text{4}}$ 

$\Rightarrow x\text{=}\dfrac{\text{5}}{\text{4}}\text{+}\dfrac{\text{7}}{\text{4}}\text{=}\dfrac{\text{12}}{\text{4}}\text{=3}\,\text{and}\,\text{x=-}\dfrac{\text{5}}{\text{4}}\text{+}\dfrac{\text{7}}{\text{4}}\text{=}\dfrac{\text{2}}{\text{4}}\text{=}\dfrac{\text{1}}{\text{2}}$ 

Therefore,$x=\dfrac{1}{2},3$ 

ii). $2{{x}^{2}}+x\text{ }4=0$ 

$\text{2}{{\text{x}}^{\text{2}}}\text{+x-- 4=0}$

Dividing equation by$2$ , 

${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{-2=0}$ 

Following procedure of completing square,

${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{-2+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=0}$ 

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-2-}\dfrac{\text{1}}{\text{16}}\text{=0}$      $\left\{ {{\left( \text{a-b} \right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{-2ab} \right\}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}\dfrac{\text{33}}{\text{16}}\text{=0}$ 

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}\dfrac{\text{33}}{\text{16}}$

Taking square root on both sides,

$\Rightarrow \text{x+}\dfrac{\text{1}}{\text{4}}\text{= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}}{\text{4}}$ 

$\Rightarrow x=\dfrac{\sqrt{33}}{4}-\dfrac{1}{4}\text{=}\dfrac{\sqrt{\text{33}}\text{-1}}{\text{4}}\,\text{and}\,\text{x=-}\dfrac{\sqrt{\text{33}}}{\text{4}}\text{-}\dfrac{\text{1}}{\text{4}}\text{=}\dfrac{\text{-}\sqrt{\text{33}}\text{-1}}{\text{4}}$ 

Therefore,$\text{x}=\dfrac{\sqrt{33}-1}{4},\dfrac{-\sqrt{33}-1}{4}$ 

$\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{\text{3}}\text{x+3=0}$

Dividing this equation by $4,$ we get

${{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{=0}$ 

By the procedure of completing square we get

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{+}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{-}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{=0}$ 

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{-}\dfrac{\text{3}}{\text{4}}\text{=0}$       $\left\{ {{\left( \text{a-b} \right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{-2ab} \right\}$

$\Rightarrow {{\left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{=0}$ 

$\Rightarrow \left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)\left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)\text{=0}$ 

$\Rightarrow \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{=0,x+}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{=0}$ 

$\Rightarrow \text{x=-}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{,-}\dfrac{\sqrt{\text{3}}}{\text{2}}$

iv). $2{{x}^{2}}+x+4=0$

$\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Dividing this equation by$2$we get

${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+2=0}$ 

By the procedure of completing square,

$\Rightarrow {{x}^{2}}+\dfrac{x}{2}+2+{{\left( \dfrac{1}{4} \right)}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}=0$ 

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+2-}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=0}$      $\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right\}$

$\Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}+2-\dfrac{1}{16}=0$ 

${{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{1}{16}-2=\dfrac{1-32}{16}$ 

Right hand side does not exist because the square root of the negative number does not exist. 

Therefore, there is no solution for quadratic equation $2{{x}^{2}}+x+4=0$

Practice Questions for Class 10 Maths Chapter 4 - Quadratic Equations

The following are some of the questions that can be taken up by students to assist them in the board preparations related to Quadratic Equations.

Question 1. A rectangular field's diagonal is 60 metres longer than the shorter side. Find the field's sides if the longer side is 30 metres more than the shorter side.

Anwer- 120m which is the correct answer.

Question 2. Is it possible to build a rectangular park with an 80 meter perimeter and a 400-square-meter area? If this is the case, determine its length and breadth.

Answer- Length and breadth both should be equal to 20 m.

Question 3- A train moving at a speed of 600 km was slowed down due to bad weather. The train's average speed dropped by 200 km/hr, and the journey time was increased by 30 minutes. Determine the train's initial duration.

Answer- 1 hour is the answer.

Vedantu's goal is to help students from all over the country prepare for exams. As a result, all of our study materials are available in PDF format, which can be downloaded for free. Our professionals answer the questions to provide students with a sample answer for each question on the Class 10 CBSE test papers . For practice, students can download and solve the question paper on their own. For exam preparation, they can use Vedantu to access critical questions, revision notes, NCERT chapter-by-chapter solutions , and other resources.

Important Related Links for CBSE Class 10 Maths

FAQs on Important Questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations

Q1: What are the roots of a quadratic equation? Explain the nature of roots of a quadratic equation.

A1: The roots of the quadratic equation are given as :  x = (-b ± √D)/2a, where D = b 2 – 4ac.

The nature of  the roots of a quadratic equation are as mentioned below: 

D > 0, roots are real and distinct (unequal)

D = 0, roots are real and equal (coincident)

D < 0, roots are imaginary and unequal

Q2: Explain in detail the nature of roots of a quadratic equation.

A2: The nature of roots of a quadratic equation is as follows:

If the value of discriminant = 0 i.e. b 2 – 4ac = 0, then the quadratic equation will have equal roots i.e. α = β = -b/2a.

If the value of discriminant < 0 i.e. b 2 – 4ac < 0, then the quadratic equation will have imaginary roots i.e α = (p + iq) and β = (p – iq). Where ‘iq’ is the imaginary part of a complex number.

If the value of discriminant (D) > 0 i.e. b 2 – 4ac > 0, then the quadratic equation will have real roots.

If the value of discriminant > 0 and D is a perfect square, then the quadratic equation will have rational roots.

If the value of discriminant (D) > 0 and D is not a perfect square, then the quadratic equation will have irrational roots i.e. α = (p + √q) and β=(p – √q).

If the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers, then the quadratic equation will have integral roots.

Q3: What is the Relationship between Coefficient and Roots of  Quadratic Equation?

A3: If α and β are roots of a Quadratic Equation ax 2 + bx + c then,

α + β = -b/a

α – β = ±√[(α + β) 2 – 4αβ]

|α + β| = √D/|a|

The relationship between the roots and coefficient of a polynomial equation can be derived by simplifying the given polynomials and substituting the above results as shown below.

α 2 β + β 2 α = αβ (α + β) = – bc/a 2

α 2 + αβ + β 2 = (α + β) 2 – αβ = (b 2 – ac)/a 2

α 2 + β 2 = (α – β) 2 – 2αβ

α 2 – β 2 = (α + β) (α – β)

α 3 + β 3 = (α + β) 3 + 3αβ(α + β)

α 3 – β 3 = (α – β) 3 + 3αβ(α – β)

(α/β) 2 + (β/α) 2 = α 4 + β 4 /α 2 β 2

Q4: What happens when a < 0 and  b 2 – 4ac > 0.

A4: When a < 0 and  b 2 – 4ac > 0, the graph of a quadratic equation will be concave downwards and will intersect the x-axis at two points α and β with α < β. The quadratic equation will have two real roots (α and β) and the curve will always lie below the x-axis.

The quadratic function f(x) will be positive i.e. f(x) > 0, for the values of x lying in the interval (α, β).

The quadratic function f(x) will be equal to zero i.e. f(x) = 0, if x = α or β

The quadratic function f(x) will be negative i.e. f(x) < 0 for the values of x lying in the interval (−∞, α) ∪ (β, ∞).

CBSE Class 10 Maths Important Questions

Cbse study materials.

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Case Study Class 10 Maths Questions

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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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Related Posts

• CBSE Class 10 Maths Sample Paper 2020-21
• Class 12 Maths Case Study Questions
• CBSE Reduced Syllabus Class 10 (2020-21)
• Class 10 Maths Basic Sample Paper 2024
• How to Revise CBSE Class 10 Maths in 3 Days
• CBSE Practice Papers 2023
• Class 10 Maths Sample Papers 2024
• Competency Based Learning in CBSE Schools

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Chapter 4 Class 10 Quadratic Equations

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Updated for Latest NCERT for 2023-2024 Boards.

Get NCERT Solutions for all exercise questions and examples of Chapter 4 Class 10 Quadratic Equations free at Teachoo. Answers to each and every question is provided video solutions. 

In this chapter, we will learn

• What is a Quadratic Equation
• What is the Standard Form of a Quadratic Equation
• Solution of a Quadratic Equation by Factorisation ( Splitting the Middle Term method)
• Solving a Quadratic Equation by Completing the Square
• Solving a Quadratic Equation using D Formula (x = -b ± √b 2 - 4ac / 2a)
• Checking if roots are real, equal or no real roots (By Checking the value of D = b 2 - 4ac)

This chapter is divided into two parts - Serial Order Wise, Concept Wise

In Serial Order Wise, the chapter is divided into exercise questions and examples.

In Concept Wise, the chapter is divided into concepts. First the concepts are explained, and then the questions of the topic are solved - from easy to difficult.

We suggest you do the Chapter from Concept Wise - it is the Teachoo (टीचू) way of learning.

Note: When you click on a link, the first question of the exercise will open. To open other question of the exercise, go to bottom of the page. There is a list with arrows. It has all the questions with Important Questions also marked.

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CBSE Class 10 Maths Case Study

CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus. 

These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.

Class 10 Maths (Formula, Case Based, MCQ, Assertion Reason Question with Solutions)

In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.

Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too. 

Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.

For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.

CBSE Class 10th Case study Questions

Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.

Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.

CBSE Class 10th Assertion Reasoning Questions

These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement? 

Moreover, to solve the problem they need to look at the given options and then answer them.

CBSE Class 10 Maths Case Based MCQ

CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.

Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.

Class 10th Mathematics Multiple Choice Questions

Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.

Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.

The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.

To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.

In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.

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Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Solutions

January 4, 2023 by Sastry CBSE

Extra Questions for Class 10 Maths Quadratic Equations with Answers

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations. According to new CBSE Exam Pattern,  MCQ Questions for Class 10 Maths  Carries 20 Marks.

You can also download Class 10 Maths to help you to revise complete syllabus and score more marks in your examinations.

