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• 1. Multiple Choice Edit 30 seconds 1 pt #1: What was the number sentence for your problem? 62-38=? 72-38=? 76-36= 37+?=73
• 2. Multiple Choice Edit 30 seconds 1 pt #2: What was the number sentence for your problem? 81+47= 36+73= 81-42= 73-36=
• 3. Multiple Choice Edit 30 seconds 1 pt #3: What was the number sentence for your problem? 81-47= 81-46= 47-81= 59-22=
• 4. Multiple Choice Edit 30 seconds 1 pt #4: What was the number sentence for your problem? 29+52 52+29 52-29= 52-49=
• 5. Multiple Choice Edit 30 seconds 1 pt #1: What was your answer? 24 34 37 23
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30 Fun Maths Questions with Answers

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Introduction

Mathematics can be fun if you treat it the right way. Maths is nothing less than a game, a game that polishes your intelligence and boosts your concentration. Compared to older times, people have a better and friendly approach to mathematics which makes it more appealing. The golden rule is to know that maths is a mindful activity rather than a task.

There is nothing like hard math problems or tricky maths questions, it’s just that you haven’t explored mathematics well enough to comprehend its easiness and relatability. Maths tricky questions and answers can be transformed into fun math problems if you look at it as if it is a brainstorming session. With the right attitude and friends and teachers, doing math can be most entertaining and delightful.

Math is interesting because a few equations and diagrams can communicate volumes of information. Treat math as a language, while moving to rigorous proof and using logical reason for performing a particular step in a proof or derivation.

Treating maths as a language totally eradicates the concept of hard math problems or tricky maths questions from your mind. Introducing children to fun maths questions can create a strong love and appreciation for maths at an early age. This way you are setting up the child’s successful future. Fun math problems will urge your child to choose to solve it over playing bingo or baking.

Apparently, there are innumerable methods to make easy maths tricky questions and answers. This includes the inception of the ideology that maths is simpler than their fear. This can be done by connecting maths with everyday life. Practising maths with the aid of dice, cards, puzzles and tables reassures that your child effectively approaches Maths.

If you wish to add some fun and excitement into educational activities, also check out

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Fun Maths Questions with answers - PDF

Here is the Downloadable PDF that consists of Fun Math questions. Click the Download button to view them.

Here are some fun, tricky and hard to solve maths problems that will challenge your thinking ability.

Answer: is 3, because ‘six’ has three letters

What is the number of parking space covered by the car?

This tricky math problem went viral a few years back after it appeared on an entrance exam in Hong Kong… for six-year-olds. Supposedly the students had just 20 seconds to solve the problem!

Believe it or not, this “math” question actually requires no math whatsoever. If you flip the image upside down, you’ll see that what you’re dealing with is a simple number sequence.

Replace the question mark in the above problem with the appropriate number.

Which number is equivalent to 3^(4)÷3^(2)

This problem comes straight from a standardized test given in New York in 2014.

There are 49 dogs signed up for a dog show. There are 36 more small dogs than large dogs. How many small dogs have signed up to compete? 

This question comes directly from a second grader's math homework.

To figure out how many small dogs are competing, you have to subtract 36 from 49 and then divide that answer, 13 by 2, to get 6.5 dogs, or the number of big dogs competing. But you’re not done yet! You then have to add 6.5 to 36 to get the number of small dogs competing, which is 42.5. Of course, it’s not actually possible for half a dog to compete in a dog show, but for the sake of this math problem let’s assume that it is.

Add 8.563 and 4.8292.

Adding two decimals together is easier than it looks. Don’t let the fact that 8.563 has fewer numbers than 4.8292 trip you up. All you have to do is add a 0 to the end of 8.563 and then add like you normally would.

I am an odd number. Take away one letter and I become even. What number am I?

Answer:  Seven (take away the ‘s’ and it becomes ‘even’).

Using only an addition, how do you add eight 8’s and get the number 1000?

Answer: 

888 + 88 + 8 + 8 + 8 = 1000

Sally is 54 years old and her mother is 80, how many years ago was Sally’s mother times her age?

41 years ago, when Sally was 13 and her mother was 39.

Which 3 numbers have the same answer whether they’re added or multiplied together?

There is a basket containing 5 apples, how do you divide the apples among 5 children so that each child has 1 apple while 1 apple remains in the basket?

