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Last modified on August 3rd, 2023

Quadratic Equation Word Problems

Here, we will solve different types of quadratic equation-based word problems. Use the appropriate method to solve them:

• By Completing the Square
• By Factoring
• By Quadratic Formula
• By graphing

For each process, follow the following typical steps:

• Make the equation
• Solve for the unknown variable using the appropriate method
• Interpret the result

The product of two consecutive integers is 462. Find the numbers?

Let the numbers be x and x + 1 According to the problem, x(x + 1) = 483 => x 2 + x – 483 = 0 => x 2 + 22x – 21x – 483 = 0 => x(x + 22) – 21(x + 22) = 0 => (x + 22)(x – 21) = 0 => x + 22 = 0 or x – 21 = 0 => x = {-22, 21} Thus, the two consecutive numbers are 21 and 22.

The product of two consecutive positive odd integers is 1 less than four times their sum. What are the two positive integers.

Let the numbers be n and n + 2 According to the problem, => n(n + 2) = 4[n + (n + 2)] – 1 => n 2 + 2n = 4[2n + 2] – 1 => n 2 + 2n = 8n + 7 => n 2 – 6n – 7 = 0 => n 2 -7n + n – 7 = 0 => n(n – 7) + 1(n – 7) = 0 => (n – 7) (n – 1) = 0 => n – 7 = 0 or n – 1 = 0 => n = {7, 1} If n = 7, then n + 2 = 9 If n = 1, then n + 1 = 2 Since 1 and 2 are not possible. The two numbers are 7 and 9

A projectile is launched vertically upwards with an initial velocity of 64 ft/s from a height of 96 feet tower. If height after t seconds is reprented by h(t) = -16t 2 + 64t + 96. Find the maximum height the projectile reaches. Also, find the time it takes to reach the highest point.

Since the graph of the given function is a parabola, it opens downward because the leading coefficient is negative. Thus, to get the maximum height, we have to find the vertex of this parabola. Given the function is in the standard form h(t) = a 2 x + bx + c, the formula to calculate the vertex is: Vertex (h, k) = ${\left\{ \left( \dfrac{-b}{2a}\right) ,h\left( -\dfrac{b}{2a}\right) \right\}}$ => ${\dfrac{-b}{2a}=\dfrac{-64}{2\times \left( -16\right) }}$ = 2 seconds Thus, the time the projectile takes to reach the highest point is 2 seconds ${h\left( \dfrac{-b}{2a}\right)}$ = h(2) = -16(2) 2 – 64(2) + 80 = 144 feet Thus, the maximum height the projectile reaches is 144 feet

The difference between the squares of two consecutive even integers is 68. Find the numbers.

Let the numbers be x and x + 2 According to the problem, (x + 2) 2 – x 2 = 68 => x 2 + 4x + 4 – x 2 = 68 => 4x + 4 = 68 => 4x = 68 – 4 => 4x = 64 => x = 16 Thus the two numbers are 16 and 18

The length of a rectangle is 5 units more than twice the number. The width is 4 unit less than the same number. Given the area of the rectangle is 15 sq. units, find the length and breadth of the rectangle.

Let the number be x Thus, Length = 2x + 5 Breadth = x – 4 According to the problem, (2x + 5)(x – 4) = 15 => 2x 2 – 8x + 5x – 20 – 15 = 0 => 2x 2 – 3x – 35 = 0 => 2x 2 – 10x + 7x – 35 = 0 => 2x(x – 5) + 7(x – 5) = 0 => (x – 5)(2x + 7) = 0 => x – 5 = 0 or 2x + 7 = 0 => x = {5, -7/2} Since we cannot have a negative measurement in mensuration, the number is 5 inches. Now, Length = 2x + 5 = 2(5) + 5 = 15 inches Breadth = x – 4 = 15 – 4 = 11 inches

A rectangular garden is 50 cm long and 34 cm wide, surrounded by a uniform boundary. Find the width of the boundary if the total area is 540 cm².

Given, Length of the garden = 50 cm Width of the garden = 34 cm Let the uniform width of the boundary be = x cm According to the problem, (50 + 2x)(34 + 2x) – 50 × 34 = 540 => 4x 2 + 168x – 540 = 0 => x 2 + 42x – 135 = 0 Since, this quadratic equation is in the standard form ax 2 + bx + c, we will use the quadratic formula, here a = 1, b = 42, c = -135 x = ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$ => ${\dfrac{-42\pm \sqrt{\left( 42\right) ^{2}-4\times 1\times \left( -135\right) }}{2\times 1}}$ => ${\dfrac{-42\pm \sqrt{1764+540}}{2}}$ => ${\dfrac{-42\pm \sqrt{2304}}{2}}$ => ${\dfrac{-42\pm 48}{2}}$ => ${\dfrac{-42+48}{2}}$ and ${\dfrac{-42-48}{2}}$ => x = {-45, 3} Since we cannot have a negative measurement in mensuration the width of the boundary is 3 cm

The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.

Let the length of the other two sides be x and x + 4 According to the problem, (x + 4) 2 + x 2 = 20 2 => x 2 + 8x + 16 + x 2 = 400 => 2x 2 + 8x + 16 = 400 => 2x 2 + 8x – 384 = 0 => x 2 + 4x – 192 = 0 => x 2 + 16x – 12x – 192 = 0 => x(x + 16) – 12(x + 16) = 0 => (x + 16)(x – 12) = 0 => x + 16 = 0 and x – 12 = 0 => x = {-16, 12} Since we cannot have a negative measurement in mensuration, the lengths of the sides are 12 and 16

Jennifer jumped off a cliff into the swimming pool. The function h can express her height as a function of time (t) = -16t 2 +16t + 480, where t is the time in seconds and h is the height in feet. a) How long did it take for Jennifer to attain a maximum length. b) What was the highest point that Jennifer reached. c) Calculate the time when Jennifer hit the water?

Comparing the given function with the given function f(x) = ax 2 + bx + c, here a = -16, b = 16, c = 480 a) Finding the vertex will give us the time taken by Jennifer to reach her maximum height  x = ${-\dfrac{b}{2a}}$ = ${\dfrac{-16}{2\left( -16\right) }}$ = 0.5 seconds Thus Jennifer took 0.5 seconds to reach her maximum height b) Putting the value of the vertex by substitution in the function, we get ${h\left( \dfrac{1}{2}\right) =-16\left( \dfrac{1}{2}\right) ^{2}+16\left( \dfrac{1}{2}\right) +480}$ => ${-16\left( \dfrac{1}{4}\right) +8+480}$ => 484 feet Thus the highest point that Jennifer reached was 484 feet c) When Jennifer hit the water, her height was 0 Thus, by substituting the value of the height in the function, we get -16t 2 +16t + 480 = 0 => -16(t 2 + t – 30) = 0 => t 2 + t – 30 = 0 => t 2 + 6t – 5t – 30 = 0 => t(t + 6) – 5(t + 6) = 0 => (t + 6)(t – 5) = 0 => t + 6 = 0 or t – 5 = 0 => x = {-6, 5} Since time cannot have any negative value, the time taken by Jennifer to hit the water is 5 seconds.

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Solving Quadratics by the Quadratic Formula

How to solve quadratic equations using the quadratic formula.

There are times when we are stuck solving a quadratic equation of the form [latex]a{x^2} + bx + c = 0[/latex] because the trinomial on the left side can’t be factored out easily. It doesn’t mean that the quadratic equation has no solution. At this point, we need to call upon the straightforward approach of the quadratic formula to find the solutions of the quadratic equation or put simply, determine the values of [latex]x[/latex] that can satisfy the equation.

In order use the quadratic formula, the quadratic equation that we are solving must be converted into the “standard form”, otherwise, all subsequent steps will not work. The goal is to transform the quadratic equation such that the quadratic expression is isolated on one side of the equation while the opposite side only contains the number zero, [latex]0[/latex].

Take a look at the diagram below.

In this convenient format, the numerical values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easily identified! Upon knowing those values, we can now substitute them into the quadratic formula then solve for the values of [latex]x[/latex].

• The Quadratic Formula

• Where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are the coefficients of an arbitrary quadratic equation in the standard form, [latex]a{x^2} + bx + c = 0[/latex].

Slow down if you need to. Be careful with every step while simplifying the expressions. This is where common mistakes usually happen because students tend to “relax” which results to errors that could have been prevented, such as in the addition, subtraction, multiplication and/or division of real numbers.

Examples of How to Solve Quadratic Equations by the Quadratic Formula

Example 1 : Solve the quadratic equation below using the Quadratic Formula.

By inspection, it’s obvious that the quadratic equation is in the standard form since the right side is just zero while the rest of the terms stay on the left side. In other words, we have something like this

This is great! What we need to do is simply identify the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] then substitute into the quadratic formula.

That’s it! Make it a habit to always check the solved values of [latex]x[/latex] back into the original equation to verify.

Example 2 : Solve the quadratic equation below using the Quadratic Formula.

This quadratic equation is absolutely not in the form that we want because the right side is NOT zero. I need to eliminate that [latex]7[/latex] on the right side by subtracting both sides by [latex]7[/latex]. That takes care of our problem. After doing so, solve for [latex]x[/latex] as usual.

The final answers are [latex]{x_1} = 1[/latex] and [latex]{x_2} = – {2 \over 3}[/latex].

Example 3 : Solve the quadratic equation below using the Quadratic Formula.

This quadratic equation looks like a “mess”. I have variable [latex]x[/latex]’s and constants on both sides of the equation. If we are faced with something like this, always stick to what we know. Yes, it’s all about the Standard Form. We have to force the right side to be equal to zero. We can do just that in two steps.

