How to Find the Solution Set of a Quadratic Inequality

College Algebra
Tutorial 23A: Quadratic Inequalities

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$desk$ Learning Objectives

After completing this tutorial, you should be able to:

Solve quadratic inequalities using a sign graph of factors.
Solve quadratic inequalities using the test-point method.

$desk$ Introduction

In this tutorial we will be looking at solving quadratic inequalities using two different methods. We will be revisiting solving quadratic equations to help solve the quadratic inequalities. If you need a review on solving quadratic equations, feel free to go to Tutorial 17: Quadratic Equations. I think we are ready to start.

$desk$ Tutorial

Quadratic Inequalities

A quadratic inequality is one that can be written in one of the following standard forms:

$quadratic$
or
$quadratic$
or
$quadratic$
or
$quadratic$

In other words, a quadratic inequality is in standard form when the inequality is set to 0.

Just like in a quadratic equation, the degree of the polynomial expression is two.

Solving Quadratic Inequalities
Using a Sign Graph of the Factors

This method of solving quadratic inequalities only works if the quadratic factors. If it doesn't factor then you will need to use the test-point method shown later on this page.

Step 1: Write the quadratic inequality in standard form.

It is VERY important that one side of the inequality is 0.

0 is our magic number. It is the only number that separates the negatives from the positives. If an expression is greater than 0, then there is no doubt that its sign is positive. Likewise, if it is less than 0, its sign is negative. You can not say this about any other number. Since we are working with inequalities, this idea will come in handy. With this technique we will be looking at the sign of a number to determine if it is a solution or not.

Step 2: Solve the quadratic equation, $quadratic equation$ , by factoring to get the boundary point(s).

The boundary point(s) will mark off where the quadratic expression is equal to 0. This is like the cross over point. 0 is neither positive or negative.

If you need a review on how to solve a quadratic equation, feel free to go to Tutorial 17: Quadratic Equations.

As mentioned above, this method of solving quadratic inequalities only works if the quadratic factors. If it doesn't factor then you will need to use the test-point method shown later on this page.

Step 3: Use the boundary points found in Step 2 to mark off test intervals on the number line and list all of the factors found in Step 2.

The boundary point(s) on the number will create test intervals.

Step 4: Find the sign of every factor in every interval.

You can choose ANY value in an interval to plug into each factor. Whatever the sign of the factor is with that value gives you the sign you need for that factor in that interval. Make sure that you find the sign of every factor in every interval.

Since the inequality will be set to 0, we are not interested in the actual value that we get when we plug in our test points, but what SIGN (positive or negative) that we get.

Step 5: Using the signs found in Step 4, determine the sign of the overall quadratic function in each interval.

Since the inequality will be set to 0, we are not interested in the actual value that we get when we plug in our test points, but what SIGN (positive or negative) that we get.

When you look at the signs of your factors in each interval, keep in mind that they represent a product of the factors that make up your overall quadratic function.

You determine the sign of the overall quadratic function by using basic multiplication sign rules:

The product of two factors that have the same sign is positive.

The product of two factors that have the opposite signs is negative.

If the quadratic expression is less than or less than or equal to 0, then we are interested in values that cause the quadratic expression to be negative.

If the quadratic expression is greater than or greater than or equal to 0, then we are interested in values that cause our quadratic expression to be positive .

Step 6: Write the solution set and graph.

$notebook$ Example 1: Solve using a sign graph of factors, write your answer in interval notation and graph the solution set: $example 1a$ .

$video$ View a video of this example

Step 1: Write the quadratic inequality in standard form.

This quadratic inequality is already in standard form.

Step 2: Solve the quadratic equation , $quadratic equation$ , by factoring to get the boundary point(s).

$example 1b$

*Set 1st factor = 0 and solve

*Set 2nd factor = 0 and solve

-5 and 3 are boundary points.

Below is a graph that marks off the boundary points -5 and 3 and shows the three sections that those points have created on the graph. Note that open holes were used on those two points since our original inequality did not include where it is equal to 0.

$example 1c$

Note that the two boundary points create three sections on the graph: $example 1d$ , $example 1e$ , and $example 1f$ .

You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get.

If we chose a number in the first interval, $example 1d$ , like -6 (I could have used -10, -25, or -10000 as long as it is in the interval), it would make both factors negative:

-6 + 5 = -1 and -6 - 3 = -9

If we chose a number in the second interval, $example 1e$ , like 0 (I could have used -4, -1, or 2 as long as it is in the interval), it would make

+ 5 positive and

- 3 negative:

If we chose a number in the third interval, $example 1f$ , like 4 (I could have used 10, 25, or 10000 as long as it is in the interval), it would make both factors positive:

$example 1k$

In the first interval, $example 1d$ , we have a negative times a negative, so the sign of the quadratic in that interval is positive.