4 children get 1 apple each while the fifth child gets the basket with the remaining apple still in it.

There is a three-digit number. The second digit is four times as big as the third digit, while the first digit is three less than the second digit. What is the number?

Fill in the question mark

Two girls were born to the same mother, at the same time, on the same day, in the same month and the same year and yet somehow they’re not twins. Why not?

Because there was a third girl, which makes them triplets!

A ship anchored in a port has a ladder which hangs over the side. The length of the ladder is 200cm, the distance between each rung in 20cm and the bottom rung touches the water. The tide rises at a rate of 10cm an hour. When will the water reach the fifth rung?

The tide raises both the water and the boat so the water will never reach the fifth rung. 

The day before yesterday I was 25. The next year I will be 28. This is true only one day in a year. What day is my Birthday?  

You have a 3-litre bottle and a 5-litre bottle. How can you measure 4 litres of water by using 3L and 5L bottles? 

Solution 1 :

First, fill 3Lt bottle and pour 3 litres into 5Lt bottle.

Again fill the 3Lt bottle. Now pour 2 litres into the 5Lt bottle until it becomes full.

Now empty 5Lt bottle.

Pour remaining 1 litre in 3Lt bottle into 5Lt bottle.

Now again fill 3Lt bottle and pour 3 litres into 5Lt bottle.

Now you have 4 litres in the 5Lt bottle. That’s it.

Solution 2 :

First, fill the 5Lt bottle and pour 3 litres into 3Lt bottle.

Empty 3Lt bottle.

Pour remaining 2 litres in  5Lt bottle into 3Lt bottle.

Again fill the 5Lt bottle and pour 1 litre into 3 Lt bottle until it becomes full.

3 Friends went to a shop and purchased 3 toys. Each person paid Rs.10 which is the cost of one toy. So, they paid Rs.30 i.e. total amount. The shop owner gave a discount of Rs.5 on the total purchase of 3 toys for Rs.30. Then, among Rs.5, Each person has taken Rs.1 and remaining Rs.2 given to the beggar beside the shop. Now, the effective amount paid by each person is Rs.9 and the amount given to the beggar is Rs.2. So, the total effective amount paid is 9*3 = 27 and the amount given to beggar is Rs.2, thus the total is Rs.29. Where has the other Rs.1 gone from the original Rs.30?

The logic is payments should be equal to receipts. We cannot add the amount paid by persons and the amount given to the beggar and compare it to Rs.30.The total amount paid is ₹27. So, from ₹27, the shop owner received 25 rupees and beggar received ₹ 2. Thus, payments are equal to receipts.

How to get a number 100 by using four sevens (7’s) and a one (1)?

Answer 1:   177 – 77 = 100 ;

Answer 2: (7+7) * (7 + (1/7)) = 100 

Move any four matches to get 3 equilateral triangles only (don’t remove matches)

Find the area of the red triangle.

To solve this fun maths question, you need to understand how the area of a parallelogram works. If you already know how the area of a parallelogram and the area of a triangle are related, then adding 79 and 10 and subsequently subtracting 72 and 8 to get 9 should make sense.

 How many feet are in a mile? 

Solve  - 15+ (-5x) =0

What is 1.92÷3

A man is climbing up a mountain which is inclined. He has to travel 100 km to reach the top of the mountain. Every day He climbs up 2 km forward in the day time. Exhausted, he then takes rest there at night time. At night, while he is asleep, he slips down 1 km backwards because the mountain is inclined. Then how many days does it take him to reach the mountain top? 

 If 72 x 96 = 6927, 58 x 87 = 7885, then 79 x 86 = ?

Answer:  

Look at this series: 36, 34, 30, 28, 24, … What number should come next?

  Look at this series: 22, 21, 23, 22, 24, 23, … What number should come next?

If 13 x 12 = 651 & 41 x 23 = 448, then, 24 x 22 =?

Look at this series: 53, 53, 40, 40, 27, 27, … What number should come next?

The ultimate goals of mathematics instruction are students understanding the material presented, applying the skills, and recalling the concepts in the future. There's little benefit in students recalling a formula or procedure to prepare for an assessment tomorrow only to forget the core concept by next week.

Teachers must focus on making sure that the students understand the material and not just memorize the procedures. After you learn the answers to a fun maths question, you begin to ask yourself how you could have missed something so easy. The truth is, most trick questions are designed to trick your mind, which is why the answers to fun maths questions are logical and easy. 