I will first subtract both sides by [latex]5x[/latex], and followed by the addition of [latex]8[/latex].

Values we need:

[latex]a = – 1[/latex], [latex]b = – \,8[/latex], and [latex]c = 2[/latex]

Example 4 : Solve the quadratic equation below using the Quadratic Formula.

Well, if you think that Example [latex]3[/latex] is a “mess” then this must be even “messier”. However, you’ll soon realize that they are really very similar.

We first need to perform some cleanup by converting this quadratic equation into standard form. Sounds familiar? Trust me, this problem is not as bad as it looks, as long as we know what to do.

Just to remind you, we want something like this

Therefore, we must do whatever it takes to make the right side of the equation equal to zero. Since we have three terms on the right side, it follows that three steps are required to make it zero.

The solution below starts by adding both sides by [latex]3{x^2}[/latex], followed by subtraction of [latex]3x[/latex], and finally the addition of [latex]5[/latex]. Done!

After making the right side equal to zero, the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are easy to identify. Plug those values into the quadratic formula, and simplify to get the final answers!

Example 5 : Solve the quadratic equation below using the Quadratic Formula.

First, we need to rewrite the given quadratic equation in Standard Form, [latex]a{x^2} + bx + c = 0[/latex].

• Eliminate the [latex]{x^2}[/latex] term on the right side.

• Eliminate the [latex]x[/latex] term on the right side.

• Eliminate the constant on the right side.

After getting the correct standard form in the previous step, it’s now time to plug the values of [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] into the quadratic formula to solve for [latex]x[/latex].

• From the converted standard form, extract the required values.

[latex]a = 1[/latex], [latex]b = – \,4[/latex], and [latex]c = – \,14[/latex]

• Then evaluate these values into the quadratic formula.

You might also like these tutorials:

• Quadratic Formula Practice Problems with Answers
• Solving Quadratic Equations by Square Root Method
• Solving Quadratic Equations by Factoring Method
• Solving Quadratic Equations by Completing the Square

2.5 Quadratic Equations

Learning objectives.

In this section, you will:

• Solve quadratic equations by factoring.
• Solve quadratic equations by the square root property.
• Solve quadratic equations by completing the square.
• Solve quadratic equations by using the quadratic formula.

The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods.

• Solving Quadratic Equations by Factoring

An equation containing a second-degree polynomial is called a quadratic equation . For example, equations such as 2 x 2 + 3 x − 1 = 0 2 x 2 + 3 x − 1 = 0 and x 2 − 4 = 0 x 2 − 4 = 0 are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.

Often the easiest method of solving a quadratic equation is factoring . Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation.

If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if a ⋅ b = 0 , a ⋅ b = 0 , then a = 0 a = 0 or b = 0 , b = 0 , where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero.

Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression ( x − 2 ) ( x + 3 ) ( x − 2 ) ( x + 3 ) by multiplying the two factors together.

The product is a quadratic expression. Set equal to zero, x 2 + x − 6 = 0 x 2 + x − 6 = 0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied.

The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , where a , b , and c are real numbers, and a ≠ 0. a ≠ 0. The equation x 2 + x − 6 = 0 x 2 + x − 6 = 0 is in standard form.

We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section.

The Zero-Product Property and Quadratic Equations

The zero-product property states

where a and b are real numbers or algebraic expressions.

A quadratic equation is an equation containing a second-degree polynomial; for example

where a , b , and c are real numbers, and if a ≠ 0 , a ≠ 0 , it is in standard form.

Solving Quadratics with a Leading Coefficient of 1

In the quadratic equation x 2 + x − 6 = 0 , x 2 + x − 6 = 0 , the leading coefficient, or the coefficient of x 2 , x 2 , is 1. We have one method of factoring quadratic equations in this form.

Given a quadratic equation with the leading coefficient of 1, factor it.

• Find two numbers whose product equals c and whose sum equals b .
• Use those numbers to write two factors of the form ( x + k ) or  ( x − k ) , ( x + k ) or  ( x − k ) , where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2 , −2 , the factors are ( x + 1 ) ( x − 2 ) . ( x + 1 ) ( x − 2 ) .
• Solve using the zero-product property by setting each factor equal to zero and solving for the variable.

Factoring and Solving a Quadratic with Leading Coefficient of 1

Factor and solve the equation: x 2 + x − 6 = 0. x 2 + x − 6 = 0.

To factor x 2 + x − 6 = 0 , x 2 + x − 6 = 0 , we look for two numbers whose product equals −6 −6 and whose sum equals 1. Begin by looking at the possible factors of −6. −6.

The last pair, 3 ⋅ ( −2 ) 3 ⋅ ( −2 ) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.

To solve this equation, we use the zero-product property. Set each factor equal to zero and solve.

The two solutions are 2 2 and −3. −3. We can see how the solutions relate to the graph in Figure 2 . The solutions are the x- intercepts of y = x 2 + x − 6 = 0. y = x 2 + x − 6 = 0.

Factor and solve the quadratic equation: x 2 − 5 x − 6 = 0. x 2 − 5 x − 6 = 0.

Solve the Quadratic Equation by Factoring

Solve the quadratic equation by factoring: x 2 + 8 x + 15 = 0. x 2 + 8 x + 15 = 0.

Find two numbers whose product equals 15 15 and whose sum equals 8. 8. List the factors of 15. 15.

The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve.

The solutions are −3 −3 and −5. −5.

Solve the quadratic equation by factoring: x 2 − 4 x − 21 = 0. x 2 − 4 x − 21 = 0.

Using the Zero-Product Property to Solve a Quadratic Equation Written as the Difference of Squares

Solve the difference of squares equation using the zero-product property: x 2 − 9 = 0. x 2 − 9 = 0.

Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property.

The solutions are 3 3 and −3. −3.

Solve by factoring: x 2 − 25 = 0. x 2 − 25 = 0.

Solving a Quadratic Equation by Factoring when the Leading Coefficient is not 1

When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures:

• With the quadratic in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , multiply a ⋅ c . a ⋅ c .
• Find two numbers whose product equals a c a c and whose sum equals b . b .
• Rewrite the equation replacing the b x b x term with two terms using the numbers found in step 2 as coefficients of x.
• Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping.
• Factor out the expression in parentheses.
• Set the expressions equal to zero and solve for the variable.

Solving a Quadratic Equation Using Grouping

Use grouping to factor and solve the quadratic equation: 4 x 2 + 15 x + 9 = 0. 4 x 2 + 15 x + 9 = 0.

First, multiply a c : 4 ( 9 ) = 36. a c : 4 ( 9 ) = 36. Then list the factors of 36. 36.

The only pair of factors that sums to 15 15 is 3 + 12. 3 + 12. Rewrite the equation replacing the b term, 15 x , 15 x , with two terms using 3 and 12 as coefficients of x . Factor the first two terms, and then factor the last two terms.

Solve using the zero-product property.

The solutions are − 3 4 , − 3 4 , and −3. −3. See Figure 3 .

Solve using factoring by grouping: 12 x 2 + 11 x + 2 = 0. 12 x 2 + 11 x + 2 = 0.

Solving a Polynomial of Higher Degree by Factoring

Solve the equation by factoring: −3 x 3 − 5 x 2 − 2 x = 0. −3 x 3 − 5 x 2 − 2 x = 0.

This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out − x − x from all of the terms and then proceed with grouping.

Use grouping on the expression in parentheses.

Now, we use the zero-product property. Notice that we have three factors.

The solutions are 0 , 0 , − 2 3 , − 2 3 , and −1. −1.

Solve by factoring: x 3 + 11 x 2 + 10 x = 0. x 3 + 11 x 2 + 10 x = 0.

Using the Square Root Property

When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property , in which we isolate the x 2 x 2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x 2 x 2 term so that the square root property can be used.

The Square Root Property

With the x 2 x 2 term isolated, the square root property states that:

where k is a nonzero real number.

Given a quadratic equation with an x 2 x 2 term but no x x term, use the square root property to solve it.

• Isolate the x 2 x 2 term on one side of the equal sign.
• Take the square root of both sides of the equation, putting a ± ± sign before the expression on the side opposite the squared term.
• Simplify the numbers on the side with the ± ± sign.

Solving a Simple Quadratic Equation Using the Square Root Property

Solve the quadratic using the square root property: x 2 = 8. x 2 = 8.

Take the square root of both sides, and then simplify the radical. Remember to use a ± ± sign before the radical symbol.

The solutions are 2 2 , 2 2 , −2 2 . −2 2 .

Solving a Quadratic Equation Using the Square Root Property

Solve the quadratic equation: 4 x 2 + 1 = 7. 4 x 2 + 1 = 7.

First, isolate the x 2 x 2 term. Then take the square root of both sides.

The solutions are 6 2 , 6 2 , and − 6 2 . − 6 2 .

Solve the quadratic equation using the square root property: 3 ( x − 4 ) 2 = 15. 3 ( x − 4 ) 2 = 15.

• Completing the Square

Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square . Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a , must equal 1. If it does not, then divide the entire equation by a . Then, we can use the following procedures to solve a quadratic equation by completing the square.

We will use the example x 2 + 4 x + 1 = 0 x 2 + 4 x + 1 = 0 to illustrate each step.

Given a quadratic equation that cannot be factored, and with a = 1 , a = 1 , first add or subtract the constant term to the right side of the equal sign.

Multiply the b term by 1 2 1 2 and square it.