Quadratic Equations Class 10 Extra Questions Very Short Answer Type

Question 2. For what values of k, the roots of the equation x 2 + 4x + k = 0 are real? Or Find the value of k for which the roots of the equation 3x 2 – 10x + k = k = 0 are reciprocal of each other. [CBSE 2019] Answer: For real roots D ≥ 0 i.e., b 2 – 4ac ≥ 0 Here, a = 1, b = 4, c = k ∴ (4) 2 – 4(1)k ≥ 0 ⇒ 16 – 4k ≥ 0 ⇒ k ≤ A Or Roots of the equation 3x 2 – 10x + k = 0 are reciprocal of each other ⇒ Product of roots = 1 ⇒ \(\frac{c}{a}=\frac{k}{3}\) = 1 ⇒ k = 3

Question 3. If the roots of quadratic equation px 2 + 6x + 1 = 0 are real, then find p. Answer: Roots are real, D ≥ 0 b 2 – 4ac ≥ 0 (6 2 ) – 4 × p × 1 ≥ o 9 ≥ p

Quadratic Equations Class 10 Extra Questions Short Answer Type-1

Question 1. Find the roots of the quadratic equation √2x 2 + 7x + 5√2 = 0 [CBSE Delhi 2017] Answer: √2x 2 + 7x + 5√2 = 0 √2x 2 + 2x + 5x + 5√2 = 0 ⇒ (√2x + 5) (x + √2) = 0 ⇒ x = \(\frac{-5}{\sqrt{2}}\), – √2 or \(\frac{-5 \sqrt{2}}{2}\), – √2

Question 3. If – 5 is a root of the quadratic equation 2x 2 + px – 15 = 0 and the quadratic equation p(x 2 + x) + k = 0 has equal roots, find the value of k. [CBSE Outside Delhi 2016] Answer: – 5 is root of the equation 2x 2 + px – 15 = 0, then 2(-5) 2 + p(-5) – 15 = 0 ⇒ 50 + (- 5)p – 15 = 0 ⇒ p = \(\frac{-35}{-5}\) = 7 ……….. (1) p(x 2 + x) + k = 0 has equal root ⇒ b 2 – 4ac = 0 Here b = p, a = p, c = k ⇒ p 2 = 4 pk ………….. (2) (7) 2 – 4 × 7k = 0 (1) and (2) gives, ∴ k = \(\frac{7}{4}\)

Question 4. For what value of k, x 2 + 4x + k is a perfect square? Answer: A quadratic expression is a perfect square, if and only if corresponding equation has equal roots. i.e., D = 0 ⇒ 16 – 4k = 0 ⇒ k = 4

Question 6. The product of two consecutive positive integers is 306. Find the integers. Answer: Let the consecutive positive integers be x and x + 1. Then, x (x + 1) = 306 or, x 2 + x – 306 = 0 or, x 2 + 18x – 17x – 306 = 0 or, x(x + 18) – 17(x + 18) = 0 or, (x + 18) (x – 17) = 0 x = – 18, 17 Neglecting x = – 18 ∴ x = 17 and x + 1 = 17 + 1 = 18 Hence, two consecutive positive integers are 17 and 18.

Quadratic Equations Class 10 Extra Questions Short Answer Type-2

Question 4. Find the value of p for which the quadratic equation (2p + 1)x 2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots. [S.A-II, 2014] Answer: (2p + 1)x 2 – (7p + 2)x + (7p – 3) = 0 …(1) For equal roots D = 0 b 2 – 4ac = 0 ⇒ [- (7p + 2)] 2 – 4(2p + 1) (7p – 3) = 0 ⇒ 49p 2 + 4 + 28p – 4 (14p 2 – 6p + 7p – 3) = 0 ⇒ 49p 2 + 28p + 4 – 56p2 + 24p – 28p + 12 = 0 ⇒ – 7p 2 + 24p + 16 = 0 ⇒ 7p 2 – 24p – 16 = 0 ⇒ 7p 2 – 28p + 4p – 16 = 0 ⇒ 7p (p – 4) + 4 (p – 4) = 0 ⇒ (p – 4) (7p + 4) = 0 ∴ p = 4, \(\frac{-4}{7}\) If p = 4, then (1) ⇒ 9x 2 – 30x + 25 = 0 ⇒ 9x 2 – 15x – 15x + 25 = 0 ⇒ 3x(3x – 5) – 5(3x – 5) = 0 ⇒ (3x – 5) (3x – 5) = 0 ⇒ (3x – 5) 2 = 0 ⇒ 3x – 5 = 0 ∴ x = \(\frac{5}{3}\) ∴ Roots are \(\frac{5}{3}\) and \(\frac{5}{3}\). If P = – \(\frac{4}{7}\), then (1) ⇒ \(\frac{-x^2}{7}\) + 2x – 7 = 0 – x 2 + 14x – 49 = 0 x 2 – 14x + 49 = 0 (x – 7) 2 = 0 ∴ x = 7, 7

Quadratic Equations Class 10 Extra Questions Long Answer Type 1

Question 3. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ? [CBSE 2018] Answer: Let original speed of train be x km/ hr Distance = 63 km; time (t 1 ) = 63/x hrs ∴ Faster speed = (x + 6) km/hr [∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)] Distance = 72 km; time (t 2 ) = 72/ (x + 6) hrs. (t 1 ) + (t 2 ) = 3 hrs \(\frac{63}{x}+\frac{72}{(x+6)}\) = 3 ⇒ 63(x + 6) + 72x = 3x (x + 6) ⇒ 63x + 378 + 72x = 3x 2 + 18x ⇒ 3x 2 – 117x – 378 = 0 ⇒ x 2 – 39x – 126 = 0 ⇒ 7x 2 – 42x + 3x + 126 = 0 ⇒ (x – 42) (x + 3) = 0 ⇒ x = -3 [Rejected as speed can’t be negative] ∴ x = 42 ∴ Original speed of train is 42 km/hr.

Question 5. A thief runs with a uniform speed of 100 m/ minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief? [CBSE 2016] Answer: Suppose the policeman catches the thief after t minutes. Uniform speed of the thief = 100 metres/minute ∴ Distance covered by thief in (t + 1) = 100 (t + 1) metres [∵ Distance = Speed × Time] Distance covered by policeman in t minutes = Sum of t terms of an AP with first term 100 and common difference 10. \(\frac{t}{2}\) [2 × 100 + (t -1) × 10] m = t[100 + 5(t – 1)] = t(5t + 95) = 5t 2 + 95t When the policeman catches the thief. 5t 2 + 95t = 100 (t + 1) ⇒ 5t 2 + 95t = 100t + 100 ⇒ 5t 2 – 5t – 100 = 0 ⇒ t 2 – t – 20 = 0 ⇒ t = – 4 and t = 5 Neglecting -ve time. We get t = 5 Thus, the policeman catches the thief after 5 minutes.

Quadratic Equations Class 10 Extra Questions HOTS

Question 3. If the equation (1 + m 2 ) x 2 + 2mcx + c 2 – a 2 = 0 has equal roots, show that c 2 = a 2 (1 + m 2 ). Answer: Given equation is, (1 + m 2 ) x 2 + 2mcx + c 2 – a 2 = 0 ∵ Roots are equal. ∴ D = 0 B 2 – 4AC = 0 (2mc) 2 – 4 (1 + m 2 ) (c 2 – a 2 ) = 0 or 4m 2 c 2 – 4 [c 2 – a 2 + m 2 c 2 – m 2 a 2 ] = 0 or 4[m 2 c 2 – c 2 + a 2 – m 2 c 2 + m 2 a 2 ] = 0 or – c 2 + a 2 (1 + m 2 ) = 0 or c 2 = a 2 (1 + m 2 ).

Question 4. A person on tour has ₹ 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by ₹ 3. Find the original duration of the tour. Answer: Let, original duration of tour = x days Expenditure on tour = ₹ 360 So, expenditure per day = ₹ \(\frac{360}{x}\) According to 1st condition, Duration of extended tour = (x + 4) days Expenditure per day = ₹ \(\frac{360}{x+4}\) According to 2nd condition, \(\frac{360}{x}-\frac{360}{x+4}\) = 3 360 (x + 4) – 360x = 3x (x + 4) 360x + 1440 – 360x = 3 (x 2 + 4x) x 2 + 4x – 480 = 0, S = 4 x 2 + 24x – 20x – 480 = 0, P = – 480 x(x + 24) – 20 (x + 24) = 0 (x + 24) (x – 20) = 0 Either x + 24 = 0 or x – 20 = 0 x = – 24 or x = 20 ∵ x cannot be negative so, x = 20 Hence, original duration of tour has 20 days.

Question 5. The hypotenuse of a right triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle. Answer: Let, shortest side of triangle = x m Hypotenuse of triangle = (2x – 1) m Third side of triangle = (x + 1) m By Pythagoras theorem, (Hyp.) 2 = (Base) 2 + (Alt.) 2 (2x – 1) 2 = (x + 1) 2 + (x) 2 4x 2 + 1 – 4x = x 2 + 1 + 2x + x 2 2x 2 – 6x = 0 2x (x – 3) = 0 Either 2x = 0 or x – 3 = 0 x = 0 or x = 3 But side cannot be zero, so . x = 3 ∴ Shortest side = x = 3 m Hypotenuse = (2x – 1) = [2 (3) – 1] = 5 m Third side = x + 1 = (3 + 1) = 4 m.