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• The equations section lets you solve an equation or system of equations. You can usually find the exact answer or, if necessary, a numerical answer to almost any accuracy you require.
• The inequalities section lets you solve an inequality or a system of inequalities for a single variable. You can also plot inequalities in two variables.
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• The graphs section contains commands for plotting equations and inequalities.
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Quadratic inequalities

Here you will learn about quadratic inequalities, including how to solve quadratic inequalities, identify solution sets using interval notation and represent solutions on a number line.

Students first learn about inequalities in 6 th grade with their work in solving linear inequalities. They expand that knowledge as they move through Algebra I and Algebra II.

What are quadratic inequalities?

Quadratic inequalities are mathematical statements that compare a quadratic expression to another value. Similar to linear inequalities, you can solve quadratic inequalities to find the range of values that makes the inequality statement true.

The strategies used to find the solutions of quadratic inequalities are the same strategies used to solve quadratic equations (factoring, using the quadratic formula, completing the square, etc).

Any quadratic equation has 0, 1 or 2 solutions only. Let’s visualize this using the quadratic equation y=x^{2}-4.

Step-by-step guide: Solving quadratic equations

Once you know the exact solutions to the quadratic equation, it is useful to plot these on a number line as shown above.

When solving any inequality, solutions are given in a range or interval, rather than an exact value.

For example, the inequality 4x>12 has the range of solutions x>3 as any value of x greater than 3 that is substituted into 4x will obtain a solution greater than 12. This can be shown on a number line:

You can see how by plotting the values for x on the number line, you divide the number line into two sets of numbers. Those that are greater than 3, and those that are less than or equal to 3. With a quadratic inequality, the number line can be divided into three sets of numbers.

Below are three alternative ways to determine the correct range of solutions for a quadratic inequality. Each refer to the example of y=x^{2}-4 outlined above.

The three alternatives are:

Visualizing solution sets using graphs

• Visualizing solution sets using the coefficient of x^2

Visualizing solution sets using substitution

[FREE] Inequalities Worksheet (Grade 1 to 7)

Use this quiz to check your grade 1 to 7 students’ understanding of inequalities. 10+ questions with answers covering a range of 1st – 7th grade inequalities topics to identify areas of strength and support!

Let’s review the example of the quadratic equation y=x^{2}-4 when y=0. The two solutions of x^{2}-4=0 are x=2 and x=- \, 2 (the points of intersection between the curve y=x^{2}-4 and the straight line y=0.

Now let’s look at the quadratic equation y=x^{2}-4 when y<0. This can be described as the quadratic inequality x^{2}-4<0.

The range of values for x that gives a solution for y<0 is between - \, 2 and 2. This is the interval (- \, 2,2). Representing this on a number line gives:

Notice that both circles are open-circles as the solutions are only less than 2 or greater than - \, 2, and not equal to.

Now let’s look at the quadratic equation y=x^{2}-4 when y\geq{0}. This can be described as the quadratic inequality x^{2}-4\geq{0}.

The range of values for x that gives a solution for y\geq{0} is less than or equal to - \, 2 and greater than or equal to 2. This is the set of solutions (- \, \infty,- \, 2]\cup[2,\infty) where is the union of the two intervals. Representing this on a number line gives:

Now the two circles are closed circles as the solutions are less than or equal to - \, 2 and greater than or equal to 2.

Visualizing solution sets using the coefficient of x²

Recall that when the coefficient of x^2 is positive, the parabola is a “ u -shape” and when the coefficient of x^2 is negative, the parabola is an “ n -shape”.

Let’s look back at the example of the quadratic equation y=x^{2}-4 when y=0. The two solutions of x^{2}-4=0 are x=2 and x=- \, 2 (the points of intersection between the curve y=x^{2}-4 and the straight line y=0.

As the quadratic y=x^{2}-4 has a positive coefficient of x^{2}\text{:}

• If y<0, then the range of solutions will be between two values. For example, x^{2}-4<0 so (- \, 2,2) .
• If y>0, then the range of solutions will be less than one value, and greater than another. For example, x^{2}-4>0 so (- \, \infty,- \, 2)\cup(2,\infty).

If the coefficient of x^{2} is negative, then the solution sets are opposite.