Add ( 1 2 b ) 2 ( 1 2 b ) 2 to both sides of the equal sign and simplify the right side. We have

The left side of the equation can now be factored as a perfect square.

Use the square root property and solve.

The solutions are −2 + 3 , −2 + 3 , and −2 − 3 . −2 − 3 .

Solving a Quadratic by Completing the Square

Solve the quadratic equation by completing the square: x 2 − 3 x − 5 = 0. x 2 − 3 x − 5 = 0.

First, move the constant term to the right side of the equal sign.

Then, take 1 2 1 2 of the b term and square it.

Add the result to both sides of the equal sign.

Factor the left side as a perfect square and simplify the right side.

The solutions are 3 + 29 2 3 + 29 2 and 3 - 29 2 3 - 29 2 .

Solve by completing the square: x 2 − 6 x = 13. x 2 − 6 x = 13.

Using the Quadratic Formula

The fourth method of solving a quadratic equation is by using the quadratic formula , a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number.

We can derive the quadratic formula by completing the square . We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 −1 and obtain a positive a . Given a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , a ≠ 0 , a ≠ 0 , we will complete the square as follows:

First, move the constant term to the right side of the equal sign:

As we want the leading coefficient to equal 1, divide through by a :

Then, find 1 2 1 2 of the middle term, and add ( 1 2 b a ) 2 = b 2 4 a 2 ( 1 2 b a ) 2 = b 2 4 a 2 to both sides of the equal sign:

Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:

Now, use the square root property, which gives

Finally, add − b 2 a − b 2 a to both sides of the equation and combine the terms on the right side. Thus,

The Quadratic Formula

Written in standard form, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , any quadratic equation can be solved using the quadratic formula :

where a , b , and c are real numbers and a ≠ 0. a ≠ 0.

Given a quadratic equation, solve it using the quadratic formula

• Make sure the equation is in standard form: a x 2 + b x + c = 0. a x 2 + b x + c = 0.
• Make note of the values of the coefficients and constant term, a , b , a , b , and c . c .
• Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
• Calculate and solve.

Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: x 2 + 5 x + 1 = 0. x 2 + 5 x + 1 = 0.

Identify the coefficients: a = 1 , b = 5 , c = 1. a = 1 , b = 5 , c = 1. Then use the quadratic formula.

Solving a Quadratic Equation with the Quadratic Formula

Use the quadratic formula to solve x 2 + x + 2 = 0. x 2 + x + 2 = 0.

First, we identify the coefficients: a = 1 , b = 1 , a = 1 , b = 1 , and c = 2. c = 2.

Substitute these values into the quadratic formula.

The solutions to the equation are − 1 + i 7 2 − 1 + i 7 2 and − 1 − i 7 2 − 1 − i 7 2

Solve the quadratic equation using the quadratic formula: 9 x 2 + 3 x − 2 = 0. 9 x 2 + 3 x − 2 = 0.

The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant , or the expression under the radical, b 2 − 4 a c . b 2 − 4 a c . The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 1 relates the value of the discriminant to the solutions of a quadratic equation.

For a x 2 + b x + c = 0 a x 2 + b x + c = 0 , where a a , b b , and c c are real numbers, the discriminant is the expression under the radical in the quadratic formula: b 2 − 4 a c . b 2 − 4 a c . It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation

Use the discriminant to find the nature of the solutions to the following quadratic equations:

• ⓐ x 2 + 4 x + 4 = 0 x 2 + 4 x + 4 = 0
• ⓑ 8 x 2 + 14 x + 3 = 0 8 x 2 + 14 x + 3 = 0
• ⓒ 3 x 2 − 5 x − 2 = 0 3 x 2 − 5 x − 2 = 0
• ⓓ 3 x 2 − 10 x + 15 = 0 3 x 2 − 10 x + 15 = 0

Calculate the discriminant b 2 − 4 a c b 2 − 4 a c for each equation and state the expected type of solutions.

x 2 + 4 x + 4 = 0 x 2 + 4 x + 4 = 0

b 2 − 4 a c = ( 4 ) 2 − 4 ( 1 ) ( 4 ) = 0. b 2 − 4 a c = ( 4 ) 2 − 4 ( 1 ) ( 4 ) = 0. There will be one rational double solution.

8 x 2 + 14 x + 3 = 0 8 x 2 + 14 x + 3 = 0

b 2 − 4 a c = ( 14 ) 2 − 4 ( 8 ) ( 3 ) = 100. b 2 − 4 a c = ( 14 ) 2 − 4 ( 8 ) ( 3 ) = 100. As 100 100 is a perfect square, there will be two rational solutions.

3 x 2 − 5 x − 2 = 0 3 x 2 − 5 x − 2 = 0

b 2 − 4 a c = ( −5 ) 2 − 4 ( 3 ) ( −2 ) = 49. b 2 − 4 a c = ( −5 ) 2 − 4 ( 3 ) ( −2 ) = 49. As 49 49 is a perfect square, there will be two rational solutions.

3 x 2 −10 x + 15 = 0 3 x 2 −10 x + 15 = 0

b 2 − 4 a c = ( −10 ) 2 − 4 ( 3 ) ( 15 ) = −80. b 2 − 4 a c = ( −10 ) 2 − 4 ( 3 ) ( 15 ) = −80. There will be two complex solutions.

Using the Pythagorean Theorem

One of the most famous formulas in mathematics is the Pythagorean Theorem . It is based on a right triangle, and states the relationship among the lengths of the sides as a 2 + b 2 = c 2 , a 2 + b 2 = c 2 , where a a and b b refer to the legs of a right triangle adjacent to the 90° 90° angle, and c c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications.

We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side.

The Pythagorean Theorem is given as

where a a and b b refer to the legs of a right triangle adjacent to the 90 ∘ 90 ∘ angle, and c c refers to the hypotenuse, as shown in Figure 4 .

Finding the Length of the Missing Side of a Right Triangle

Find the length of the missing side of the right triangle in Figure 5 .

As we have measurements for side b and the hypotenuse, the missing side is a.

Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse.

Access these online resources for additional instruction and practice with quadratic equations.

• The Zero-Product Property
• Quadratic Formula with Two Rational Solutions
• Length of a leg of a right triangle

2.5 Section Exercises

How do we recognize when an equation is quadratic?

When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 we may graph the equation y = a x 2 + b x + c y = a x 2 + b x + c and have no zeroes ( x -intercepts).

When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side?

In the quadratic formula, what is the name of the expression under the radical sign b 2 − 4 a c , b 2 − 4 a c , and how does it determine the number of and nature of our solutions?

Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method.

For the following exercises, solve the quadratic equation by factoring.

x 2 + 4 x − 21 = 0 x 2 + 4 x − 21 = 0

x 2 − 9 x + 18 = 0 x 2 − 9 x + 18 = 0

2 x 2 + 9 x − 5 = 0 2 x 2 + 9 x − 5 = 0

6 x 2 + 17 x + 5 = 0 6 x 2 + 17 x + 5 = 0

4 x 2 − 12 x + 8 = 0 4 x 2 − 12 x + 8 = 0

3 x 2 − 75 = 0 3 x 2 − 75 = 0

8 x 2 + 6 x − 9 = 0 8 x 2 + 6 x − 9 = 0

4 x 2 = 9 4 x 2 = 9

2 x 2 + 14 x = 36 2 x 2 + 14 x = 36

5 x 2 = 5 x + 30 5 x 2 = 5 x + 30

4 x 2 = 5 x 4 x 2 = 5 x

7 x 2 + 3 x = 0 7 x 2 + 3 x = 0

x 3 − 9 x = 2 x 3 − 9 x = 2

For the following exercises, solve the quadratic equation by using the square root property.

x 2 = 36 x 2 = 36

x 2 = 49 x 2 = 49

( x − 1 ) 2 = 25 ( x − 1 ) 2 = 25

( x − 3 ) 2 = 7 ( x − 3 ) 2 = 7

( 2 x + 1 ) 2 = 9 ( 2 x + 1 ) 2 = 9

( x − 5 ) 2 = 4 ( x − 5 ) 2 = 4

For the following exercises, solve the quadratic equation by completing the square. Show each step.

x 2 − 9 x − 22 = 0 x 2 − 9 x − 22 = 0

2 x 2 − 8 x − 5 = 0 2 x 2 − 8 x − 5 = 0

x 2 − 6 x = 13 x 2 − 6 x = 13

x 2 + 2 3 x − 1 3 = 0 x 2 + 2 3 x − 1 3 = 0

2 + z = 6 z 2 2 + z = 6 z 2

6 p 2 + 7 p − 20 = 0 6 p 2 + 7 p − 20 = 0

2 x 2 − 3 x − 1 = 0 2 x 2 − 3 x − 1 = 0

For the following exercises, determine the discriminant, and then state how many solutions there are and the nature of the solutions. Do not solve.

2 x 2 − 6 x + 7 = 0 2 x 2 − 6 x + 7 = 0

x 2 + 4 x + 7 = 0 x 2 + 4 x + 7 = 0

3 x 2 + 5 x − 8 = 0 3 x 2 + 5 x − 8 = 0

9 x 2 − 30 x + 25 = 0 9 x 2 − 30 x + 25 = 0

2 x 2 − 3 x − 7 = 0 2 x 2 − 3 x − 7 = 0

6 x 2 − x − 2 = 0 6 x 2 − x − 2 = 0

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution .

2 x 2 + 5 x + 3 = 0 2 x 2 + 5 x + 3 = 0

x 2 + x = 4 x 2 + x = 4

3 x 2 − 5 x + 1 = 0 3 x 2 − 5 x + 1 = 0

x 2 + 4 x + 2 = 0 x 2 + 4 x + 2 = 0

4 + 1 x − 1 x 2 = 0 4 + 1 x − 1 x 2 = 0

For the following exercises, enter the expressions into your graphing utility and find the zeroes to the equation (the x -intercepts) by using 2 nd CALC 2:zero . Recall finding zeroes will ask left bound (move your cursor to the left of the zero,enter), then right bound (move your cursor to the right of the zero,enter), then guess (move your cursor between the bounds near the zero, enter). Round your answers to the nearest thousandth.

Y 1 = 4 x 2 + 3 x − 2 Y 1 = 4 x 2 + 3 x − 2

Y 1 = −3 x 2 + 8 x − 1 Y 1 = −3 x 2 + 8 x − 1

Y 1 = 0.5 x 2 + x − 7 Y 1 = 0.5 x 2 + x − 7

To solve the quadratic equation x 2 + 5 x − 7 = 4 , x 2 + 5 x − 7 = 4 , we can graph these two equations

Y 1 = x 2 + 5 x − 7 Y 2 = 4 Y 1 = x 2 + 5 x − 7 Y 2 = 4

and find the points of intersection. Recall 2 nd CALC 5:intersection. Do this and find the solutions to the nearest tenth.

To solve the quadratic equation 0.3 x 2 + 2 x − 4 = 2 , 0.3 x 2 + 2 x − 4 = 2 , we can graph these two equations

Y 1 = 0.3 x 2 + 2 x − 4 Y 2 = 2 Y 1 = 0.3 x 2 + 2 x − 4 Y 2 = 2

Beginning with the general form of a quadratic equation, a x 2 + b x + c = 0 , a x 2 + b x + c = 0 , solve for x by using the completing the square method, thus deriving the quadratic formula.

Show that the sum of the two solutions to the quadratic equation is − b a − b a .

A person has a garden that has a length 10 feet longer than the width. Set up a quadratic equation to find the dimensions of the garden if its area is 119 ft. 2 . Solve the quadratic equation to find the length and width.

Abercrombie and Fitch stock had a price given as P = 0.2 t 2 − 5.6 t + 50.2 , P = 0.2 t 2 − 5.6 t + 50.2 , where t t is the time in months from 1999 to 2001. ( t = 1 t = 1 is January 1999). Find the two months in which the price of the stock was $30.

Suppose that an equation is given p = −2 x 2 + 280 x − 1000 , p = −2 x 2 + 280 x − 1000 , where x x represents the number of items sold at an auction and p p is the profit made by the business that ran the auction. How many items sold would make this profit a maximum? Solve this by graphing the expression in your graphing utility and finding the maximum using 2 nd CALC maximum. To obtain a good window for the curve, set x x [0,200] and y y [0,10000].

Real-World Applications

A formula for the normal systolic blood pressure for a man age A , A , measured in mmHg, is given as P = 0.006 A 2 − 0.02 A + 120. P = 0.006 A 2 − 0.02 A + 120. Find the age to the nearest year of a man whose normal blood pressure measures 125 mmHg.

The cost function for a certain company is C = 60 x + 300 C = 60 x + 300 and the revenue is given by R = 100 x − 0.5 x 2 . R = 100 x − 0.5 x 2 . Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $300.

A falling object travels a distance given by the formula d = 5 t + 16 t 2 d = 5 t + 16 t 2 ft, where t t is measured in seconds. How long will it take for the object to travel 74 ft?

A vacant lot is being converted into a community garden. The garden and the walkway around its perimeter have an area of 378 ft 2 . Find the width of the walkway if the garden is 12 ft. wide by 15 ft. long.

An epidemiological study of the spread of a certain influenza strain that hit a small school population found that the total number of students, P P , who contracted the flu t t days after it broke out is given by the model P = − t 2 + 13 t + 130 , P = − t 2 + 13 t + 130 , where 1 ≤ t ≤ 6. 1 ≤ t ≤ 6. Find the day that 160 students had the flu. Recall that the restriction on t t is at most 6.

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Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Access for free at https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites

• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra 2e
• Publication date: Dec 21, 2021
• Location: Houston, Texas
• Book URL: https://openstax.org/books/college-algebra-2e/pages/1-introduction-to-prerequisites
• Section URL: https://openstax.org/books/college-algebra-2e/pages/2-5-quadratic-equations

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Quadratic Equations

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• Sravanth C.
• Akshay Yadav
• Yash Dev Lamba
• Mehul Arora
• Omkar Kulkarni
• Skanda Prasad
• A Former Brilliant Member

A quadratic equation is a polynomial equation with degree two. In other words, it is an equation of the form \( ax^2 + bx + c =0 \), where \( a \), \(b\) and \(c\) are real numbers and \(a\neq 0\).

Solving by Factoring

Finding quadratic equation from roots, solving by completing the square, solving by quadratic formula, nature of roots of quadratic equation, word problems - basic, biquadratic equations, quadratic equations - problem solving.

Main Article: Factoring Polynomials

We can solve quadratics using factoring and the zero product property . In general, we can rewrite a quadratic as the product of two linear factors such that \( ax^2 + bx + c = a(x+p)(x+q) \). By the zero product property,

\[ \text{if }\ ax^2 + bx + c = a(x+p)(x+q) = 0, \ \text{then either }\ x = -p \ \text{ or }\ x = -q.\]

Now, to factorise a quadratic equation, follow these steps.

\(1)\) We have to break \(b\) (the coefficient of \(x\)) into two terms in such a way that their sum is \(b\) and their product is \(ac:\) \[ax^2+bx+c=0 \implies ax^2+(b_1+b_2)x+c=0 \text{ such that }b_1+b_2=b \text{ and } b_1 \times b_2=ac.\] \(2)\) Next we need to group \(ax^2 + b_1x \text{ and } b_2x+c\) and factorize them in a way that they both have one factor common.

Now we will have the equation transformed into factors. From here on, the solution is easy. We use the zero product property and equate each factor to \(0\), i.e. \((x-\alpha)=0 \implies x=\alpha\) and \((x-\beta)=0 \implies x=\beta\).

Solve \(x^2+5x+6=0\) for \(x\) by the method of factoring. Following the steps mentioned above, we first break the coefficient of \(x\) in two terms such that their sum equals \(5\) and their product equals \(1 \times 6=6\): \[x^2+(2+3)x +6=0,\] where we can observe that \(2+3=5\) and \(2 \times 3=6\): \[\begin{align} x^2+2x+3x+6&=0 \\ x(x+2)+3(x+2)&=0. \end{align}\] Taking out \((x+2)\) as a common factor, we have \[\begin{align} (x+3)(x+2)&=0 \\ x+3&=0 \implies x=-3 \\ x+2&=0 \implies x=-2. \end{align}\] Therefore the two roots of the given equation are \(-3\) and \(-2\). \(_\square\)

Method of Solving a Quadratic Equation by Factorizing: Step 1. Make the given equation free from fractions and radicals and put it into the standard form \(ax^2+bx+c=0.\) Step 2. Factorize \(ax^2+bx+c\) into two linear factors. Step 3. Put each linear factor equal to \(0\) (to apply the zero product rule). Step 4. Solve these linear equations and get two roots of the given quadratic equation.

Solve \(x^2 - x - 6 =0 \) by the method of factoring. We have \[ x^2 - x - 6 = (x-3)(x+2),\] which gives \( x = 3 \) or \( x = -2 \). \(_\square\) Note that the factors of \( x^2 - x - 6 \) are \(1, x^2 - x - 6, x-3,\) and \(x+2\).

Solve the equation \( x^2+3x+2=0 \) for \(x\). We have \[\begin{align} x^2+3x+2 &=0 \\ x^2+2x+x+2 &=0\\ x(x+2)+1(x+2) & =0\\ (x+2)(x+1) & =0. \end{align}\] So, \[\begin{align} (x+2)=0 \text{ or } (x+1)=0 \\ x=-2 \text{ or } x =-1.\ _\square \end{align}\]

Note: We cannot always factor into linear factors using only real numbers . For some quadratics \((\)e.g., \( x^2 + 1 )\), the linear factors require complex numbers:

\[ x^2 + 1 = (x+i)(x-i) .\]

Try the following problems:

What are the solutions to the equation

\[ x^2 = 4 ? \]

Find the positive root of the equation \(x^{2}+x-20=0\).

If the value of \( a^2 + 6a -6 \) is \(a\), then find the minimum value of \(a\).

When two places of the variable are given, we have to write them of the form \(\text{(variable - value = 0)}\).