Multiple Choice Questions

Choose the correct option for each of the following:

Question 1. The condition on ‘a’ for ax 2 + bx + c = 0 to represent a quadratic equation is: (a) a = 0 (b) a > 0 (c) a < 0 (d) a ≠ 0 Answer: (d) a ≠ 0

Question 2. The value of k for which the expression x 2 + 4x + k is a perfect square is : (a) 4 (b) – 4 (c) 16 (d) 3 Answer: (a) 4

Question 3. Which of the following equation has 3 as a root? (a) x 2 – 4x + 2 = 0 (b) x 2 + x – 3 = 0 (c) 2x 2 – 8x + 6 = 0 (d) 3x 2 – 6x – 2 = 0 Answer: (c) 2x 2 – 8x + 6 = 0

Question 4. If \(\frac{1}{3}\) is a root of x 2 + kx – 1 = 0, then value of k is: (a) \(\frac{4}{3}\) (b) \(\frac{8}{3}\) (c) \(\frac{1}{3}\) (d) – 3 Answer: (b) \(\frac{8}{3}\)

Question 5. Which of the following is true about the equation \(\frac{1}{x^2-3 x+5}\) = 1 (a) Its discriminant is not defined (b) Its discriminant is – 7 (c) Its discriminant is -11 (d) Its discriminant is same as coefficient of x Answer: (b) Its discriminant is – 7

Question 6. Which of the following is not a quadratic equation? (a) \(\frac{1}{2 x^2-x+1}\) = – 2 (b) (x 2 – 1) 2 = x 4& + 31 + 4x 2 (c) 2(x – 1) 2 = 4x 2 – 3x + 1 (d) (√3x + √5) 2 + x 2 = 4x 2 – 5x Answer: (d) (√3x + √5) 2 + x 2 = 4x 2 – 5x

Question 7. The roots of the equation ax 2 + bx + c = 0 are non- real, if (a) b 2 – 4ac = 0 (b) b 2 – 4ac < 0 (c) b 2 – 4ac > 0 (d) coefficient of x is zero Answer: (b) b 2 – 4ac < 0

Question 8. The roots of equation ax 2 + bx + c = 0, a ≠ 0 are real, if b 2 – 4ac is (a) = 0 (b) ≥ 0 (c) ≤ 0 (d) none of these Answer: (b) ≥ 0

Question 9. If n denotes the number of roots of a quadratic equation, then (a) n is always equal to 2 (b) n < 2 (c) n ≤ 2 (d) n > 2 Answer: (c) n ≤ 2

Question 10. The positive value of /c for which the equation x 2 + kx + 9 = 0 and x 2 – 12x + k = 0 will both have real roots (a) 36 (b) 6 (c) 16 (d) 12 Answer: (b) 6

Question 11. The equation (a 2 + b 2 )x 2 – 2 (ac + bd) x + c 2 + d 2 = 0 has equal roots, then (a) ab = cd (b) ad = bc (c) ad = √bc (d) ad = √cd Answer: (d) ad = √cd

Question 12. The sum of a number and its reciprocal is 10/3, then the number is: (a) 1 (b) 2 (c) 3 (d) 5 Answer: (c) 3

Question 13. The value of k for which – 5 is a root of 2x 2 + px – 15 = 0 and the quadratic equation p(x 2 + x) + k = 0 has equal roots is (a) \(\frac{3}{2}\) (b) \(\frac{3}{7}\) (c) \(\frac{7}{4}\) (d) \(\frac{1}{4}\) Answer: (c) \(\frac{7}{4}\)

Question 14. What constant should be added or subtracted to solve the quadratic equation x 2 – \(\frac{\sqrt{5}}{2}\)x – 4 = 0 by the method of completing the square? (a) \(\frac{5}{4}\) (b) \(\frac{5}{16}\) (c) \(\frac{\sqrt{5}}{4}\) (d) \(\frac{25}{4}\) Answer: (b) \(\frac{5}{16}\)

Question 15. The value of k for which roots of kx 2 – 3x + 1 = 0 are real is . (a) k ≥ \(\frac{9}{4}\) (b) k ≤ \(\frac{- 9}{4}\) (c) k ≤ \(\frac{9}{4}\) (d) k ≤ \(\frac{3}{2}\) Answer: (c) k ≤ \(\frac{9}{4}\)

Fill in the blanks:

Question 1. Quadratic equation ax 2 + bx + c = 0 has two _____________ and _____________ roots if b 2 – 4ac > 0. Answer: real, unequal

Question 2. Quadratic equation ax 2 + bx + c = 0 has two real and _____________ roots if b 2 – 4ac = 0. Answer: equal

Question 3. For a quadratic equation ax 2 + bx + c = 0, the quantity b 2 – 4ac is called _____________ . Answer: discriminant

Question 4. Quadratic equation ax 2 + bx + c = 0 has two _____________ roots if b 2 – 4ac < 0. Answer: imaginary Question 5. If D > 0 and is a perfect square then roots are _____________ (rational/irrational/imaginary). Answer: rational

Question 6. If D > 0 and it is not a perfect square then roots are _____________ (rational/irrational/imaginary). Answer: irrational

Question 7. If the value of mα 2 + nα + p is zero, then a is said to be the _____________ of equation mx 2 + nx + p = 0. Answer: root

Question 8. The quadratic formula for finding roots of equation ax 2 + bx + c = 0 is given by _____________ . Answer: x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Question 9. A quadratic equation can have at most _____________ roots. Answer: two

Question 10. The quadratic equation x 2 – 3x + 5 = 0 has _____________ roots. Answer: imaginary

Extra Questions for Class 10 Maths

Ncert solutions for class 10 maths, free resources.

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Quick Resources

Important Questions for Class 10 Maths Chapter 4 Quadratic Equations

Chapter 4 quadratic equations important questions for cbse class 10 maths board exams.

Important Questions for Chapter 4 Quadratic Equations Class 10 Maths

Quadratic equations class 10 maths important questions very short answer (1 mark).

Let one root be α and other root be 6α.

Since k = 0 is not possible, therefore k = 3.

4. If 1 is a root of the equations ay 2 + ay + 3 = 0 and y 2  + y + b = 0, then find the value of ab.

ay 2 + ay + 3 = 0 ⇒ a(1) 2 + a(1) + 3 = 0 ⇒ 2a = -3 ⇒ a = −32

y 2 + y + b = 0 ⇒ 1 2 + 1 + b = 0 ⇒ b = -2 ∴ ab = (−32)(−2) = 3

5. Find the roots of the equation x 2  – 3x – m (m + 3) = 0, where m is a constant.

x 2 – 3x – m(m + 3) = 0 ⇒ D = b 2 – 4ac ⇒ D = (- 3) 2 – 4(1) [-m(m + 3)] = 9 + 4m (m + 3) = 4m 2 + 12m + 9 = (2m + 3) 2

∴ x = m + 3 or -m

Quadratic Equations Class 10 Maths Important Questions Short Answer (2 Marks)

6. For what values of k, the roots of the equation x 2 + 4x + k = 0 are real?

Comparing the given equation with ax 2 + bx + c = 0, we get a = 1, b = 4, c = k.

Since, given the equation has real roots,

D ≥ 0

⇒ b 2 - 4ac ≥ 0

⇒ 4 2 - 4 × 1 × k ≥ 0

⇒ 4k ≤ 16

⇒ k ≤ 4

7. Find the value of m so that the quadratic equation mx (x – 7) + 49 = 0 has two equal roots.

We have, mx (x – 7) + 49 = 0 mx 2 – 7mx + 49 = 0 Here, a = m, b = – 7m, c = 49 D = b 2 – 4ac = 0 …[For equal roots] ⇒ (-7m) 2 – 4(m) (49) = 0 ⇒ 49m 2 – 4m (49) = 0 ⇒ 49m (m – 4) = 0 ⇒ 49m = 0 or m – 4 = 0 m = 0 (rejected) or m = 4 ∴ m = 4

8. Find the value of k for which the roots of the equations 3x 2 - 10x + k = 0 are reciprocal of each other.

Comparing the given equation with ax 2  + bx + c = 0 we get a = 3, b = -10, c = k

Let one root be α so other root is 1/α

product of roots, α × 1/α = c/a

Hence, the value of k is 3.

9. Find the roots of the following quadratic equation :

15x 2  - 10√6x + 10 = 0

We have 15x 2  - 10√6x + 10 = 0

We have, px (x – 3) + 9 = 0 px 2 – 3px + 9 = 0 Here a = p, b = -3p, D = 0 b 2 – 4ac = 0 ⇒ (-3p) 2 – 4(p)(9) = 0 ⇒ 9p 2 – 36p = 0 ⇒ 9p (p – 4) = 0 ⇒ 9p = 0 or p – 4= 0 p = 0 (rejected) or p = 4 ∴ p = 4 …(∵ Coefficient of x 2 cannot be zero)

11. Solve for x:

36x 2 – 12ax + (a 2 – b 2 ) = 0

We have, 36x 2 – 12ax + (a 2 – b 2 ) = 0 ⇒ (36x 2 – 12ax + a 2 ) – b 2 = 0 ⇒ [(6x) 2 – 2(6x)(a) + (a) 2 ] – b 2 = 0 ⇒ (6x – a) 2 – (b) 2 = 0 …[∵ x 2 – 2xy + y 2 = (x – y) 2 ] ⇒ (6x – a + b) (6x – a – b) = 0 ...[∵ x 2 – y 2 = (x + y)(x – y)] ⇒ 6x – a + b = 0 or 6x – a – b = 0 ⇒ 6x = a – b or 6x = a + b ⇒ x = a− b/6 or a+ b/6

12. Solve for x:

4√3x 2  + 5x - 2√3 = 0

Here 4x 2 + 3x + 5 = 0

But (2x+34) 2  cannot be negative for any real value of x.

14. Solve for x : x 2 - (√3+1)x + √3 = 0

x 2 - (√3+1)x + √3 = 0

⇒ x 2 -√3 x - 1x + √3 = 0

⇒ x(x-√3) - 1(x-√3) = 0

⇒ (x-√3) (x-1) = 0

Thus, x = √3, x = 1

15. Find the roots of the following quadratic equation :

(x+3)(x-1) = 3(x - 1/3)

9x 2 - 6b 2 x - (a 4 - b 4 ) = 0

Comparing with Ax 2 + Bx + C

A = 9, B = -6b 2 , C = -(a 4 - b 4 )

We have, ax 2 + 7x + b = 0 Here ‘a’ = a, ‘b’ = 7, ‘c’ = b Now, α = 23 and β = -3 …[Given]

The given quadratic equation can be written as (9x 2 – 6b 2 x + b 4 ) – a 4 = 0 ⇒ (3x – b 2 ) 2 – (a 2 ) 2 = 0 ⇒ (3x – b 2 + a 2 ) (3x – b 2 – a 2 ) = 0 …[x 2 – y 2 = (x + y) (x – y)] ⇒ 3x – b 2 + a 2 = 0 or 3x – b 2 – a 2 = 0 ⇒ 3x = b 2 – a 2 or 3x = b 2 + a 2

We have, 2x 2 + px – 15 =0 Since (-5) is a root of the given equation ∴ 2(-5) 2 + p(-5) – 15 = 0 ⇒ 2(25) – 5p – 15 = 0 ⇒ 50 – 15 = 5p ⇒ 35 = 5p ⇒ p = 7 …(i)

Now, p(x 2 + x) + k ⇒ px 2 + px + k = 0 7x 2 + 7x + k = 0 …[From (i)] Here, a = 7, b = 7, c = k

D = 0 …[Roots are equal] ⇒ b 2 – 4ac = 0 ⇒ (7) 2 – 4(7)k = 0 ⇒ 49 – 28k = 0 ⇒ 49 = 28k ∴ k = 49/28 = 7/4

13√3 x 2 + 10x + √3 = 0

Comparing with ax2 + bx + c = 0, we get

a = 13√3, b = 10, c = √3

b 2 - 4ac = 10 2 - 4(13√3)(√3)

As D < 0, the equation has no real roots.