In the second interval, $example 1e$ , we have a positive times a negative, so the sign of the quadratic in that interval is negative.

In the third interval, $example 1f$ , we have two positives, so the sign of the quadratic in that interval is positive.

Keep in mind that our original problem is $example 1a$ . Since we are looking for the quadratic expression to be LESS THAN 0 , that means we need our sign to be NEGATIVE .

$example 1l$

It looks like the only interval that this quadratic is negative is the second interval, $example 1e$ .

Interval notation: $example 1e$

Graph:
$example 1j$

*Open interval indicating all values between -5 and 3

*Visual showing all numbers between -5 and 3 on the number line

$notebook$ Example 2: Solve using a sign graph of factors, write your answer in interval notation and graph the solution set: $example 2a$ .

$video$ View a video of this example

Step 1: Write the quadratic inequality in standard form.

$example 2b$

*Inv. of sub. 6 x squared is add. 6 x squared

Step 2: Solve the quadratic equation , $quadratic equation$ , by factoring to get the boundary point(s).

$example 2c$

*Factor

*Set 1st factor = 0 and solve

*Set 2nd factor = 0 and solve

-5/2 and -1/3 are boundary points.

Below is a graph that marks off the boundary points -5/2 and -1/3 and shows the three sections that those points have created on the graph. Note that closed holes were used on those two points since our original inequality includes where it is equal to 0.

$example 2d$

Note that the two boundary points create three sections on the graph: $example 2e$ , $example 2f$ , and $example 2g$ .

You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get.

If we chose a number in the first interval, $example 2e$ , like -4 (I could have used -10, -25, or -10000 as long as it is in the interval), it would make both factors negative:

2(-4) + 5 = -3 and 3(-4) + 1 = -11

If we chose a number in the second interval, $example 2f$ , like -1 (I could have used -2, -3/2, or -1/2 as long as it is in the interval), it would make 2

+ 5 positive and 3

+ 1 negative:

2(-1) + 5 = 3 and 3(-1) + 1 = -2

If we chose a number in the third interval, $example 2g$ , like 0 (I could have used 10, 25, or 10000 as long as it is in the interval), it would make both factors positive:

2(0) + 5 = 5 and 3(0) + 1 = 1

$example 2o$

In the first interval, $example 2e$ , we have a negative times a negative, so the sign of the quadratic in that interval is positive.

In the second interval, $example 2f$ , we have a positive times a negative, so the sign of the quadratic in that interval is negative.

In the third interval, $example 2g$ , we have two positives, so the sign of the quadratic in that interval is positive.

Keep in mind that our inequality is $example 2m$ . Since we are looking for the quadratic expression to be GREATER THAN OR EQUAL TO 0 , that means we need our sign to be POSITIVE (OR O) .

$example 2n$

It looks like there are two intervals where the quadratic is positive, the first, $example 2e$ , and the third, $example 2g$ .

Interval notation: $example 2k$

Graph:
$example 2l$

*Closed intervals indicating all values less than or equal to -5/2 or greater than or equal to -1/3

*Visual showing all numbers less than or equal to -5/2 or greater than or equal to -1/3

Solving Quadratic Inequalities
Using the Test-Point Method

The test-point method for solving quadratic inequalities works for any quadratic that has a real number solution, whether it factors or not.

Step 1: Write the quadratic inequality in standard form.

It is VERY important that one side of the inequality is 0.

Step 2: Solve the quadratic equation, $quadratic equation$ , to get the boundary point(s).

The boundary point(s) will mark off where the quadratic expression is equal to 0. This is like the cross over point. 0 is neither positive or negative.

If you need a review on how to solve a quadratic equation, feel free to go to Tutorial 17: Quadratic Equations.

Step 3: Use the boundary point(s) found in step 2 to mark off test intervals on the number line.

The boundary point(s) on the number will create test intervals.

Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.

You can choose ANY point in an interval to represent it. You need to make sure that you test one point from each interval. Sometimes more than one interval can be part of the solution set.

Since the inequality will be set to 0, we are not interested in the actual value that we get when we plug in our test points, but what SIGN (positive or negative) that we get.

If the quadratic expression is less than or less than or equal to 0, then we are interested in values that cause the quadratic expression to be negative.

If the quadratic expression is greater than or greater than or equal to 0, then we are interested in values that cause our quadratic expression to be positive.

Step 5: Write the solution set and graph.

$notebook$ Example 3: Solve using the test-point method , write your answer in interval notation and graph the solution set: $example 3a$ .

$video$ View a video of this example

Step 1: Write the quadratic inequality in standard form.

This quadratic inequality is already in standard form.

Step 2: Solve the quadratic equation , $quadratic equation$ , to get the boundary point(s).

$example 3b$

*Set 1st factor = 0 and solve

*Set 2nd factor = 0 and solve

-2/5 and 0 are boundary points.