To find the equation from the roots: Step 1. If the variable \(x\) is given, and two values \(x=a\) and \(x=b\) are given, then we have to simplify them to \[x-a=0 \quad\text{ and }\quad x-b=0.\] Step 2. Multiplying the equations and simplifying them, we arrive at this: \[\begin{align} (x-a)(x-b)&=0\\ x^2 -(a+b)x+ab&= 0. \end{align}\]

Find the quadratic equation whose roots are \(2\) and \(-3\) Considering the equation in variable \(x\), we have the following: \[\begin{align} x=2&\implies (x-2)=0\\ x=-3&\implies (x+3)=0. \end{align}\] Multiplying both the equations, we have \[\begin{align} (x-2)(x+3) & =0\\ x(x+3)-2(x+3) & =0\\ x^2+3x-2x-6 & =0\\ x^2+x-6 & =0.\ _\square \end{align}\]

Find the quadratic equation whose roots are \(5\) and \(6\) Considering the equation in variable \(x\), we have the following: \[\begin{align} x&=5 \implies x-5=0\\ x&=6\implies x-6=0. \end{align}\] Multiplying both the equations gives \[\begin{align} (x-5)(x-6) & =0\\ x(x-6)-5(x-6) & =0\\ x^2-6x-5x+30 & =0\\ x^2-11x+30 & =0.\ _\square \end{align}\]

Main Article: Completing The Square

For a quadratic polynomial \(f(x) = ax^2 + bx +c\), completing the square means finding an expression of the form

\[f(x) = a(x-b)^2 + c.\]

Complete the square for the quadratic \( x^2 + 8x + 10 \). Since our middle term is \( 8x \), we know that we will want a perfect square with the form \( (x+4)^2 \), which expands to \( x^2 + 8x + 16 \). Thus, we can do the following: \[\begin{align} x^2 + 8x + 10 &= x^2 + 8x + 16 - 16 + 10\\ &= (x^2 + 8x + 16) - 6 \\ &= ( x+ 4 )^2 -6. \ _\square \end{align} \]

Solve this equation \(2x^{2}+3x+1=0\) by method of completing square. First take \(2\) as common: \(2\left(x^{2}+\frac{3x}{2}\right)+1=0\). Since our middle term is \(\frac 32 x,\) we know that we will want a perfect square with the form \[\left(x+\dfrac{3}{4}\right)^2=x^{2}+\dfrac{3x}{2}+\dfrac{9}{16}.\] So rewrite the whole equation as \[\begin{align} 2\left(x+\dfrac{3x}2 +\dfrac{9}{16}-\dfrac{9}{16}\right)+1 & =0\\ 2\left(x+\dfrac{3x}2 + \left(\dfrac{3}{4}\right)^{2}-\dfrac{9}{16}\right)+1 & =0\\ 2\left(x+\dfrac{3}{4}\right)^{2}-\dfrac{1}{8} & =0\\ \left(x+\dfrac{3}{4}\right)^{2} & =\dfrac{1}{16}. \end{align}\] Thus we have \[\begin{align} x+\dfrac{3}{4} &=\pm \dfrac{1}{4}\\ \Rightarrow x &=-\dfrac{1}{2} \, \text{ or }\, x=-1.\ _\square \end{align}\]

Find the minimum value of \(4x^{2}+8x+16\) for real \(x\).

Main Article: Quadratic Formula

The quadratic formula states that for the equation \( ax^2 + bx + c =0 \), the values of \( x\) are given by the following:

\[ x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}. \]

To see how this formula is derived via completing the square, see Quadratic Formula .

Solve \(5x^2-2x-3=0\) for \(x\). Here, \(a=5, b=-2, c=-3\). Using the quadratic formula, we get \[\begin{align} x &= \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ & = \dfrac {-(-2) \pm \sqrt{(-2)^2 - 4×5×-3}}{2×5}\\ & = \dfrac {2 \pm \sqrt{4 +60}}{10} = \dfrac {2 \pm \sqrt{64}}{10}\\ & = \dfrac {2 \pm 8}{10}\\ \Rightarrow x & = -0.6 \, \text{ or }\, x=1.\ _\square \end{align}\]

Solve \(x^2-4x+1\) for \(x\). Here, \(a=1, b=-4, c=1\). Using the quadratic formula, we get \[\begin{align} x & = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}\\ & = \dfrac {-(-4) \pm \sqrt{(-4)^2 - 4×1×1}}{2×1}\\ & = \dfrac {4 \pm \sqrt{16 -4}}{2} =\dfrac {4 \pm \sqrt{12}}{2}\\ &=\dfrac {4 \pm 2×\sqrt {3}}{2}\\ \Rightarrow x & =2+\sqrt3 \, \text{ or }\, x=2-\sqrt3.\ _\square \end{align}\]

Solve \(x^2-20x-69 = 0\) for \(x.\) Substituting the values \(a=1, b=-20, c=-69\) in the quadratic formula, we get \[\begin{align} x & = \dfrac {-(-20) \pm \sqrt {(-20)^2 -4×1×-69}}{2×1}\\ & = \dfrac {20 \pm \sqrt {400 +276}}{2}\\ & = \dfrac {20 \pm \sqrt {676}}{2}\\ & = \dfrac {20 \pm 26}{2}\\ \Rightarrow x & = 23 \, \text{ or }\, x=-3.\ _\square \end{align}\]

\[ x= \dfrac { -b\pm \sqrt { b ^2 - 4ac } }{ 2a } \]

Using the quadratic formula above, find the roots of the equation

\[ x^2−20x−69=0.\]

Solve the quadratic equation

\[ (5x-3)^2 = 32 . \]

Check out the set: 2016 Problems.

Main Article: Parabolas

Here is an example illustrating the above.

Find the equation of a parabola with vertex at \((0,0)\) if its axis of symmetry is the \(y\)-axis and its graph contains the point \(\left(\frac {-1}{2}, 2 \right)\). We write the vertex form of the parabola as \(y = A(x^2)\). Plug in the coordinates of the given point to find \(A\) \[\begin{align} 2 &= A \times\left (\dfrac {-1}{2}\right) ^2 \\ A &= 8\\ \Rightarrow y &= 8x^2.\ _\square \end{align}\]

If the McDonald's logo were stored as a set of pixels, enlargement would quickly result in distorted or pixelated images, which are an eyesore. As such, companies often make vector images of their logos, in which the information is stored as mathematical formulae. Such vector images are easily scaled while maintaining sharp, crisp images.

As a first approximation, the logo is deconstructed and approximated as 2 parabolic curves of the form \( y = -A(x-5)^2 \) and \( y = - A (x+5)^2 \). The McDonald's logo has a height to length ratio of 1.05. What is \(A\)?

As \(x\) ranges over all real values, what is the minimum of

\[ x^2 + ( x + 1) ^2 ? \]

The nature of roots of a quadratic equation can be determined by observing the quadratic formula closely. It basically consists of a discriminant which actually makes the difference in formula and leads us two roots.

We know the quadratic formula is

\[ x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} \]

for any quadratic equation written in standard form of \(ax^2+bx+c=0\). The discriminant \(D\) for the quadratic equation is

\[D=b^2-4ac,\]

\[\begin{cases} b^2-4ac \gt 0: & \text{two distinct real roots} \\ b^2-4ac=0: & \text{equal and real roots} \\ b^2-4ac \lt 0: & \text{imaginary roots}. \end{cases}\]

Determine the nature of roots of the following two quadratic equations: \[\begin{align} 2x^2+x-1&=0 \\ x^2-4x+4 &=0. \end{align}\] For the quadratic equation \( 2x^2+x-1=0\): Since \(a=2,b=1,c=-1,\) \[\begin{align} b^2-4ac & =1^2-4 \times 2 \times -1\\ & =9 >0, \end{align}\] which implies that the roots are real and distinct. For the quadratic equation \(x^2-4x+4=0\): Since \(a=1,b=-4,c=4,\) \[\begin{align} b^2-4ac & =(-4)^2-4 \times 1 \times 4\\ & =0, \end{align}\] which implies that the roots are real and repeated. \(_\square\)

Find the value of \(k\) for which the following quadratic polynomial has repeated roots: \[x^2+4x+k.\] We know that if \(D=0,\) then the quadratic polynomial has repeated roots. So, \[\begin{align} b^2-4ac&=0\\ (4)^2-4(1)(k)&=0\\ k&=4.\ _\square \end{align}\]

Show that the equation \(x^2+dx-1=0\) has real and distinct roots for all real values of \(d\). Here, \(a=1, b=d,\) and \(c=-1\). So the discriminant would be \[D=d^2-4×1×-1=d^2+4.\] Since \(d^2\) is a perfect square, it is always greater than or equal to \(0\). So, \[D=d^2+4\geq 4.\] Thus, the discriminant is always greater than \(0\), implying that this equation has distinct real roots for any real value of \(d\). \(_\square\)

Two years ago, a man's age was three times the square of his son's age. In three years, his age will be four times his son's age. Find their present ages. Let the present age of the son be \(x\). Then the son's age two years ago was \(x-2\), and his father's age two years ago was \(3\times (x-2)^2\). This implies the present age of the father is \(\big[3×(x-2)^2\big]+2\), and hence in three years his age will be \(\big[3×(x-2)^2\big] +2+ 3= \big[3×(x-2)^2\big] +5\). The son's age in 3 years will be \(x+3\). According to given conditions, the following holds: \[ \begin{array}{rl} 3(x-2)^2+5 & = 4(x+3) \\ \Rightarrow 3x^2-16x+5 & =0 \\ \Rightarrow (3x-1)(x-5) & =0 \\ \Rightarrow x & =\frac 13, 5. \end{array} \] If \(x\) = \( \frac {1}{3}\), then the son's age 2 years ago would become negative, which is impossible. So, the son's present age is \(x=5\), which implies that the present age of the man is \[\begin{align} 3\times (x-2)^2+2 &= 3\times (5-2)^2+2\\ &= 3\times 3^2+2\\ &= 3×9+2\\ &= 27+2\\ &= 29.\ _\square \end{align}\]

Find two numbers whose sum is \(40\), and product \(375\). Let one number be \(x\). Then, according to the first condition, the second number is \(40-x\). Substituting, the value in the second condition, we get \[\begin{align} x(40-x) &= 375\\ 40x-x^2 &= 375\\ x^2-40x+375 &=0\\ x^2-25x-15x+375 &=0\\ x(x-25)-15(x-25) &=0\\ (x-15)(x-25) &=0\\ \Rightarrow x&=15, x=25. \end{align}\] Therefore, the smaller number is \(15\) and the larger one is \(25\). \(_\square\)

The product of two consecutive positive integers is 90. What is their sum? Since the integers are consecutive, we can rewrite the expression above as \( n(n+1) = 90 \). This gives us the following quadratic equation: \( n^2 +n -90 = 0 \). Factoring, we can see that \[ n^2 +n -90 = (n - 9)(n+10) =0, \] which implies \( n = 9 \). Then the two numbers are 9 and 10, and their sum is 19. \(_\square\)

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

The difference of the cubes of two consecutive odd positive integers is 400 more than the sum of their squares. Find the sum of the two integers.