22. Find the positive value of k for which x 2 - 8x + k = 0, will have real roots.

x 2 - 8x + k = 0

Comparing with Ax 2 + Bx + C = 0, we get

A = 1, B = -8, C= k

Since the given equation has real roots,

B 2 - 4AC > 0

⇒ (-8) 2  - 4(1)(k) ≥ 0

⇒ 64 - 4k ≥ 0

⇒ 16 - k ≥ 0

⇒ 16 ≥ k

Thus, k ≤ 16

23. If 2 is a root of the equation x 2 + kx + 12 = 0 and the equation x 2 + kx + q = 0 has equal roots, find the value of q.

x 2 + kx + 12 = 0

If 2 is the root of above equation, it must satisfy it.

22 + 2k + 12 = 0

⇒ 2k + 16 = 0

⇒ k = -8

Substituting k = -8 in x 2 + kx + q = 0 we have

x 2 - 8x + q = 0

For equal roots,

(-8) 2 - 4(1)q = 0

⇒ 64 - 4q = 0

⇒ 4q = 64

⇒ q = 16

24. Solve for x : √3 x2 + 10x + 7√3 = 0

2x 2 + kx + 8 = 0

Comparing with ax 2 + bx + c = 0, we get

a = 2, b = k and c = 8

For equal roots, D = 0,

b 2 - 4ac = 0

k 2 - 4×2×8 = 0

⇒ k 2 = 64

⇒ k = ±√64

Thus k = ±8

Quadratic Equations Class 10 Maths Important Questions Short Answer-II (3 Marks)

26. Find the values of k for which the quadratic equation x 2 + 2√2 kx + 18 = 0 has equal roots.

x 2 + 2√2 kx + 18 = 0

Comparing it by ax 2 + bx + c, we get a = 1, b = 2√2 k and c = 18.

Given that,

Equation x 2 + 2√2 kx + 18 = 0 has equal roots.

⇒ (2√2 k) 2 -4118 = 0

⇒ 8k 2 - 72 = 0

⇒ 8k 2 = 72

⇒ k 2 = 72/8 = 9

⇒ k = ±3

27. If α and β are the zeroes of the polynomial f(x) = x 2 - 4x - 5 then find the value of α 2 + β 2 .

p(x) = x 2 - 4x - 5

Comparing it by ax 2 + bx + c, we get a = 1, b = -4 and c = -5.

Since, given α and β are the zeroes of the polynomial,

28. Find the quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. Hence find the zeroes.

Sum of zeroes,

α + β = -3 ...(1)

and product of zeroes, αβ = 2

Thus quadratic equation is

x 2 - (α+β)x + αβ = 0

⇒ x 2 - (-3)x + 2 = 0

⇒ x 2 + 3x + 2 = 0

Thus quadratic equation is x 2 + 3x + 2 = 0

Now, above equation can be written as

x 2 + 2x + x + 2 = 0

⇒ x(x+2) + (x+2) = 0

⇒ (x+2) (x+1) = 0

Hence, zeroes are -2 and -1.

29. Find the roots of the following quadratic equation: 2√3 x 2 – 5x + √3 = 0

We have, 2√3 x 2  – 5x + √3 = 0

Here, a = 2√3, b = -5, c = √3

D = b 2 – 4ac ∴ D = (-5) 2 – 4 (2√3)(√3) = 25 – 24 = 1

31. Solve for x: 4x 2 – 4ax + (a 2 – b 2 ) = 0

33. Find the value(s) of k so that the quadratic equation 3x 2 – 2kx + 12 = 0 has equal roots.

Given: 3x 2 – 2kx + 12 = 0 Here a = 3, b = -2k, c = 12 D = 0 …[Since roots are equal as b 2 – 4ac = 0] ∴ (-2k) 2 – 4(3) (12) = 0 ⇒ 4k 2 – 144 = 0 ⇒ k 2 = 144/4 = 36 ∴ k = ±√36 ∴ k = ±6

34. Find the quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. Hence find the zeroes.

Sum of zeroes α + β = -3

and product of zeroes αβ = 2

Thus, quadratic equation is x 2 + 3x + 2 = 0

35. Find the zeroes of the quadratic polynomial 6x 2  - 3 - 7x and verify the relationship between the zeroes and the coefficients.

We have, p(x) = 6x 2  - 3 - 7x For zeroes of polynomial, p(x) = 0 6x 2  - 7x - 3 = 0 ⇒ 6x 2  - 9x + 2x - 3 = 0 ⇒ 3x(2x - 3) + 1(2x-3) = 0 ⇒ (2x-3) (3x+1) = 0 Thus, 2x - 3 = 0 and 3x + 1 = 0 Hence, x = 3/2 and x = -1/3 Therefore, α = 3/2 and β = -1/3 are the zeroes of the given polynomial. Verification:

We have, (k + 4) x 2 + (k + 1) x + 1 = 0 Here, a = k + 4, b = k + 1, c = 1 D =0 …[∵ Roots are equal] b 2 – 4ac = 0 ∴ (k + 1) 2 – 4(k + 4)(1) = 0 k 2 + 2k + 1 – 4k – 16 = 0 ⇒ k 2 – 2k – 15 = 0 ⇒ k 2 – 5k + 3k – 15 = 0 ⇒ k(k – 5) + 3(k – 5) = 0 ⇒ (k – 5)(k + 3) = 0 ⇒ k – 5 = 0 or k + 3= 0 ⇒ k = 5 or k = -3 ∴ k = 5 and -3

38. Find the zeroes of the quadratic polynomial x 2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.

p(x) = x 2 + 7x + 10

For zeroes of polynomial p(x) = 0,

x 2 + 7x + 10 = 0

⇒ x 2 + 5x + 2x + 10 = 0

⇒ x(x+5) + 2(x+5) = 0

⇒ (x+5) (X+2) = 0

x = -2 and x = -5

Therefore, α = -2 and β = -5 are the zeroes of the given polynomial.

Verification:

α + β = -2 + (-5)

y 2 + k 2 = 2(k + 1)y y 2 – 2(k + 1)y + k 2 = 0 Here a = 1, b = -2(k + 1), c = k 2 D = 0 …[Roots are equal] b 2 – 4ac = 0 ∴ [-2(k + 1)] 2 – 4 × (1) × (k 2 ) = 0 ⇒ 4(k 2 + 2k + 1) – 4k 2 = 0 ⇒ 4k 2 + 8k + 4 – 4k 2 = 0 ⇒ 8k + 4 = 0 ⇒ 8k = -4 ∴ k = -4/8 = -1/2

41. Solve the equation 3/x+1 - 1/2 = 2/3x-1, x≠ -1, x≠ 1/3 for x.

⇒ 2(2x + 2) = (5 – x)(3x – 1) ⇒ 4x + 4 = 15x – 5 – 3x 2 + x ⇒ 4x + 4 – 15x + 5 + 3x 2 – x = 0 ⇒ 3x 2 – 12x + 9 = 0 ⇒ x 2 – 4x + 3 = 0 …[Dividing by 3] ⇒ x 2 – 3x – x + 3 = 0 ⇒ x(x – 3) – 1(x – 3) = 0 ⇒ (x – 1) (x – 3) = 0 ⇒ x – 1 = 0 or x – 3 = 0 ∴ x = 1 or x = 3

42. Solve for x:

2x/x-3 + 1/2x+3 + 3x+9/(x+9)(2x+3) = 0, x ≠ 3, -3/2

2x(2x+3) + (x-3) + (3x+9) = 0 ⇒ 4x 2 + 6x + x - 3 + 3x + 9 = 0 ⇒ 4x 2 + 10x + 6 = 0 ⇒ 2x 2 + 5x + 3 =0 ⇒ (x+1) (2x+3) = 0 Thus, x = -1, x = -3/2

43. Solve the following quadratic equation for x :

x 2 2 + (a/a+b + a+b/a)x + 1 = 0

x 2 - (2b-1)x + (b 2 -b-20) = 0

Comparing with Ax 2 + Bx + C = 0 we have

A = 1, B = -(2b-1), C = (b 2 -b-20)

Thus x = b+4 and x = b-5

46. Find the roots of the equation 2x 2  + x - 4 = 0, by the method of completing the squares.

2x 2  + x - 4 = 0

Let three consecutive natural numbers are x, x + 1, x + 2. According to the question, (x + 1) 2 – [(x + 2) 2 – x 2 ] = 60 ⇒ x 2 + 2x + 1 – (x 2 + 4x + 4 – x 2 ) = 60 ⇒ x 2 + 2x + 1 – 4x – 4 – 60 = 0 ⇒ x 2 – 2x – 63 = 0 ⇒ x 2 – 9x + 7x – 63 = 0 ⇒ x(x – 9) + 7(x – 9) = 0 ⇒ (x – 9) (x + 7) = 0 ⇒ x – 9 = 0 or x + 7 = 0 ⇒ x = 9 or x = -7 Natural nos. can not be negative, ∴ x = 9 ∴ Numbers are 9, 10, 11.