Below is a graph that marks off the boundary points -2/5 and 0 and shows the three sections that those points have created on the graph. Note that closed holes were used on those two points since our original inequality includes where it is equal to 0.

$example 3c$

Note that the two boundary points create three sections on the graph: $example 3d$ , $example 3e$ , and $example 3f$ .

Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.

You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get.

Keep in mind that our original problem is $example 3a$ . Since we are looking for the quadratic expression to be LESS THAN OR EQUAL TO 0 , that means we need our sign to be NEGATIVE (OR 0) .

From the interval $example 3d$ , I choose to use -1 to test this interval:
(I could have used -10, -25, or -10000 as long as it is in the interval)

$example 3g$

*Chose -1 from 1st interval to plug in for x

Since 3 is positive and we are looking for values that cause our quadratic expression to be less than or equal to 0 (negative or 0), $example 3d$ would not be part of the solution.

From the interval $example 3e$ , I choose to use -1/5 to test this interval.
(I could have used -1/6, -1/7, or -1/8 as long as it is in the interval)

$example 3h$

*Chose -1/5 from 2nd interval to plug in for x

Since -1/5 is negative and we are looking for values that cause our expression to be less than or equal to 0 (negative or 0), $example 3e$ would be part of the solution.

From the interval $example 3f$ , I choose 1 to use to test this interval.
(I could have used 10, 25, or 10000 as long as it is in the interval)

$example 3i$

*Chose 1 from 3rd interval to plug in for x

Since 7 is positive and we are looking for values that cause our quadratic expression to be less than or equal to 0 (negative or 0), $example 3f$ would not be part of the solution.

Interval notation: $example 3e$

Graph:
$example 3j$

*Closed interval indicating all values between -2/5 and 0, inclusive

*Visual showing all numbers between -2/5 and 0, inclusive

$notebook$ Example 4: Solve using the test-point method , write your answer in interval notation and graph the solution set: $example 4b$ .

$video$ View a video of this example

Step 1: Write the quadratic inequality in standard form.

$example 4b2$

*Inv. of sub. 2 is add. 2

�

Step 2: Solve the quadratic equation , $quadratic equation$ , to get the boundary point(s).

Since $example 4b15$ cannot be solved by factoring, how can we find the solution?

How about using the quadratic formula?

$example 4b3$

*Identify a , b , and c

*Quadratic formula

*Plug in values for a , b , and c into formula

$example 4b4$ and $example 4b5$ are boundary points.

Below is a graph that marks off the boundary points $example 4b4$ and $example 4b5$ and shows the three sections that those points have created on the graph. Note that open holes were used on those two points since our original inequality does not include where it is equal to 0.

$example 4b6$

Note that the two boundary points create three sections on the graph: $example 4b7$ , $example 4b8$ , and $example 4b9$ .

Step 4: Test a point in each test interval found in step 3 to see which interval(s) is part of the solution set.

You can choose ANY point in an interval to represent that interval. Remember that we are not interested in the actual value that we get, but what SIGN (positive or negative) that we get.

Keep in mind that our original problem is $example 4b10$ . Since we are looking for the quadratic expression to be GREATER THAN 0 , that means we need our sign to be POSITIVE.

From the interval $example 4b7$ , I choose to use 0 to test this interval:
(I could have used -10, -25, or -10000 as long as it is in the interval)
Note that $example 4b5$ is approximately .35.

$example 4b11$

*Chose 0 from 1st interval to plug in for x

Since 2 is positive and we are looking for values that cause our quadratic expression to be greater than 0, $example 4b7$ would be part of the solution.

$example 4b12$

*Chose 1 from 2nd interval to plug in for x

Since -3 is negative and we are looking for values that cause our expression to be greater than 0, $example 4b8$ would not be part of the solution.

From the interval $example 4b9$ , I choose 6 to use to test this interval.
(I could have used 10, 25, or 10000 as long as it is in the interval)
Note that $example 4b4$ is approximately 5.65.

$example 4b13$

*Chose 6 from 3rd interval to plug in for x

Since 2 is positive and we are looking for values that cause our quadratic expression to be greater than 0, $example 4b9$ would be part of the solution.

Interval notation: $example 4b7$ $example 4b16$ $example 4b9$

Graph:
$example 4b14$

*Open intervals indicating all values less than $example 4b5$ or greater than $example 4b4$ .

*Visual showing all numbers less than $example 4b5$ or greater than $example 4b4$ .

$desk$ Practice Problems

These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice.

To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.

$pencil$ Practice Problems 1a - 1b: Solve (using any method), write your answer in interval notation and graph the solution set.

$desk$ Need Extra Help on these Topics?

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Last revised on Dec. 30, 2009 by Kim Seward.
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How to Find the Solution Set of a Quadratic Inequality

Source: https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut23_quadineq.htm

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How to Find the Solution Set of a Quadratic Inequality

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