Clarification : The odd positive integers are \( 1, 3, 5, 7, 9, \ldots.\) Two consecutive odd positive integers refer to two consecutive numbers in this sequence. It does not refer to two consecutive integers (of which one will not be odd).

Sometimes, the quadratic formula could be useful in solving equations of larger degree.

Solve \( x^{4} - 3x^{2} + 1 = 0 \). That equation isn't something you'd want to factor. So, you could make the substitution \( u = x^{2} \). Then the equation would read \[ u^{2} - 3u + 1 = 0. \] We can solve that with the quadratic formula: \[ u = \dfrac{3 \pm \sqrt{5}}{2}. \] But we're not done yet. We want \( x\), not \(u\). Since \( u = x^{2} = \frac{3 \pm \sqrt{5}}{2} \), solving that equation for \(x\) gives \[x=\pm\sqrt{\dfrac{3 \pm \sqrt{5}}{2}}.\ _\square\]

This section contains miscellaneous problems on quadratic equations for you to try, which will eventually enhance your problem solving skills.

\[ x^2 - 5ax + 100 = \ 0 \\ x^2 - 100x + 5a = \ 0 \]

Find the sum of all possible values of \(x\) that satisfy the equations above. Note that \(a\) is an arbitrary constant.

\[ x(x+1)(x+2)(x+3)=120\]

Find the sum of all the real roots of the equation above.

• Factoring Polynomials
• Completing The Square
• Quadratic Formula

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Quadratic Equations

An example of a Quadratic Equation :

The function can make nice curves like this one:

The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x 2 ).

It is also called an "Equation of Degree 2" (because of the "2" on the x )

Standard Form

The Standard Form of a Quadratic Equation looks like this:

• a , b and c are known values. a can't be 0
• x is the variable or unknown (we don't know it yet)

Here are some examples:

Have a Play With It

Play with the Quadratic Equation Explorer so you can see:

• the function's graph, and
• the solutions (called "roots").

Hidden Quadratic Equations!

As we saw before, the Standard Form of a Quadratic Equation is

But sometimes a quadratic equation does not look like that!

For example:

How To Solve Them?

The " solutions " to the Quadratic Equation are where it is equal to zero .

They are also called " roots ", or sometimes " zeros "

There are usually 2 solutions (as shown in this graph).

And there are a few different ways to find the solutions:

Just plug in the values of a, b and c, and do the calculations.

We will look at this method in more detail now.

About the Quadratic Formula

First of all what is that plus/minus thing that looks like ± ?

The ± means there are TWO answers:

x = −b + √(b 2 − 4ac) 2a

x = −b − √(b 2 − 4ac) 2a

Here is an example with two answers:

But it does not always work out like that!

• Imagine if the curve "just touches" the x-axis.
• Or imagine the curve is so high it doesn't even cross the x-axis!

This is where the "Discriminant" helps us ...

Discriminant

Do you see b 2 − 4ac in the formula above? It is called the Discriminant , because it can "discriminate" between the possible types of answer:

• when b 2 − 4ac is positive, we get two Real solutions
• when it is zero we get just ONE real solution (both answers are the same)
• when it is negative we get a pair of Complex solutions

Complex solutions? Let's talk about them after we see how to use the formula.

Using the Quadratic Formula

Just put the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x 2 + 6x + 1 = 0

Answer: x = −0.2 or x = −1

Let's check the answers:

Remembering The Formula

A kind reader suggested singing it to "Pop Goes the Weasel":

Try singing it a few times and it will get stuck in your head!

Or you can remember this story:

x = −b ± √(b 2 − 4ac) 2a

"A negative boy was thinking yes or no about going to a party, at the party he talked to a square boy but not to the 4 awesome chicks. It was all over at 2 am. "

Complex Solutions?

When the Discriminant (the value b 2 − 4ac ) is negative we get a pair of Complex solutions ... what does that mean?

It means our answer will include Imaginary Numbers . Wow!

Example: Solve 5x 2 + 2x + 1 = 0

Answer: x = −0.2 ± 0.4 i

The graph does not cross the x-axis. That is why we ended up with complex numbers.

In a way it is easier: we don't need more calculation, we leave it as −0.2 ± 0.4 i .

Example: Solve x 2 − 4x + 6.25 = 0

Answer: x = 2 ± 1.5 i

BUT an upside-down mirror image of our equation does cross the x-axis at 2 ± 1.5 (note: missing the i ).

Just an interesting fact for you!

• Quadratic Equation in Standard Form: ax 2 + bx + c = 0
• Quadratic Equations can be factored
• Quadratic Formula: x = −b ± √(b 2 − 4ac) 2a
• positive, there are 2 real solutions
• zero, there is one real solution
• negative, there are 2 complex solutions

Quadratic Equation

Quadratic equations are second-degree algebraic expressions and are of the form ax 2 + bx + c = 0. The term "quadratic" comes from the Latin word "quadratus" meaning square, which refers to the fact that the variable x is squared in the equation. In other words, a quadratic equation is an “equation of degree 2.” There are many scenarios where a quadratic equation is used. Did you know that when a rocket is launched, its path is described by a quadratic equation? Further, a quadratic equation has numerous applications in physics, engineering, astronomy, etc.

Quadratic equations have maximum of two solutions, which can be real or complex numbers. These two solutions (values of x) are also called the roots of the quadratic equations and are designated as (α, β). We shall learn more about the roots of a quadratic equation in the below content.

What is Quadratic Equation?

A quadratic equation is an algebraic equation of the second degree in x. The quadratic equation in its standard form is ax 2 + bx + c = 0, where a and b are the coefficients, x is the variable, and c is the constant term. The important condition for an equation to be a quadratic equation is the coefficient of x 2 is a non-zero term (a ≠ 0). For writing a quadratic equation in standard form, the x 2 term is written first, followed by the x term, and finally, the constant term is written.

Further, in real math problems the quadratic equations are presented in different forms: (x - 1)(x + 2) = 0, -x 2 = -3x + 1, 5x(x + 3) = 12x, x 3 = x(x 2 + x - 3). All of these equations need to be transformed into standard form of the quadratic equation before performing further operations.

Roots of a Quadratic Equation

The roots of a quadratic equation are the two values of x, which are obtained by solving the quadratic equation. These roots of the quadratic equation are also called the zeros of the equation. For example, the roots of the equation x 2 - 3x - 4 = 0 are x = -1 and x = 4 because each of them satisfies the equation. i.e.,

• At x = -1, (-1) 2 - 3(-1) - 4 = 1 + 3 - 4 = 0
• At x = 4, (4) 2 - 3(4) - 4 = 16 - 12 - 4 = 0

There are various methods to find the roots of a quadratic equation. The usage of the quadratic formula is one of them.

Quadratic Formula

Quadratic formula is the simplest way to find the roots of a quadratic equation . There are certain quadratic equations that cannot be easily factorized, and here we can conveniently use this quadratic formula to find the roots in the quickest possible way. The two roots in the quadratic formula are presented as a single expression. The positive sign and the negative sign can be alternatively used to obtain the two distinct roots of the equation.

Quadratic Formula: The roots of a quadratic equation ax 2 + bx + c = 0 are given by x = [-b ± √(b 2 - 4ac)]/2a.

This formula is also known as the Sridharacharya formula .

Example: Let us find the roots of the same equation that was mentioned in the earlier section x 2 - 3x - 4 = 0 using the quadratic formula.

a = 1, b = -3, and c = -4.

x = [-b ± √(b 2 - 4ac)]/2a = [-(-3) ± √((-3) 2 - 4(1)(-4))]/2(1) = [3 ± √25] / 2 = [3 ± 5] / 2 = (3 + 5)/2 or (3 - 5)/2 = 8/2 or -2/2 = 4 or -1 are the roots.

Proof of Quadratic Formula

Consider an arbitrary quadratic equation: ax 2 + bx + c = 0, a ≠ 0

To determine the roots of this equation, we proceed as follows:

ax 2 + bx = -c ⇒ x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square , by introducing a new term (b/2a) 2 on both sides:

• x 2 + bx/a + (b/2a) 2 = -c/a + (b/2a) 2

The left-hand side is now a perfect square:

(x + b/2a) 2 = -c/a + b 2 /4a 2 ⇒ (x + b/2a) 2 = (b 2 - 4ac)/4a 2

This is good for us, because now we can take square roots to obtain:

x + b/2a = ±√(b 2 - 4ac)/2a

x = (-b ± √(b 2 - 4ac))/2a

Thus, by completing the squares, we were able to isolate x and obtain the two roots of the equation.

Nature of Roots of the Quadratic Equation

The roots of a quadratic equation are usually represented to by the symbols alpha (α), and beta (β). Here we shall learn more about how to find the nature of roots of a quadratic equation without actually finding the roots of the equation.