48. If the sum of two natural numbers is 8 and their product is 15, find the numbers.

Let the numbers be x and (8 – x). According to the Question, x(8 – x) = 15 ⇒ 8x – x 2 = 15 ⇒ 0 = x 2 – 8x + 15 ⇒ x 2 – 5x – 3x + 15 = 0 ⇒ x(x – 5) – 3(x – 5) = 0 ⇒ (x – 3)(x – 5) = 0 x – 3 = 0 or x – 5 = 0 x = 3 or x = 5 When x = 3, numbers are 3 and 5. When x = 5, numbers are 5 and 3.

49. If 2 is a root of the quadratic equation 3x 2 + px - 8 = 0 and the quadratic equation 4x 2 - 2px + k = 0 has equal roots, find k.

3x 2 + px - 8 = 0

Since 2 is a root of above equation, it must satisfy it.

Substituting, x= 2 in we have

12 + 2p - 8 = 0

⇒ p = -2

4x 2 - 2px + k = 0 has equal roots.

or, 4x 2 + 4x + k = 0 has equal roots.

D = b 2 - 4ac = 0

4 2 - 4(4)(k) = 0

⇒ 16 - 16k = 0

⇒ 16k = 16

⇒ k = 1

50. If -3 is a root of quadratic equation 2x 2 + px - 15 = 0, while the quadratic equation x 2 - 4px + k = 0 has equal roots. Find the value of k.

Given -3 is a root of quadratic equation.

2x 2 + px - 15

Since 3 is a root of above equation, it must satisfy it.

Substituting x 3 = in above equation we have,

2(-3) 2 + p(-3) - 15 = 0

⇒ 2×9 - 3p - 15 = 0

⇒ p = 1

x 2 - 4px + k = 0 has equal roots

or, x 2 - 4x + k  = 0 has equal roots.

⇒ (-4) 2 - k = 0

⇒ 16 - 4k = 0

⇒ 4k = 16

⇒ k = 4

51. Solve 1/(a+b+x) = 1/a + 1/b + 1/x, a+b ≠ 0

Quadratic Equations Class 10 Maths Important Questions Long Answer (4 Marks)

⇒ 2x = -8x + 40

⇒ 10x = 40

⇒ x = 4

Hence, x = 15, 4

53. Solve for x: 1/x+1 + 2/x+2 = 4/x+4, x≠ -1,-2,-4

(3k + 1)x 2 + 2(k + 1) + 1 = 0 Here, a = 3k + 1, b = 2(k + 1), c = 1 D = 0 …[∵ Roots are equal] As b 2 – 4ac = 0 ∴ [2(k + 1)] 2 – 4(3k + 1)(1) = 0 ⇒ 4(k + 1) 2 – 4(3k + 1) = 0 ⇒ 4(k 2 + 2k + 1 – 3k – 1) = 0 ⇒ (k 2 – k) = 0/4 ⇒ k(k – 1) = 0 ⇒ k = 0 or k – 1 = 0 ∴ k = 0 or k = 1 Roots are x = −b/2a ...[ Given, equal roots]

(2p + 1)x 2 – (7p + 2)x + (7p – 3) = 0 Here, a = 2p + 1, b = -(7p + 2), c = 7p – 3 D = 0 …[∵ Equal roots As h2 – 4ac = 0] ∴ [-(7p + 2)] 2 – 4(2p + 1)(7p – 3) = 0 ⇒ (7p + 2) 2 – 4(14p 2 – 6p + 7p – 3) = 0 ⇒ 49p 2 + 28p + 4 – 56p 2 + 24p – 28p + 12 = 0 ⇒ -7p 2 + 24p + 16 = 0 ⇒ 7p 2 – 24p – 16 = 0 …[Dividing both sides by -1] ⇒ 7p 2 – 28p + 4p – 16 = 0 ⇒ 7p(p – 4) + 4(p – 4) = 0 ⇒ (p – 4) (7p + 4) = 0 ⇒ p – 4 = 0 or 7p + 4 = 0 ⇒ p = 4 or p = −4/7

59. Check whether the equation 5x 2  - 6x - 2 = 0 has real roots if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.

5x 2  - 6x - 2 = 0

Comparing with ax 2  + bx + c = 0 we get,

a = 5, b = -6 and c = -2

b 2  - 4ac = (-6) 2  - 4×5×(-2)

= 36 + 40 = 76 > 0

So the equation has real and two distinct roots.

5x 2  - 6x = 2

Dividing both the sides by 5 we get

Let the number of books he bought = x Increased number of books he had bought = x +4 Total amount = ₹80 According to the problem,

⇒ x(x + 4) = 320 ⇒ x 2 + 4x – 320 = 0 ⇒ x 2 + 20x – 16x – 320 = 0 ⇒ x(x + 20) – 16(x + 20) = 0 ⇒ (x + 20) (x – 16) = 0 ⇒ x + 20 = 0 or x – 16 = 0 ⇒ x = -20 …(neglected) or x = 16 ∴ Number of books he bought = 16

62.  Find the positive values of k for which quadratic equations x 2  + kx + 64 = 0 and x 2  - 8x + k = 0 both will have the real roots.

(i) For x 2  + kx + 64 = 0 to have real roots

k 2  - 256 ≥ 0

⇒ k 2  ≥ 256

⇒ k ≥ 16 or k < -16

(ii) For x 2  - 8x + k = 0 to have real roots

64 - 4k ≥ 0

Therefore, For (i) and (ii) to hold simultaneously

63. Sum of the areas of two squares is 400 cm 2 . If the difference of their perimeters is 16 cm, find the sides of the two squares.

Let the side of Large square = x cm Let the side of small square = y cm According to the Question, x 2 + y 2 = 400 …(i) …[∵ area of square = (side) 2 ] 4x – 4y = 16 …[∵ Perimeter of square = 4 sides] ⇒ x – y = 4 …[Dividing both sides by 4] ⇒ x = 4 + y …(ii) Putting the value of x in equation (i), (4 + y) 2 + y2 2 = 400 ⇒ y 2 + 8y + 16 + y 2 – 400 = 0 ⇒ 2y 2 + 8y – 384 = 0 ⇒ y 2 + 4y – 192 = 0 …[Dividing both sides by 2] ⇒ y 2 + 16y – 12y – 192 = 0 ⇒ y(y + 16) – 12(y + 16) = 0 ⇒ (y – 12)(y + 16) = 0 ⇒ y – 12 = 0 or y + 16 = 0 ⇒ y = 12 or y = -16 …[Neglecting negative value] ∴ Side of small square = y = 12 cm and Side of large square = x = 4 + 12 = 16 cm

64. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Let the length of shorter side be x m. ∴ length of diagonal = (x + 16) m and length of longer side = (x + 14) m Using pythagoras theorem, (l) 2 + (b) 2 = (h) 2 ∴ x 2 + (x + 14)2 2 = (x + 16) 2 ⇒ x 2 + x 2 + 196 + 28x = x 2 + 256 + 32x ⇒ x 2 – 4x – 60 = 0 ⇒ x 2 – 10x + 6x – 60 = 0 ⇒ x(x – 10) + 6(x – 10) = 0 ⇒ (x – 10) (x + 6) = 0 ⇒ x – 10 = 0 or x + 6 = 0 ⇒ x = 10 or x = -6 (Neglect as length cannot be negative]) ⇒ x = 10 m 

Length of shorter side = x = 10 m Length of diagonal = (x + 16) m = 26 m Length of longer side = (x + 14)m = 24m ∴ Length of sides are 10 m and 24 m.

65. The sum of two numbers is 9 and the sum of their reciprocals is 1/2. Find the numbers.

Let the numbers be x and 9 – x. According to the Question,

When x = 3, numbers are 3 and 6. When x = 6, numbers. are 6 and 3.

66. Find the nature of the roots of the quadratic equation 4x 2  + 4√3x + 3 = 0.

4x 2  + 4√3x + 3 = 0

Comparing the given equation with ax 2  + bx + c = 0, we get

a = 4, b = 4√3 and c = 3.

Since, b 2  - 4ac = 0, then roots of the given equation are real and equal.

67. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.

Let three numbers in A.P. are a – d, a, a + d. a – d + a + a + d = 12 ⇒ 3a = 12 ⇒ a = 4 (a – d) 3 + (a) 3 + (a + d) 3 = 288 ⇒ a 3 – 3a 2 d + 3ad 2 – d 3 + a 3 + a 3 + 3a 2 d + 3ad 2 + d 3 = 288 ⇒ 3a 3 + 6ad 2 = 288 ⇒ 3a(a 2 + 2d 2 ) = 288 ⇒ 3 × 4(4 2 + 2d 2 ) = 288 ⇒ (16 + 2d 2 ) = 288/12 ⇒ 2d 2 = 24 – 16 = 8 ⇒ d 2 = 4 ⇒ d = ± 2 When, a = 4, d = 2, numbers are: a – d, a, a + d, i.e., 2, 4, 6 When, a = 4, d = -2, numbers are: a – d, a, a + d, i.e., 6, 4, 2

68. Find the values of k for which the quadratic equations (k+4)x 2  + (k+1)x + 1 = 0 has equal roots. Also, find the roots.

(k+4)x 2  + (k+1)x + 1 = 0

Comparing with Ax 2  + Bx + C = 0, we get

A = (k+4), B = (k+1), C = 1

If roots are equal, B 2  - 4AC = 0

(k+1) 2  - 4(k+4)(1) = 0

⇒ k 2  + 1 + 2k - 4k - 16 = 0

⇒ k 2  - 2k -15 = 0

⇒ (k-5) (k+3) = 0

⇒ k = 5, -3

For k = 5, equation becomes

9x 2  + 6x + 1 = 0

⇒ (3x+1) 2  = 0

⇒ x = -1/3

For k = -3, equation becomes

x 2  - 2x + 1 = 0

⇒ (x-1) 2  = 0

⇒ x = 1

Hence, roots are 1 and -1/3

69. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.