As the quadratic y=- \, x^{2}+4 has a negative coefficient of x^{2}\text{:}

• If y<0, then the range of solutions will be less than one value, and greater than another. For example, - \, x^{2}-4<0 so (- \, \infty,- \, 2)\cup(2,\infty).
• If y>0, then the range of solutions will be between two values. For example, - \, x^{2}-4>0 so (- \, 2,2) .

This is the same if the quadratic inequalities are also “less than or equal to”, and “greater than or equal to”.

The intervals of the two quadratics y=x^{2}-4 and y=- \, x^{2}+4 are given below when y>0 and when y<0.

You can use substitution to determine whether the solution is between two values, or the union of two intervals. Using a number line can help to visualize the correct interval(s).

Let’s look again at the example of the quadratic equation y=x^{2}-4 when y=0. The two solutions of x^{2}-4=0 are x=2 and x=- \, 2 (the points of intersection between the curve y=x^{2}-4 and the straight line y=0.

Now let’s look at the quadratic equation y=x^{2}-4 when y<0. This is the quadratic inequality x^{2}-4<0.

First, solve the quadratic equation x^{2}-4=0.

At this stage, it is not clear whether the solution is the interval (- \, 2,2) or (- \, \infty,- \, 2]\cup[2,\infty) so use substitution to find out which values of x satisfy the inequality x^{2}-4<0.

Substitute x=3 into x^{2}-4<0 to test the validity of the interval (- \, \infty,- \, 2]\cup[2,\infty).

5\nleq{0} and so this value is not within the valid range.

Conversely, substitute x=0 into x^{2}-4<0 to test its validity.

- \, 4\leq{0} and so this value is within the valid range.

The solution set is therefore (- \, 2,2).

Common Core State Standards

How does this relate to high school math?

• High School Algebra – Creating Equations (HSA-CED.A.1) Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions

How to solve quadratic inequalities

In order to solve quadratic inequalities:

Use a strategy to solve quadratic equations to find values of \textbf{x}.

Place interval limits on the number line.

Write the intervals formed by the values of \textbf{x}.

Select a test point from the bounded interval and substitute into the quadratic.

Determine which intervals represent the solution set to the inequality.

Solving quadratic equations examples

Example 1: closed points.

Determine the solution set to the quadratic inequality, x^{2}+4x+3\leq0.

This quadratic factors, so using the factoring method for solving quadratics, factor the quadratic to find the values of x.

The two solutions to the quadratic equation x^{2}+4x+3=0 are x=- \, 3 and x=- \, 1.

2 Place interval limits on the number line.

x=- \, 3 and x=- \, 1 are the limits to the solution set for x^{2}+4x+3\leq{0}. Since the inequality is “less than or equal to” use closed points to plot the values for x.

3 Write the intervals formed by the values of \textbf{x}.

The three intervals are:

4 Select a test point from the bounded interval and substitute into the quadratic.

Let x=- \, 2\text{:}

As - \, 1\leq{0}, this is a true comparison so the interval is part of solution.

5 Determine which intervals represent the solution set to the inequality.

The solution set to the quadratic inequality is [- \, 3,- \, 1].

Note: Graph with number line for reference.

Example 2: open points

Solve the quadratic inequality, x^2+7 x+10>0.

The quadratic can be factored so use the factoring strategy of solving quadratics to find the values of x.

\begin{aligned}&x^{2}+7x+10=0 \\\\ &(x+5)(x+2)=0 \\\\ &x+5=0\text{ so }x=- \, 5 \\\\ &\text{And }x+2=0\text{ so }x=- \,2 \end{aligned}

Since the inequality is “greater than”, use open points to plot the limits for x.

The intervals are:

\begin{array}{c}(- \, \infty,- \, 5) \\\\ (- \, 5,- \, 2) \\\\ (- \, 2,\infty) \end{array}

Let x=- \, 3\text{:}

x^{2}+7x+10=(- \, 3)^{2}+7(- \, 3)+10=9-21+10=- \, 2

As - \, 2\ngeq{0}, this is a false comparison so the interval is not part of solution.

The solution set to the quadratic inequality is (- \, \infty,- \, 5)\cup(- \, 2,\infty).

Example 3: rearrange quadratic

Solve the quadratic inequality, 5x^{2}-27x\leq{- \, 10}.

Before solving the quadratic for the x values, first rearrange the quadratic so that one side has a value of 0. In this case, add 10 to both sides of the inequality.