The nature of roots of a quadratic equation can be found without actually finding the roots (α, β) of the equation. This is possible by taking the discriminant value, which is part of the formula to solve the quadratic equation. The value b 2 - 4ac is called the discriminant of a quadratic equation and is designated as 'D'. Based on the discriminant value the nature of the roots of the quadratic equation can be predicted.

Discriminant: D = b 2 - 4ac

• D > 0, the roots are real and distinct
• D = 0, the roots are real and equal.
• D < 0, the roots do not exist or the roots are imaginary .

Now, check out the formulas to find the sum and the product of the roots of the equation.

Sum and Product of Roots of Quadratic Equation

The coefficient of x 2 , x term, and the constant term of the quadratic equation ax 2 + bx + c = 0 are useful in determining the sum and product of the roots of the quadratic equation. The sum and product of the roots of a quadratic equation can be directly calculated from the equation, without actually finding the roots of the quadratic equation. For a quadratic equation ax 2 + bx + c = 0, the sum and product of the roots are as follows.

• Sum of the Roots: α + β = -b/a = - Coefficient of x/ Coefficient of x 2
• Product of the Roots: αβ = c/a = Constant term/ Coefficient of x 2

Writing Quadratic Equations Using Roots

The quadratic equation can also be formed for the given roots of the equation. If α, β, are the roots of the quadratic equation, then the quadratic equation is as follows.

x 2 - (α + β)x + αβ = 0

Example: What is the quadratic equation whose roots are 4 and -1?

Solution: It is given that α = 4 and β = -1. The corresponding quadratic equation is found by:

x 2 - (α + β)x + αβ = 0 x 2 - (α + β)x + αβ = 0 x 2 - (4 - 1)x + (4)(-1) = 0 x 2 - 3x - 4 = 0

Formulas Related to Quadratic Equations

The following list of important formulas is helpful to solve quadratic equations.

• The quadratic equation in its standard form is ax 2 + bx + c = 0
• For D > 0 the roots are real and distinct.
• For D = 0 the roots are real and equal.
• For D < 0 the real roots do not exist, or the roots are imaginary.
• The formula to find the roots of the quadratic equation is x = [-b ± √(b 2 - 4ac)]/2a.
• The sum of the roots of a quadratic equation is α + β = -b/a.
• The product of the Root of the quadratic equation is αβ = c/a.
• The quadratic equation whose roots are α, β, is x 2 - (α + β)x + αβ = 0.
• The condition for the quadratic equations a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0 having the same roots is (a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2 .
• When a > 0, the quadratic expression f(x) = ax 2 + bx + c has a minimum value at x = -b/2a.
• When a < 0, the quadratic expression f(x) = ax 2 + bx + c has a maximum value at x = -b/2a.
• The domain of any quadratic function is the set of all real numbers.

Methods to Solve Quadratic Equations

A quadratic equation can be solved to obtain two values of x or the two roots of the equation. There are four different methods to find the roots of the quadratic equation. The four methods of solving the quadratic equations are as follows.

• Factorizing of Quadratic Equation
• Using quadratic formula (which we have seen already)

Method of Completing the Square

• Graphing Method to Find the Roots

Let us look in detail at each of the above methods to understand how to use these methods, their applications, and their uses.

Solving Quadratic Equations by Factorization

Factorization of quadratic equation follows a sequence of steps. For a general form of the quadratic equation ax 2 + bx + c = 0, we need to first split the middle term into two terms, such that the product of the terms is equal to the constant term. Further, we can take the common terms from the available term, to finally obtain the required factors as follows:

• x 2 + (a + b)x + ab = 0
• x 2 + ax + bx + ab = 0
• x(x + a) + b(x + a)
• (x + a)(x + b) = 0

Here is an example to understand the factorization process.

• x 2 + 5x + 6 = 0
• x 2 + 2x + 3x + 6 = 0
• x(x + 2) + 3(x + 2) = 0
• (x + 2)(x + 3) = 0

Thus the two obtained factors of the quadratic equation are (x + 2) and (x + 3). To find its roots, just set each factor to zero and solve for x. i.e., x + 2 = 0 and x + 3 = 0 which gives x = -2 and x = -3. Thus, x = -2 and x = -3 are the roots of x 2 + 5x + 6 = 0.

Further, there is another important method of solving a quadratic equation. The method of completing the square for a quadratic equation is also useful to find the roots of the equation.

The method of completing the square in a quadratic equation is to algebraically square and simplify, to obtain the required roots of the equation. Consider a quadratic equation ax 2 + bx + c = 0, a ≠ 0. To determine the roots of this equation, we simplify it as follows:

• ax 2 + bx + c = 0
• ax 2 + bx = -c
• x 2 + bx/a = -c/a

Now, we express the left-hand side as a perfect square, by introducing a new term (b/2a) 2 on both sides:

• (x + b/2a) 2 = -c/a + b 2 /4a 2
• (x + b/2a) 2 = (b 2 - 4ac)/4a 2
• x + b/2a = + √(b 2 - 4ac)/2a
• x = - b/2a + √(b 2 - 4ac)/2a
• x = [-b ± √(b 2 - 4ac)]/2a

Here the '+' sign gives one root and the '-' sign gives another root of the quadratic equation. Generally, this detailed method is avoided, and only the quadratic formula is used to obtain the required roots.

Graphing a Quadratic Equation

The point(s) where the graph cuts the horizontal x-axis (typically the x-intercepts ) is the solution of the quadratic equation. These points can also be algebraically obtained by equalizing the y value to 0 in the function y = ax 2 + bx + c and solving for x.

Quadratic Equations Having Common Roots

Consider two quadratic equations having common roots a 1 x 2 + b 1 x + c 1 = 0, and a 2 x 2 + b 2 x + c 2 = 0. Let us solve these two equations to find the conditions for which these equations have a common root. The two equations are solved for x 2 and x respectively.

(x 2 )(b 1 c 2 - b 2 c 1 ) = (-x)/(a 1 c 2 - a 2 c 1 ) = 1/(a 1 b 2 - a 2 b 1 )

x 2 = (b 1 c 2 - b 2 c 1 ) / (a 1 b 2 - a 2 b 1 )

x = (a 2 c 1 - a 1 c 2 ) / (a 1 b 2 - a 2 b 1 )

Hence, by simplifying the above two expressions we have the following condition for the two equations having the common root.

(a 1 b 2 - a 2 b 1 ) (b 1 c 2 - b 2 c 1 ) = (a 2 c 1 - a 1 c 2 ) 2

Maximum and Minimum Value of Quadratic Expression

The maximum and minimum values of the quadratic expressions are of further help to find the range of the quadratic expression: The range of the quadratic expressions also depends on the value of a. For positive values of a( a > 0), the range is [ F(-b/2a), ∞), and for negative values of a ( a < 0), the range is (-∞, F(-b/2a)].

• For a > 0, Range: [ f(-b/2a), ∞)
• For a < 0, Range: (-∞, f(-b/2a)]

Note that the domain of a quadratic function is the set of all real numbers, i.e., (-∞, ∞).

Tips and Tricks on Quadratic Equation:

Some of the below-given tips and tricks on quadratic equations are helpful to more easily solve quadratic equations.

• The quadratic equations are generally solved through factorization. But in instances when it cannot be solved by factorization, the quadratic formula is used.
• The roots of a quadratic equation are also called the zeroes of the equation.
• For quadratic equations having negative discriminant values, the roots are represented by complex numbers.
• The sum and product of the roots of a quadratic equation can be used to find higher algebraic expressions involving these roots.

☛Related Topics:

• Roots Calculator
• Quadratic Factoring Calculator
• Roots of Quadratic Equation Calculator

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Quadratic Equations Examples

Example 1: James is a fitness enthusiast and goes for a jog every morning. The park where he jogs is rectangular in shape and measures 12m × 8m. An environmentalist group plans to revamp the park and decides to build a pathway surrounding the park. This would increase the total area to 140 sq m. What will be the width of the pathway?

Let’s denote the width of the pathway by x.

Then, the length and breadth of the outer rectangle is (12+2x) m and (8+2x) m.

Given that, area = 140

(12 + 2x)(8 + 2x) = 140

2(6 + x) 2(4 + x) = 140

(6 + x)(4 + x) = 35

24 + 6x + 4x + x 2 = 35

x 2 + 10x -11 = 0

x 2 + 11x - x - 11 = 0

x(x + 11) - 1(x + 11) = 0

(x + 11)(x - 1) = 0

(x + 11) =0 and (x - 1) = 0

x = -11 and x = 1

Since length can’t be negative, we take x = 1.

Answer: Therefore the width of the pathway is 1 m.

Example 2: Rita throws a ball upwards from a platform that is 20m above the ground. The height of the ball from the ground at a time 't', is denoted by 'h'. Suppose h = -4t 2 + 16t + 20. Find the maximum height attained by the ball.

We can rearrange the terms of the quadratic equation

h = -4t 2 + 16t + 20

in such a way that it is easy to find the maximum value of this equation.

= -4t 2 + 16t + 20

= -4(t 2 - 4t - 5)

= -4((t - 2) 2 - 9)

= -4(t - 2) 2 + 36

We should keep the value of (t - 2) 2 minimum in order to find the maximum value of h.

So, the minimum value (t - 2) 2 can take is 0.

Answer: Therefore the maximum height attained is 36m.

Example 3: Find the quadratic equation having the roots 5 and 8 respectively.

The quadratic equation having the roots α, β, is x 2 - (α + β)x + αβ = 0.