Perimeter of right ∆ = 60 cm …[Given] a + b + c = 60 ⇒ a + b + 25 = 60 ⇒ a + b = 60 – 25 = 35 …(i) In rt. ∆ACB, AC 2 + BC 2 = AB 2 b 2 + a 2 = (25) 2 …[Pythagoras’ theorem] ⇒ a 2 + b 2 = 625 …(ii) From (i), a + b = 35 (a + b) 2 = (35) …[Squaring both sides] ⇒ a 2 + b 2 + 2ab = 1225 ⇒ 625 + 2ab = 1225 …[From (ii)] ⇒ 2ab = 1225 – 625 = 600 ⇒ ab = 300 …(iii) Area of ∆ = 1/2 × base × corresponding altitude = 1/2 × b × a = 1/2 (300) ...[From (iii)] = 150 cm 2

70. The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 1/15. Find the fraction.

Let the denominator be x and the numerator be x – 3. ∴ Fraction =x−3/x New denominator = x + 1 According to the Question,

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations  contain all the solutions to the problems provided in the Class 10 Maths NCERT textbook for CBSE exam preparations. The questions from every section are framed and solved accurately by the subject experts. NCERT Solutions for Class 10 are detailed and step-by-step guides to all the queries of the students. The exercises present in the chapter should be dealt with utmost sincerity if one wants to score well in the examinations. Maths is a subject that requires a good understanding and a lot of practice. The tips and tricks to solve the problems easily are also provided here. A quadratic equation in the variable x is an equation of the form ax 2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0. That is, ax 2 + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

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Download most important questions for class 10 maths chapter – 4 quadratic equations.

Quadratic equations arise in several situations around us. Hence, students should give special attention to learning the concepts related to this chapter of the latest CBSE Syllabus for 2023-24 thoroughly to excel in Class 10 Maths examinations. NCERT Solutions help the students in learning these concepts as well as in evaluating themselves. Practising these solutions repeatedly is bound to help the students in overcoming their shortcomings. Maths has either a correct answer or a wrong one. Therefore, it is imperative to concentrate while solving the questions to score full marks.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Exercise 4.1
• Exercise 4.2
• Exercise 4.3
• Exercise 4.4

NCERT Solutions Class 10 Maths Chapter 4 – Quadratic Equations

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Exercise 4.1 page: 73.

1. Check whether the following are quadratic equations:

(i) (x + 1) 2 = 2(x – 3)

(ii) x 2 – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x 2 + 3x + 1 = (x – 2) 2

(vii) (x + 2) 3 = 2x (x 2 – 1)

(viii) x 3 – 4x 2 – x + 1 = (x – 2) 3

(x + 1) 2 = 2(x – 3)

By using the formula for (a+b) 2 = a 2 +2ab+b 2

⇒ x 2  + 2x + 1 = 2x – 6

⇒ x 2  + 7 = 0

The above equation is in the form of ax 2 + bx + c = 0

Therefore, the given equation is a quadratic equation.

(ii) Given, x 2 – 2x = (–2) (3 – x)

⇒ x 2  –   2x = -6 + 2x

⇒ x 2  – 4x + 6 = 0

(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)

By multiplication,

⇒ x 2  – x – 2 = x 2  + 2x – 3

⇒ 3x – 1 = 0

The above equation is not in the form of ax 2 + bx + c = 0

Therefore, the given equation is not a quadratic equation.

(iv) Given, (x – 3)(2x +1) = x(x + 5)

⇒ 2x 2  – 5x – 3 = x 2  + 5x

⇒  x 2  – 10x – 3 = 0

(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

⇒ 2x 2  – 7x + 3 = x 2  + 4x – 5

⇒ x 2  – 11x + 8 = 0

The above equation is in the form of ax 2  + bx + c = 0.

(vi) Given, x 2  + 3x + 1 = (x – 2) 2

By using the formula for (a-b) 2 =a 2 -2ab+b 2

⇒ x 2  + 3x + 1 = x 2  + 4 – 4x

⇒ 7x – 3 = 0

(vii) Given, (x + 2) 3  = 2x(x 2  – 1)

By using the formula for (a+b) 3  = a 3 +b 3 +3ab(a+b)

⇒ x 3  + 8 + x 2  + 12x = 2x 3  – 2x

⇒ x 3  + 14x – 6x 2  – 8 = 0

(viii) Given, x 3  – 4x 2  – x + 1 = (x – 2) 3

By using the formula for (a-b) 3  = a 3 -b 3 -3ab(a-b)

⇒  x 3  – 4x 2  – x + 1 = x 3  – 8 – 6x 2   + 12x

⇒ 2x 2  – 13x + 9 = 0

2. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m 2 . The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken

(i) Let us consider,

The breadth of the rectangular plot = x m

Thus, the length of the plot = (2x + 1) m

As we know,

Area of rectangle = length × breadth = 528 m 2

Putting the value of the length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2x 2  + x =528

⇒ 2x 2  + x – 528 = 0

Therefore, the length and breadth of the plot satisfy the quadratic equation, 2x 2  + x – 528 = 0, which is the required representation of the problem mathematically.

(ii) Let us consider,

The first integer number = x

Thus, the next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1) = 306

⇒ x 2  + x = 306

⇒ x 2  + x – 306 = 0

Therefore, the two integers x and x+1 satisfy the quadratic equation, x 2  + x – 306 = 0, which is the required representation of the problem mathematically.

(iii) Let us consider,

Age of Rohan’s = x  years

Therefore, as per the given question,

Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

(x + 3)(x + 29) = 360

⇒ x 2  + 29x + 3x + 87 = 360

⇒ x 2  + 32x + 87 – 360 = 0

⇒ x 2  + 32x – 273 = 0

Therefore, the age of Rohan and his mother satisfies the quadratic equation, x 2  + 32x – 273 = 0, which is the required representation of the problem mathematically.

(iv) Let us consider,

The speed of the train = x   km/h

Time taken to travel 480 km = 480/x km/hr

As per second condition, the speed of train = ( x  – 8) km/h

Also given, the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = (480/x)+3 km/h

Speed × Time = Distance

( x  – 8)(480/ x  )+ 3 = 480

⇒ 480 + 3 x  – 3840/ x  – 24 = 480

⇒ 3 x  – 3840/ x  = 24

⇒ x 2  – 8 x  – 1280 = 0

Therefore, the speed of the train satisfies the quadratic equation, x 2  – 8 x  – 1280 = 0, which is the required representation of the problem mathematically.

Exercise 4.2 Page: 76

1. Find the roots of the following quadratic equations by factorisation:

(i) x 2  – 3x – 10 = 0 (ii) 2x 2  + x – 6 = 0 (iii) √2 x 2  + 7x + 5√2 = 0 (iv) 2x 2  – x +1/8 = 0 (v) 100x 2  – 20x + 1 = 0

(i) Given,  x 2  – 3 x  – 10 =0

Taking L.H.S.,

=> x 2  – 5 x  + 2 x  – 10

=> x ( x  – 5) + 2( x  – 5)

=>( x  – 5)( x  + 2)

The roots of this equation, x 2  – 3 x  – 10 = 0 are the values of x for which ( x  – 5)( x  + 2) = 0

Therefore,  x  – 5 = 0 or  x  + 2 = 0

=>  x  = 5 or  x  = -2

(ii) Given, 2 x 2  +  x  – 6 = 0

=> 2 x 2  + 4 x  – 3 x  – 6

=> 2 x ( x  + 2) – 3( x  + 2)

=> ( x  + 2)(2 x  – 3)

The roots of this equation, 2 x 2  +  x  – 6=0 are the values of x for which ( x x  + 2)(2 x  – 3) = 0

Therefore,  x  + 2 = 0 or 2 x  – 3 = 0

=>  x  = -2 or  x  = 3/2

(iii) √2  x 2  + 7 x  + 5√2=0

=> √2  x 2  + 5 x  + 2 x  + 5√2

=>  x  (√2 x  + 5) + √2(√2 x  + 5)= (√2 x  + 5)( x  + √2)

The roots of this equation, √2  x 2  + 7 x  + 5√2=0 are the values of x for which (√2 x  + 5)( x  + √2) = 0

Therefore, √2 x  + 5 = 0 or  x  + √2 = 0

=>  x  = -5/√2 or  x  = -√2

(iv) 2 x 2  –  x  +1/8 = 0

=1/8 (16 x 2   – 8 x  + 1)

= 1/8 (16 x 2   – 4 x  -4 x  + 1)

= 1/8 (4 x (4 x    – 1) -1(4 x  – 1))

= 1/8 (4 x  – 1) 2

The roots of this equation, 2 x 2  –  x  + 1/8 = 0, are the values of x for which (4 x  – 1) 2 = 0

Therefore, (4 x  – 1) = 0 or (4 x  – 1) = 0

⇒  x  = 1/4 or  x  = 1/4

(v) Given, 100x 2  – 20x + 1=0

= 100x 2  – 10x – 10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1) 2

The roots of this equation, 100x 2  – 20x + 1=0, are the values of x for which (10x – 1) 2 = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

2. Solve the problems given in Example 1.

Represent the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

(i) Let us say the number of marbles John has = x

Therefore, the number of marble Jivanti has = 45 – x

After losing 5 marbles each,

Number of marbles John has = x  – 5

Number of marble Jivanti has = 45 – x  – 5 = 40 –  x

Given that the product of their marbles is 124.

∴ ( x  – 5)(40 –  x ) = 124

⇒  x 2  – 45 x  + 324 = 0

⇒  x 2  – 36 x  – 9 x  + 324 = 0

⇒  x ( x  – 36) -9( x  – 36) = 0

⇒ ( x  – 36)( x  – 9) = 0

Thus, we can say,

x  – 36 = 0 or  x  – 9 = 0

⇒  x  = 36 or  x  = 9

If John’s marbles = 36

Then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9

Then, Jivanti’s marbles = 45 – 9 = 36

(ii) Let us say the number of toys produced in a day is x .

Therefore, cost of production of each toy = Rs(55 –  x )

Given the total cost of production of the toys = Rs 750

∴  x (55 –  x ) = 750

⇒  x 2  – 55 x  + 750 = 0

⇒  x 2  – 25 x  – 30 x  + 750 = 0

⇒  x ( x  – 25) -30( x  – 25) = 0

⇒ ( x  – 25)( x  – 30) = 0

Thus, either x  -25 = 0 or  x  – 30 = 0

⇒  x  = 25 or  x  = 30

Hence, the number of toys produced in a day will be either 25 or 30.

3. Find two numbers whose sum is 27 and product is 182.

Let us say the first number is x, and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x 2  – 27x – 182 = 0

⇒ x 2  – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

4. Find two consecutive positive integers, the sum of whose squares is 365.

Let us say the two consecutive positive integers are x  and  x  + 1.