5x^{2}-27x+10\leq{0}

The quadratic can be factored to find the values of x. So use the strategy of factoring to find the values of x.

\begin{aligned}&5x^{2}-27x+10=0 \\\\ &(5x-2)(x-5)=0 \\\\ &5x-2=0\text{ so }x= \, \cfrac{2}{5} \\\\ &\text{And }x-5=0\text{ so }x=5 \end{aligned}

Since the inequality is “less than or equal to” use solid points to plot the limits of x.

Let x=1\text{:}

5x^{2}-27x+10=5(1)^{2}-27(1)+10=5-27+10=- \, 12

As - \, 12\leq{0}, this is a true comparison so the interval is part of solution.

The solution set to the quadratic inequality is \left[\cfrac{2}{5}, 5\right].

Example 4: two intervals in solution set

Find the solution to the quadratic inequality, 2x^{2}+4x+4>- \, 5x.

Before solving the quadratic for the x values, first rearrange the quadratic so that one side has a value of 0. In this case, add 5x to both sides of the inequality.

\begin{aligned}2x^{2}+4x+4&>- \, 5x \\\\ 2x^{2}+9x+4&>0 \end{aligned}

\begin{aligned}&2x^{2}+9x+4=0 \\\\ &(2x+1)(x+4)=0 \\\\ &2x+1=0\text{ so }x=- \, \cfrac{1}{2} \\\\ &\text{And }x+4=0\text{ so }x=- \, 4 \end{aligned}

Use open points to plot the values of x.

2x^{2}+9x+4=2(- \, 2)^{2}+9(- \, 2)+4=8-18+4=- \, 6

As - \, 6\ngtr{0}, this is a false comparison so the interval is not part of solution.

The solution set to the quadratic inequality is (- \, \infty,- \, 4)\cup\left(- \, \cfrac{1}{2},\infty\right).

Example 5: no c term

Find the solution to the quadratic inequality, 4x^2\geq- \, 28x.

Before solving, make sure one side of the equation is 0. In this case, add 28x to both sides.

\begin{aligned}&4x^2\geq- \, 28x \\\\ &4x^2+28x\geq{0} \end{aligned}

Now you can use one of the strategies for solving quadratics. This quadratic factors, so use that strategy.

\begin{aligned}&4x^{2}+28x=0 \\\\ &4x(x+7)=0 \\\\ &4x=0\text{ so }x=0 \\\\ &\text{And }x+7=0\text{ so }x=- \, 7 \end{aligned}

Use closed points to plot the numbers on the number line since the original inequality is “less than and equal to.”

\begin{array}{c}(- \, \infty,- \, 7] \\\\ {[- \, 7,0]} \\\\ {[0,\infty)} \end{array}

4x^{2}+28x=4(- \, 2)^{2}+28(- \, 2)=16-56=- \, 40

As - \, 40\ngeq{0}, this is a false comparison so the interval is not part of solution.

Since there are two intervals that represent the solution set to this quadratic inequality, use the union symbol in between intervals.

(- \, \infty,- \, 7]\cup[- \, 1,\infty)

Example 6: quadratic formula

Solve the quadratic inequality, 5x^2+6x-12<0.

The inequality is ready to solve, but notice that it does not factor. So, you can use the quadratic formula to find the values of x.

The general form of a quadratic equation is ax^2+bx+c=0.

The quadratic formula is used to solve any quadratic equation.

The quadratic formula is x=\cfrac{- \, b\pm\sqrt{b^2-4ac}}{2a}.

In this question, a=5, \, b=6, and c=- \, 12

Substitute those values into the quadratic formula to find the values for x\text{:}

\begin{aligned}x&=\cfrac{- \, 6\pm\sqrt{6^2-4\times5\times-12}}{2\times5} \\\\ &=\cfrac{- \, 6\pm\sqrt{36-(- \, 240)}}{10} \\\\ &=\cfrac{- \, 6\pm\sqrt{276}}{10} \\\\ x&=\cfrac{- \, 6+2\sqrt{69}}{10}\text{ and }x=\cfrac{- \, 6-2\sqrt{69}}{10} \end{aligned}

At this point, calculate decimals values rounded to the nearest hundredth.

x=\cfrac{- \, 6+2\sqrt{69}}{10}=1.06

x=\cfrac{- \, 6-2\sqrt{69}}{10}=- \, 2.26

Step-by-step guide: Quadratic formula

Use solid points when plotting the values of x on the number line.