Given α = 5, and β = 8.

Therefore the quadratic equation is:

x 2 - (5 + 8)x + 5×8 = 0

x 2 - 13x + 40 = 0

Answer: Hence the required quadratic equation is x 2 - 13x + 40 = 0

Example 4: The quad equation 2x 2 + 9x + 7 = 0 has roots α, β. Find the quadratic equation having the roots 1/α, and 1/β.

The quadratic equation having roots that are reciprocal to the roots of the equation ax 2 + bx + c = 0, is cx 2 + bx + a = 0.

The given quadratic equation is 2x 2 + 9x + 7 = 0.

Hence the required equation having reciprocal roots is 7x 2 + 9x + 2 = 0.

Method 2: From the given equation,

α + β = -9/2 and α β = 2/7.

The new equation should have its roots to be 1/α and 1/β.

Their sum = 1/α + 1/β = (α + β) / α β = -9/7 Their product = 1/α β = 2/7 Thus, the required equation is, x 2 - (1/α + 1/β)x + 1/α β = 0 x 2 - (-9/7)x + 2/7 = 0 Multiplying both sides by 7, 7x 2 + 9x + 2 = 0

Answer: Therefore the equation is 7x 2 + 9x + 2 = 0.

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Practice Questions on Quadratic Equation

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FAQs on Quadratic Equation

What is the definition of a quadratic equation.

A quadratic equation in math is a second-degree equation of the form ax 2 + bx + c = 0. Here a and b are the coefficients, c is the constant term, and x is the variable. Since the variable x is of the second degree, there are two roots or answers for this quadratic equation. The roots of the quadratic equation can be found by either solving by factorizing or through the use of the quadratic formula.

What is the Quadratic Formula?

The quadratic equation formula to solve the equation ax 2 + bx + c = 0 is x = [-b ± √(b 2 - 4ac)]/2a. Here we obtain the two values of x, by applying the plus and minus symbols in this formula. Hence the two possible values of x are [-b + √(b 2 - 4ac)]/2a, and [-b - √(b 2 - 4ac)]/2a.

How do You Solve a Quadratic Equation?

There are several methods to solve quadratic equations, but the most common ones are factoring, using the quadratic formula, and completing the square.

• Factoring involves finding two numbers that multiply to equal the constant term, c, and add up to the coefficient of x, b.
• The quadratic formula is used when factoring is not possible, and it is given by x = [-b ± √(b 2 - 4ac)]/2a.
• Completing the square involves rewriting the quadratic equation in a different form that allows you to easily solve for x.

What is Determinant in Quadratic Formula?

The value b 2 - 4ac is called the discriminant and is designated as D. The discriminant is part of the quadratic formula. The discriminants help us to find the nature of the roots of the quadratic equation, without actually finding the roots of the quadratic equation.

What are Some Real-Life Applications of Quadratic Equations?

Quadratic equations are used to find the zeroes of the parabola and its axis of symmetry . There are many real-world applications of quadratic equations.

• They can be used in running time problems to evaluate the speed, distance or time while traveling by car, train or plane.
• Quadratic equations describe the relationship between quantity and the price of a commodity.
• Similarly, demand and cost calculations are also considered quadratic equation problems.
• It can also be noted that a satellite dish or a reflecting telescope has a shape that is defined by a quadratic equation.

How are Quadratic Equations Different From Linear Equations?

A linear degree is an equation of a single degree and one variable, and a quadratic equation is an equation in two degrees and a single variable. A linear equation is of the the form ax + b = 0 and a quadratic equation is of the form ax 2 + bx + c = 0. A linear equation has a single root and a quadratic equation has two roots or two answers. Also, a quadratic equation is a product of two linear equations.

What Are the 4 Ways To Solve A Quadratic Equation?

The four ways of solving a quadratic equation are as follows.

• Factorizing method
• Roots of Quadratic Equation Formula Method
• Method of Completing Squares
• Graphing Method

How to Solve a Quadratic Equation by Completing the Square?

The quadratic equation is solved by the method of completing the square and it uses the formula (a + b)^2 = a 2 + 2ab + b 2 (or) (a - b)^2 = a 2 - 2ab + b 2 .

How to Find the Value of the Discriminant?

The value of the discriminant in a quadratic equation can be found from the variables and constant terms of the standard form of the quadratic equation ax 2 + bx + c = 0. The value of the discriminant is D = b 2 - 4ac, and it helps to predict the nature of roots of the quadratic equation, without actually finding the roots of the equation.

How Do You Solve Quadratic Equations With Graphing?

The quadratic equation can be solved similarly to a linear equal by graphing. Let us take the quadratic equation ax 2 + bx + c = 0 as y = ax 2 + bx + c . Here we take the set of values of x and y and plot the graph. The two points where this graph meets the x-axis, are the solutions of this quadratic equation.

How Important Is the Discriminant of a Quadratic Equation?

The discriminant is very much needed to easily find the nature of the roots of the quadratic equation. Without the discriminant, finding the nature of the roots of the equation is a long process, as we first need to solve the equation to find both the roots. Hence the discriminant is an important and needed quantity, which helps to easily find the nature of the roots of the quadratic equation.

Where Can I Find Quadratic Equation Solver?

To get the quadratic equation solver, click here . Here, we can enter the values of a, b, and c for the quadratic equation ax 2 + bx + c = 0, then it will give you the roots along with a step-by-step procedure.

What is the Use of Discriminants in Quadratic Formula?

The discriminant (D = b 2 - 4ac) is useful to predict the nature of the roots of the quadratic equation. For D > 0, the roots are real and distinct, for D = 0 the roots are real and equal, and for D < 0, the roots do not exist or the roots are imaginary complex numbers . With the help of this discriminant and with the least calculations, we can find the nature of the roots of the quadratic equation.

How do you Solve a Quadratic Equation without Using the Quadratic Formula?

There are two alternative methods to the quadratic formula. One method is to solve the quadratic equation through factorization, and another method is by completing the squares. In total there are three methods to find the roots of a quadratic equation.

How to Derive Quadratic Formula?

The algebra formula (a + b) 2 = a 2 + 2ab + b 2 is used to solve the quadratic equation and derive the quadratic formula. This algebraic formula is used to manipulate the quadratic equation and derive the quadratic formula to find the roots of the equation.

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How to Solve a Quadratic Equation: A Step-by-Step Guide

Last Updated: September 6, 2024 Approved

Factoring the Equation

Using the quadratic formula, completing the square, practice problems and answers, expert q&a.

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 11 references cited in this article, which can be found at the bottom of the page. wikiHow marks an article as reader-approved once it receives enough positive feedback. This article has 16 testimonials from our readers, earning it our reader-approved status. This article has been viewed 1,437,198 times.

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

Quadradic Formula for Solving Equations

• Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. [4] X Research source

• 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

• So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.

• 4x 2 - 5x - 13 = x 2 -5
• 4x 2 - x 2 - 5x - 13 +5 = 0
• 3x 2 - 5x - 8 = 0

• {-b +/-√ (b 2 - 4ac)}/2
• {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)

• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6

• (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6

• x = (-1, 8/3)

• 2x 2 - 9 = 12x =
• In this equation, the a term is 2, the b term is -12, and the c term is -9.

• 2x 2 - 12x - 9 = 0
• 2x 2 - 12x = 9

• 2x 2 /2 - 12x/2 = 9/2 =
• x 2 - 6x = 9/2

• -6/2 = -3 =
• (-3) 2 = 9 =
• x 2 - 6x + 9 = 9/2 + 9

• x = 3 + 3(√6)/2
• x = 3 - 3(√6)/2)

• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [19] X Research source Thanks Helpful 0 Not Helpful 0
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 3 Not Helpful 0
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
• ↑ https://www.mathsisfun.com/algebra/factoring-quadratics.html
• ↑ https://flexbooks.ck12.org/cbook/ck-12-cbse-math-class-10/section/4.3/primary/lesson/solution-of-a-quadratic-equation-by-factorization/
• ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
• ↑ https://www.cuemath.com/algebra/quadratic-equations/
• ↑ https://www.purplemath.com/modules/solvquad4.htm
• ↑ https://www.purplemath.com/modules/quadform.htm
• ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
• ↑ https://pressbooks.bccampus.ca/algebraintermediate/chapter/solve-quadratic-equations-using-the-quadratic-formula/
• ↑ https://www.mathsisfun.com/algebra/completing-square.html
• ↑ https://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm

About This Article

To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No

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• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

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• High School Math Solutions – Quadratic Equations Calculator, Part 2 Solving quadratics by factorizing (link to previous post) usually works just fine. But what if the quadratic equation...

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Solving Quadratic Equations

Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

Step I: Denote the unknown quantities by x, y etc.

Step II: use the conditions of the problem to establish in unknown quantities.

Step III: Use the equations to establish one quadratic equation in one unknown.

Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

Now we will learn how to frame the equations from word problem:

1.  The product of two consecutive integers is 132. Frame an equation for the statement. What is the degree of the equation?

Method I: Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

Therefore, x(x + 1) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

Method II: Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

From (i), y = 1 + x.

Putting y = 1 + x in (ii),

x(1 + x) = 132

Solving the quadratic equation, we get the value of x. Then the value of y can be determined by substituting the value of x in y = 1 + x.

2. The length of a rectangle is greater than its breadth by 3m. If its area be 10 sq. m, find the perimeter.

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

⟹ x\(^{2}\) + 3x - 10 = 0

⟹ (x + 5)(x - 2) = 0

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m

and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = -5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring

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