Therefore, as per the given questions,

x 2  + ( x  + 1) 2  = 365

⇒  x 2  +  x 2  + 1 + 2 x  = 365

⇒ 2 x 2  + 2x – 364 = 0

⇒  x 2  +  x  – 182 = 0

⇒  x 2  + 14 x  – 13 x  – 182 = 0

⇒  x ( x  + 14) -13( x  + 14) = 0

⇒ ( x  + 14)( x  – 13) = 0

Thus, either, x  + 14 = 0 or  x  – 13 = 0,

⇒  x  = – 14 or  x  = 13

Since the integers are positive, x can be 13 only.

∴  x  + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let us say the base of the right triangle is x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras’ theorem, we know,

Base 2  + Altitude 2  = Hypotenuse 2

∴ x 2  + (x – 7) 2  = 13 2

⇒ x 2  + x 2  + 49 – 14x = 169

⇒ 2x 2  – 14x – 120 = 0

⇒ x 2  – 7x – 60 = 0

⇒ x 2  – 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm, and the altitude of this triangle will be (12 – 7) cm = 5 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.

Let us say the number of articles produced is x .

Therefore, cost of production of each article = Rs (2 x  + 3)

Given the total cost of production is Rs.90

∴  x (2 x  + 3) = 90

⇒ 2 x 2  + 3 x  – 90 = 0

⇒ 2 x 2  + 15 x  -12 x  – 90 = 0

⇒  x (2 x  + 15) -6(2 x  + 15) = 0

⇒ (2 x  + 15)( x  – 6) = 0

Thus, either 2 x  + 15 = 0 or  x  – 6 = 0

⇒  x  = -15/2 or  x  = 6

As the number of articles produced can only be a positive integer, x can only be 6.

Hence, the number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15

Exercise 4.3 Page: 87

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2 x 2  – 7 x  +3 = 0

(ii) 2 x 2  +  x  – 4 = 0 (iii) 4 x 2  + 4√3 x  + 3 = 0

(iv) 2 x 2  +  x  + 4 = 0

(i) 2 x 2  – 7 x  + 3 = 0

⇒ 2 x 2  – 7 x  = – 3

Dividing by 2 on both sides, we get

⇒ x 2 -7x/2 = -3/2

⇒ x 2 -2 × x ×7/4 = -3/2

On adding (7/4) 2 to both sides of the equation, we get

⇒ (x) 2 -2×x×7/4 +(7/4) 2 = (7/4) 2 -3/2

⇒ (x-7/4) 2 = (49/16) – (3/2)

⇒(x-7/4) 2 = 25/16

⇒(x-7/4) 2 = ±5/4

⇒  x  = 7/4 ± 5/4

⇒  x  = 7/4 + 5/4 or x = 7/4 – 5/4

⇒ x = 12/4 or x = 2/4

⇒ x = 3 or x = 1/2

(ii) 2x 2  + x – 4 = 0

⇒ 2x 2  + x = 4

Dividing both sides of the equation by 2, we get

⇒ x 2  +x/2 = 2

Now on adding (1/4) 2  to both sides of the equation, we get,

⇒ (x) 2  + 2 × x × 1/4 + (1/4) 2  = 2 + (1/4) 2

⇒ (x + 1/4) 2  = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 – 1/4

⇒ x = (± √33-1)/4

Therefore, either x = (√33-1)/4 or x = (-√33-1)/4

(iii) 4x 2  + 4√3x + 3 = 0

Converting the equation into a 2 +2ab+b 2 form, we get,

⇒ (2x) 2  + 2 × 2x × √3 + (√3) 2  = 0

⇒ (2x + √3) 2  = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

Therefore, either x = -√3/2 or x = -√3/2

(iv) 2x 2  + x + 4 = 0

⇒ 2x 2  + x = -4

⇒ x 2  + 1/2x = 2

⇒ x 2  + 2 × x × 1/4 = -2

By adding (1/4) 2  to both sides of the equation, we get

⇒ (x) 2  + 2 × x × 1/4 + (1/4) 2  = (1/4) 2  – 2

⇒ (x + 1/4) 2  = 1/16 – 2

⇒ (x + 1/4) 2  = -31/16

As we know, the square of numbers cannot be negative.

Therefore, there is no real root for the given equation, 2x 2 + x + 4 = 0

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

(i) 2x 2  – 7x + 3 = 0

On comparing the given equation with ax 2  + bx + c = 0, we get,

a = 2, b = -7 and c = 3

By using the quadratic formula, we get,

⇒ x = (7±√(49 – 24))/4

⇒ x = (7±√25)/4

⇒ x = (7±5)/4

⇒ x = (7+5)/4 or x = (7-5)/4

⇒ x = 12/4 or 2/4

∴  x = 3 or 1/2

(ii) 2x 2  + x – 4 = 0

a = 2, b = 1 and c = -4

⇒x = (-1±√1+32)/4

⇒x = (-1±√33)/4

∴ x = (-1+√33)/4 or x = (-1-√33)/4

(iii) 4x 2  + 4√3x + 3 = 0

On comparing the given equation with ax 2  + bx + c = 0, we get

a = 4, b = 4√3 and c = 3

⇒ x = (-4√3±√48-48)/8

⇒ x = (-4√3±0)/8

∴ x = -√3/2 or x = -√3/2

(iv) 2x 2  + x + 4 = 0

a = 2, b = 1 and c = 4

By using the quadratic formula, we get

⇒ x = (-1±√1-32)/4

⇒ x = (-1±√-31)/4

As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.

3. Find the roots of the following equations:

(i) x-1/x = 3, x ≠ 0 (ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7

(i) x-1/x = 3

⇒ x 2  – 3x -1 = 0

a = 1, b = -3 and c = -1

⇒ x = (3±√9+4)/2

⇒ x = (3±√13)/2

∴ x = (3+√13)/2 or x = (3-√13)/2

(ii) 1/x+4 – 1/x-7 = 11/30 ⇒ x-7-x-4/(x+4)(x-7) = 11/30

⇒ -11/(x+4)(x-7) = 11/30

⇒ (x+4)(x-7) = -30

⇒ x 2  – 3x – 28 = 30

⇒ x 2  – 3x + 2 = 0

We can solve this equation by factorisation method now.

⇒ x 2  – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(x – 1) = 0

⇒ x = 1 or 2

4. The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Let us say the present age of Rahman is x  years.

Three years ago, Rehman’s age was ( x  – 3) years.

Five years after, his age will be ( x  + 5) years.

Given the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/ x -3 + 1/ x -5 = 1/3

(x +5+ x -3)/( x -3)( x +5) = 1/3

(2 x +2)/( x -3)( x +5) = 1/3

⇒ 3(2 x  + 2) = ( x -3)( x +5)

⇒ 6 x  + 6 =  x 2  + 2 x  – 15

⇒  x 2  – 4 x  – 21 = 0

⇒  x 2  – 7 x  + 3 x  – 21 = 0

⇒  x ( x  – 7) + 3( x  – 7) = 0

⇒ ( x  – 7)( x  + 3) = 0

⇒  x  = 7, -3

As we know, age cannot be negative.

Therefore, Rahman’s present age is 7 years.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Let us say the marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

As per the given question,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

⇒ -x 2  + 25x + 54 = 210

⇒ x 2  – 25x + 156 = 0

⇒ x 2  – 12x – 13x + 156 = 0

⇒ x(x – 12) -13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

⇒ x = 12, 13

Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18, and if the marks in Maths are 13, then marks in English will be 30 – 13 = 17 .

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Let us say the shorter side of the rectangle is x  m.

Then, larger side of the rectangle = ( x  + 30) m

As given, the length of the diagonal is = x + 30 m

⇒  x 2  + ( x  + 30) 2  = ( x  + 60) 2

⇒  x 2  +  x 2  + 900 + 60 x  =  x 2  + 3600 + 120 x

⇒  x 2  – 60 x  – 2700 = 0

⇒  x 2  – 90 x  + 30 x  – 2700 = 0

⇒  x ( x  – 90) + 30( x  -90) = 0

⇒ ( x  – 90)( x  + 30) = 0

⇒  x  = 90, -30

However, the side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m.

And the length of the larger side will be (90 + 30) m = 120 m.

7. The difference of the squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let us say the larger and smaller number be x  and  y,  respectively.

As per the question given,

x 2  –  y 2  = 180 and  y 2  = 8 x

⇒  x 2  – 8 x  = 180

⇒  x 2  – 8 x  – 180 = 0

⇒  x 2  – 18 x  + 10 x  – 180 = 0

⇒  x ( x  – 18) +10( x  – 18) = 0

⇒ ( x  – 18)( x  + 10) = 0

⇒  x  = 18, -10

However, the larger number cannot be considered a negative number, as 8 times the larger number will be negative, and hence, the square of the smaller number will be negative, which is not possible.

Therefore, the larger number will be 18 only.

∴  y 2  = 8x = 8 × 18 = 144

⇒  y  = ±√144 = ±12

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and -12.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

It is given that

Distance = 360 km

Consider x as the speed, then the time taken

If the speed is increased by 5 km/h, the speed will be (x + 5) km/h.

Distance will be the same.

t = 360/(x + 5)

We know that

Time with original speed – Time with increased speed = 1

360/x – 360/(x + 5) = 1

LCM = x (x + 5)

360 x + 1800 – 360x = x (x + 5)

x 2  + 5x = 1800

x 2  + 5x – 1800 = 0

x 2  + 45x – 40x – 1800 = 0

x (x + 45) – 40 (x + 45) = 0

(x – 40) (x + 45) = 0

x = 40 km/hr

As we know, the value of speed cannot be negative.

Therefore, the speed of the train is 40 km/h.

Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = ( x  – 10) hr

Part of the tank filled by smaller pipe in 1 hour = 1/ x

Part of the tank filled by larger pipe in 1 hour = 1/( x  – 10)

1/ x  + 1/ x -10 = 8/75

x -10+ x / x ( x -10) = 8/75

⇒ 2 x -10/ x ( x -10) = 8/75

⇒ 75(2 x  – 10) = 8 x 2  – 80 x

⇒ 150 x  – 750 = 8 x 2  – 80 x

⇒ 8 x 2  – 230 x  +750 = 0

⇒ 8 x 2  – 200 x  – 30 x  + 750 = 0

⇒ 8 x ( x  – 25) -30( x  – 25) = 0

⇒ ( x  – 25)(8 x  -30) = 0

⇒  x  = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

Therefore, the time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours, respectively.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Let us say the average speed of the passenger train =   x  km/h.

Average speed of express train = ( x  + 11) km/h

Given the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,

(132/x) – (132/(x+11)) = 1

132(x+11-x)/(x(x+11)) = 1

132 × 11 /(x(x+11)) = 1

⇒ 132 × 11 =  x ( x  + 11)

⇒  x 2  + 11 x  – 1452 = 0

⇒  x 2  +  44 x  -33 x  -1452 = 0

⇒  x ( x  + 44) -33( x  + 44) = 0

⇒ ( x  + 44)( x  – 33) = 0

⇒  x  = – 44, 33

As we know, speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

11. Sum of the areas of two squares is 468 m 2 . If the difference between their perimeters is 24 m, find the sides of the two squares.

Let the sides of the two squares be  x  m and  y  m.

Therefore, their perimeter will be 4 x  and 4 y,  respectively

And the area of the squares will be x 2  and  y 2,  respectively.

4 x  – 4 y  = 24

x  –  y  = 6

x  =  y  + 6

Also,  x 2  +   y 2  = 468

⇒ (6 +  y 2 ) +  y 2  = 468

⇒ 36 +  y 2  + 12 y  +  y 2  = 468

⇒ 2 y 2  + 12 y  + 432 = 0

⇒  y 2  + 6y – 216 = 0

⇒  y 2  + 18 y  – 12 y  – 216 = 0

⇒  y ( y  +18) -12( y  + 18) = 0

⇒ ( y  + 18)( y  – 12) = 0

⇒  y  = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

Exercise 4.4 Page: 91

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them. (i) 2 x 2  – 3 x  + 5 = 0 (ii) 3 x 2  – 4√3 x  + 4 = 0 (iii) 2 x 2  – 6 x  + 3 = 0

2x 2  – 3 x  + 5 = 0

Comparing the equation with  ax 2  +  bx  +  c  = 0, we get

a  = 2,  b  = -3 and  c  = 5

We know, Discriminant =  b 2  – 4 ac

=  ( – 3) 2  – 4 (2) (5) = 9 – 40

= – 31

As you can see, b 2  – 4ac < 0

Therefore, no real root is possible for the given equation, 2x 2  – 3 x + 5 = 0

(ii) 3 x 2  – 4√3 x  + 4 = 0

a  = 3,  b  = -4√3 and  c  = 4

= (-4√3) 2  – 4(3)(4)

= 48 – 48 = 0

As  b 2  – 4 ac  = 0,

Real roots exist for the given equation, and they are equal to each other.

Hence, the roots will be – b /2 a  and – b /2 a .

– b /2 a  = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.

(iii) 2 x 2  – 6 x  + 3 = 0

a  = 2,  b  = -6,  c  = 3

As we know, Discriminant =  b 2  – 4 ac

= (-6) 2  – 4 (2) (3)

= 36 – 24 = 12

As  b 2  – 4 ac  > 0,

Therefore, there are distinct real roots that exist for this equation, 2 x 2  – 6 x + 3 = 0

= (-(-6) ± √(-6 2 -4(2)(3)) )/ 2(2)

= (6±2√3 )/4

Therefore, the roots for the given equation are (3+√3)/2 and (3-√3)/2

2. Find the values of  k for each of the following quadratic equations so that they have two equal roots. (i) 2 x 2  +  kx  + 3 = 0 (ii)  kx  ( x  – 2) + 6 = 0

(i) 2 x 2  +  kx  + 3 = 0

Comparing the given equation with  ax 2  +  bx  +  c  = 0, we get,

a  = 2,  b  = k and  c  = 3

= ( k ) 2  – 4(2) (3)

=  k 2  – 24

For equal roots, we know,

Discriminant = 0

k 2  – 24 = 0

k = ±√24 = ±2√6

(ii)  kx ( x  – 2) + 6 = 0

or  kx 2  – 2 kx  + 6 = 0

Comparing the given equation with  ax 2  +  bx  +  c  = 0, we get

a  =  k ,  b  = – 2 k  and  c  = 6

= ( – 2 k ) 2 – 4 ( k ) (6)

= 4 k 2 – 24 k

b 2 – 4 ac  = 0

4 k 2 – 24 k  = 0

4 k  ( k  – 6) = 0

Either 4 k  = 0 or  k  = 6 = 0

k  = 0 or  k  = 6

However, if  k  = 0, then the equation will not have the terms ‘ x 2 ‘ and ‘ x ‘.

Therefore, if this equation has two equal roots,  k  should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m 2 ? If so, find its length and breadth.

Let the breadth of the mango grove be l .

The length of the mango grove will be 2 l .

Area of the mango grove = (2 l ) ( l )= 2 l 2

2 l 2  = 800

l 2  = 800/2 = 400

l 2  – 400 =0

a  = 1,  b  = 0,  c  = 400

=> (0) 2  – 4 × (1) × ( – 400) = 1600

Here,  b 2  – 4 ac  > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

As we know, the value of length cannot be negative.

Therefore, the breadth of the mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.

Let’s say the age of one friend is x years.

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of First friend = ( x  – 4) years

Age of Second friend = (20 –  x  – 4) = (16 –  x ) years

As per the given question, we can write,

( x  – 4) (16 –  x ) = 48

16 x – x 2  – 64 + 4 x  = 48

 – x 2  + 20 x –  112 = 0

x 2  – 20 x +  112 = 0

a  =  1 ,  b  = -2 0  and  c  = 112

Discriminant =  b 2  – 4 ac

= (- 20 ) 2  – 4 × 112

= 400 – 448 = -48

b 2  – 4 ac  < 0

Therefore, there will be no real solution possible for the equations. Hence, the condition doesn’t exist.

5. Is it possible to design a rectangular park of perimeter 80 and an area of 400 m2? If so, find its length and breadth.

Let the length and breadth of the park be l and b.

Perimeter of the rectangular park = 2 ( l + b ) = 80

So, l + b  = 40

Or,  b  = 40 –  l

Area of the rectangular park =  l×b = l(40 – l) =  40 l  –  l 2 = 400

l 2   –    40 l  + 400   = 0, which is a quadratic equation.

a  = 1,  b  = -40,  c  = 400

Since, Discriminant =  b 2  – 4 ac

=(- 40 ) 2  – 4 × 400

= 1600 – 1600 = 0

Thus, b 2  – 4 ac  = 0

Therefore, this equation has equal real roots. Hence, the situation is possible.

The root of the equation,

l  = – b /2 a

l  = -(-40)/2(1) = 40/2 = 20

Therefore, the length of the rectangular park, l  = 20 m

And the breadth of the park, b  = 40 –  l  = 40 – 20 = 20 m.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

A 1-mark question was asked from Chapter 4 Quadratic Equations in the year 2018. However, in the year 2017, a total of 13 marks were asked from the topic Quadratic Equations. Therefore, students need to have a thorough understanding of the topic. The topics and sub-topics provided in this chapter include:

4.1 Introduction

If we equate the polynomial  ax 2 + bx + c, a ≠ 0 to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. In this chapter, students will study quadratic equations and various ways of finding their roots. They will also see some applications of quadratic equations in daily life situations.

4.2 Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax 2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0. In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax 2 + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

4.3 Solution of Quadratic Equations by Factorisation

A real number α is called a root of the quadratic equation ax 2  + bx + c = 0, a ≠ 0 if a α 2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax 2 + bx + c and the roots of the quadratic equation ax 2  + bx + c = 0 are the same.

4.4 Solution of a Quadratic Equation by Completing the Square

Finding the value that makes a quadratic equation a square trinomial is called  completing the square. The square trinomial can then be solved easily by factorising.

4.5 Nature of Roots

If b 2  – 4ac < 0, then there is no real number whose square is b 2  – 4ac. Therefore, there are no real roots for the given quadratic equation in this case. Since b 2  – 4ac determines whether the quadratic equation ax 2  + bx + c = 0 has real roots or not, b 2  – 4ac is called the discriminant of this quadratic equation. So, a quadratic equation ax 2  + bx + c = 0 has (i) two distinct real roots, if b 2  – 4ac > 0, (ii) two equal real roots, if b 2 – 4ac = 0, (iii) no real roots, if b 2 – 4ac < 0. 4.6

List of exercises we covered in NCERT Solutions for Class 10 Maths Chapter 4

Exercise 4.1 Solutions – 2 Questions Exercise 4.2 Solutions – 6 Questions Exercise 4.3 Solutions – 11 Questions Exercise 4.4 Solutions – 5 Questions

In a quadratic equation, x represents an unknown form, and a, b, and c are the known values. An equation to be quadratic “a” should not be equal to 0. The equation is of the form ax 2 + bx + c = 0. The values of a, b, and c are always real numbers. A quadratic equation can be calculated by completing the square. A quadratic equation has

• Two different real roots
• No real roots
• Two equal roots

Key Features of NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

• The highly experienced faculty at BYJU’S designs NCERT Solutions with utmost care.
• The solutions are 100% accurate and can be used by the students while preparing for their CBSE board exams.
• All the minute concepts are also covered to help students face other competitive exams more confidently.
• The exercise questions present in the NCERT Textbook have been answered in a step-wise manner so that students attain good scores not only on the final answer but also on each step.

For more questions to practise, students can refer to the other study materials which are given at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations

Disclaimer – 

Dropped Topics –  4.4 Solution of a quadratic equation by completing the squares

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Home > Class 10 Maths Important Questions

Most Important Questions Class 10 Maths Quadratic Equation with Solutions

Class 10 mathematics contains various questions like Quadratic Equation Most Important Questions that are useful for the students even in their competitive examinations like JEE & NEET. Quadratic equations are the equations in which at least one of the variables is squared. We use quadratic equations to determine the profit of the product and formulate the object’s speed. 

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