\begin{array}{c}(- \, \infty,- \, 2.26) \\\\ (- \, 2.26,1.06) \\\\ (1.06,\infty) \end{array}

Let x=0\text{:}

5x^{2}+6x-12=5(0)^{2}+6(0)-12=0+0-12=- \, 12

The intervals that are the solution set are:

(- \, 2.26,1.06)

Teaching tips for quadratic inequalities

• Start the lesson by assessing if students can recall how to solve linear inequalities, including how to graph them on the number line. This will help guide the lesson.
• Infuse activities such as gallery walks so that students have an opportunity to showcase their work as well as have open dialogue with classmates about how to solve the problems.
• Instead of worksheets, have students practice skills by game-playing.
• Have access to the plot of each quadratic equation so that students can visualize the solution.

Easy mistakes to make

• Forgetting the negative value when taking the square root of a number A common error when calculating the square root of a number is only writing the positive solution. A negative number squared will also give a positive solution. For example: \sqrt{16}=4 and - \, 4.4\times{4}=16 and (- \, 4)\times(- \, 4)=16.
• Not using the original inequality symbol when testing points When substituting test points to check if the intervals work as the solution, substitute those values for x into the original inequality.
• Mixing up way to plot numbers in the number line For example, when the inequality is \leq or \geq, use solid points on the number line because the solution set includes those values of x. When the inequality is < or >, use open points on the number line because the solution set does not include the values for x.

Related quadratic inequalites lessons

• Inequalities
• Greater than sign
• Less than sign
• Linear inequalities
• Solving inequalities
• Graphing inequalities

Practice quadratic inequalities questions

1. Find the solution set to the quadratic inequality, x^{2}+13x+30<0.

The quadratic inequality factors, so use factoring to find the limit values for x.

Plot the values of x on the number line and determine the intervals and the test point you will use.

Let x=- \, 5\text{:}

– \,10<0 so this is a true comparison and this interval is part of the solution.

The solution is the interval: (- \, 10, – \, 3)

Note: Graph and number line for reference.

2. Solve the quadratic inequality, x^{2}+3x-10\leq{0}.

The quadratic inequality factors, so use factoring to find the values for x.

– \, 10\leq{0} so this is a true comparison and this interval is part of the solution.

The solution is: [- \, 5, 2]

3. Find the solution to the quadratic inequality, x^2+6>7 x.

First, make sure one side of the inequality has a value of 0. In this case, subtract 7x from both sides of the inequality.

The quadratic factors, so factor it to find the values of x.

Let x=2\text{:}

– \, 6\ngtr{0} so this is a false comparison and this interval is not part of the solution.

The solution is:

4. Find the solution to the quadratic inequality, 2x^{2}-11x+5<0.

The quadratic factors, so find the values of x by factoring.

– \, 9<0 so this is a true comparison and this interval is part of the solution.

The solution is \left(\cfrac{1}{2}, 5\right)

5. Find the solution to the quadratic inequality, x^2-25>0.

The quadratic is the difference of two perfect squares so it factors.

– \, 25\ngtr{0} so this is a false comparison and this interval is not part of the solution.

6. Find the solution to the quadratic inequality, 1-x^2\leq{0}.

Factor the quadratic to find the two limiting values of x\text{:}

– \, 1\nleq{0} so this is a false comparison and this interval is not part of the solution.

7. Solve the quadratic inequality: 3x^{2}+4x-2\leq0.

The quadratic does not factor, so use the quadratic formula to find the values for x.

Identify the values of a, b, and c and then substitute into the formula.

Substitute these values into the quadratic formula to find the values for x\text{:}

Plot the values of x on the number line with closed points.

– \, 2\leq{0} so this is a true comparison and this interval is part of the solution.

Quadratic inequalities FAQs

The discriminant is helpful when determining the type of solutions you will have for a quadratic equation. So, it can be helpful when solving a quadratic inequality.

Yes, you can sketch the parabola using a table of values and vertex point to determine which points make the quadratic inequality true and not true.

If the quadratic does not factor, you can use the quadratic formula or the strategy of completing the square. Remember, you can use the quadratic formula and completing the square regardless if the quadratic factors or not.

The next lessons are

• Math formulas
• Coordinate plane
• Types of graphs

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