quadratic inequalities homework

Solving Quadratic Inequalities

... and more ...

A Quadratic Equation (in Standard Form) looks like:

The above is an equation (=) but sometimes we need to solve inequalities like these:

Solving inequalities is very like solving equations ... we do most of the same things.

So this is what we do:

• find the "=0" points
• greater than zero (>0), or
• less than zero (<0)
• then pick a test value to find out which it is (>0 or <0)

Here is an example:

Example: x 2 − x − 6 < 0

x 2 − x − 6 has these simple factors (what luck!):

(x+2)(x−3) < 0

Firstly , let us find where it is equal to zero:

(x+2)(x−3) = 0

It is equal to zero when x = −2 or x = +3 because when x = −2, then (x+2) is zero or when x = +3, then (x−3) is zero

So between −2 and +3, the function will either be

• always greater than zero, or
• always less than zero

We don't know which ... yet!

Let's pick a value in-between (say x=0) and test it:

So between x=−2 and x=+3, the function is less than zero.

And that is the region we want, so...

x 2 − x − 6 < 0 in the interval (−2, 3)

Note: x 2 − x − 6 > 0   in the interval (−∞,−2) and (3, +∞)

Also try the Inequality Grapher .

What If It Doesn't Go Through Zero?

A "Real World" Example

A stuntman will jump off a 20 m building. a high-speed camera is ready to film him between 15 m and 10 m above the ground..

When should the camera film him?

We can use this formula for distance and time:

d = 20 − 5t 2

• d = distance above ground (m), and
• t = time from jump (seconds)

(Note: if you are curious about the formula, it is simplified from d = d 0 + v 0 t + ½a 0 t 2 , where d 0 =20 , v 0 =0 , and a 0 =−9.81 the acceleration due to gravity.)

OK, let's go.

First , let us sketch the question:

The distance we want is from 10 m to 15 m :

10 < d < 15

And we know the formula for d :

10 < 20 − 5t 2 < 15

Now let's solve it!

First, let's subtract 20 from both sides:

−10 < −5t 2 <−5

 Now multiply both sides by −(1/5). But because we are multiplying by a negative number, the inequalities will change direction ... read Solving Inequalities to see why.

2 > t 2 > 1

To be neat, the smaller number should be on the left, and the larger on the right. So let's swap them over (and make sure the inequalities still point correctly):

1 < t 2 < 2

 Lastly, we can safely take square roots, since all values are greater then zero:

√1 < t < √2

We can tell the film crew:

"Film from 1.0 to 1.4 seconds after jumping"

Higher Than Quadratic

The same ideas can help us solve more complicated inequalities:

Example: x 3 + 4 ≥ 3x 2 + x

First, let's put it in standard form:

x 3 − 3x 2 − x + 4 ≥ 0

This is a cubic equation (the highest exponent is a cube, i.e. x 3 ), and is hard to solve, so let us graph it instead:

The zero points are approximately :

And from the graph we can see the intervals where it is greater than (or equal to) zero:

• From −1.1 to 1.3, and
• From 2.9 on

In interval notation we can write:

Approximately: [−1.1, 1.3] U [2.9, +∞)

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Quadratic Inequalities Worksheets

Supercharge your high school students' solving skills with our printable quadratic inequalities worksheets. Represent the inequality as an equation, moving the terms to one side and equating it to zero, factor the equation and find the zeros to obtain break points or critical points, graph them on a number line, and determine the interval. Solving quadratic inequalities algebraically, graphically, and completing the table of signs, graphing the parabola and shading the solution region based on the inequality are the exercises presented in these pdfs. You now have the chance to test your skills with our free quadratic inequalities worksheets.

Printing Help - Please do not print quadratic inequalities worksheets directly from the browser. Kindly download them and print.

Solving Quadratic Inequalities Algebraically | Level 1

With either -1 or 1 as leading coefficients and integer break points, these printable quadratic inequalities worksheets get high school students transforming quadratic inequalities into quadratic equations and factoring them to solve the inequality.

• Download the set

Solving Quadratic Inequalities Algebraically | Level 2

Triumph in solving quadratic inequalities whose leading coefficients are integers by finding the factors, equating them to zero, determining their roots, and finding the solution range in this compilation of quadratic inequalities worksheet pdfs.

Solving Quadratic Inequalities Algebraically | Level 3

Instruct high school students to express the quadratic inequalities in standard form by moving the expressions on one side and equating to 0, factor them, determine their roots, to find the critical point and figure out the solution interval.

Solving Quadratic Inequalities | Table Method

Factor the quadratic inequalities and list out the intervals in the table of signs. Take a test value from each interval and apply on both the factors. Complete the table with the resultant plus or minus sign. Select the solution range that satisfies the inequality.

Graphing and Solving Inequalities

Check if the quadratic inequality is inclusive or strict. Graph the parabola y = f(x) for the quadratic inequality f(x) ≤ 0 or f(x) ≥ 0. Find the vertex and identify the values of x for which the part of the parabola will either be negative or positive depending on the inequalities.

Graphing Quadratic Inequalities

High school students plot x-intercepts, figure out the axis of symmetry and the vertex of the parabola, determine the direction, and illustrate the inequality using dotted or solid lines. Shade the parabola below or above the x-axis, inside or outside the parabola based on the solution.

Related Worksheets

» One Step Inequalities

» Two Step Inequalities

» Multi Step Inequalities

» Graphing Linear Equations

» Compound Inequalities

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To solve a quadratic inequality, follow these steps:

Solve the inequality as though it were an equation.

The real solutions to the equation become boundary points for the solution to the inequality.

Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

If a test point satisfies the original inequality, then the region that contains that test point is part of the solution.

Represent the solution in graphic form and in solution set form.

Solve ( x – 3)( x + 2) > 0.

Make the boundary points. Here, the boundary points are open circles because the original inequality does not include equality (see Figure 1).

Select points from the different regions created (see Figure 2).

See whether the test points satisfy the original inequality.

Since x = –3 satisfies the original inequality, the region x < –2 is part of the solution. Since x = 0 does not satisfy the original inequality, the region –2 < x < 3 is not part of the solution. Since x = 4 satisfies the original inequality, the region x > 3 is part of the solution.

Represent the solution in graphic form and in solution set form. The graphic form is shown in Figure 3.

The solution set form is { x | x < –2 or x > 3}.

Figure 1. Boundary points.

Figure 2. Three regions are created.

Figure 3. Solution to Example

Solve 9 x 2 – 2 ≤ –3 x .

Mark the boundary points using solid circles, as shown in Figure 4, since the original inequality includes equality.

Select points from the regions created (see Figure 5).

Represent the solution in graphic form and in solution set form. The graphic form is shown in Figure 6.

Figure 4. Solid dots mean inclusion.

Figure 5. Regions to test for Example

Figure 6. Solution to Example.

Solve 4 t 2 – 9 < –4 t .

Since this quadratic is not easily factorable, the quadratic formula is used to solve it.

Reduce by dividing out the common factor of 4.

Mark the boundary points using open circles, as shown in Figure 7, since the original inequality does not include equality.

Select points from the different regions created (see Figure 8).

Represent the solution in graphic form and in solution set form. The graphic form is shown in Figure 9.

Figure 7. Open dots mean exclusion.

Figure 8. Regions to test for Example.

Figure 9. Solution to Example.

Since this quadratic is not factorable using rational numbers, the quadratic formula will be used to solve it.

These are imaginary answers and cannot be graphed on a real number line. Therefore, the inequality x 2 + 2 x + 5 < 0 has no real solutions.

Previous Quiz: Solving Radical Equations

Next Quiz: Solving Quadratic Inequalities

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How to Solve Quadratic Inequalities

Last Updated: June 4, 2020 References

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 11 references cited in this article, which can be found at the bottom of the page. This article has been viewed 186,089 times.

Factoring the Inequality

Determining the Roots of the Inequality

Plotting the Solution Set on a Number Line

Plotting the Solution Set on a Coordinate Plane

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• ↑ http://www.mathsisfun.com/algebra/quadratic-equation.html
• ↑ http://www.mathwarehouse.com/dictionary/B-words/what-is-a-binomial.php
• ↑ https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratic-inequalities/v/quadratic-inequality-example-2
• ↑ http://www.purplemath.com/modules/ineqsolv.htm
• ↑ http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/inequalitiesrev4.shtml
• ↑ http://www.themathpage.com/aprecalc/roots-zeros-polynomial.htm
• ↑ http://www.virtualnerd.com/algebra-2/quadratics/inequalities/graphing-solving-inequalities/graph-inequality
• ↑ http://www.dummies.com/test-prep/act/act-trick-for-quadratics-how-to-quickly-find-the-direction-of-a-parabola/
• ↑ http://www.varsitytutors.com/hotmath/hotmath_help/topics/graphing-quadratic-inequalities
• ↑ https://www.khanacademy.org/math/algebra-home/alg-quadratics/alg-quadratic-inequalities/v/quadratic-inequalities-visual-explanation
• ↑ http://www.purplemath.com/modules/ineqquad.htm

About This Article

To solve a quadratic inequality, first write it as ax^2 + bx + c is less than 0. Then find 2 factors whose product is its first term and 2 factors whose product is its third term. Be sure the 2 factors whose product is its third term also have a sum that’s equal to its second term. Now determine whether your factors have the same or opposite signs by seeing if the product of the factors is greater or less than 0. Finally, turn each factor into an inequality, simplify, and check the validity of the roots for each option. If you want to learn how to show the solutions on a number line, keep reading the article! Did this summary help you? Yes No

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9.8 Solve Quadratic Inequalities

Learning objectives.

By the end of this section, you will be able to:

• Solve quadratic inequalities graphically
• Solve quadratic inequalities algebraically

Be Prepared 9.22

Before you get started, take this readiness quiz.

Solve: 2 x − 3 = 0 . 2 x − 3 = 0 . If you missed this problem, review Example 2.2 .

Be Prepared 9.23

Solve: 2 y 2 + y = 15 2 y 2 + y = 15 . If you missed this problem, review Example 6.45 .

Be Prepared 9.24

Solve 1 x 2 + 2 x − 8 > 0 1 x 2 + 2 x − 8 > 0 If you missed this problem, review Example 7.56 .

We have learned how to solve linear inequalities and rational inequalities previously. Some of the techniques we used to solve them were the same and some were different.

We will now learn to solve inequalities that have a quadratic expression. We will use some of the techniques from solving linear and rational inequalities as well as quadratic equations.

We will solve quadratic inequalities two ways—both graphically and algebraically.

Solve Quadratic Inequalities Graphically

A quadratic equation is in standard form when written as ax 2 + bx + c = 0. If we replace the equal sign with an inequality sign, we have a quadratic inequality in standard form.

Quadratic Inequality

A quadratic inequality is an inequality that contains a quadratic expression.

The standard form of a quadratic inequality is written:

The graph of a quadratic function f ( x ) = ax 2 + bx + c = 0 is a parabola. When we ask when is ax 2 + bx + c < 0, we are asking when is f( x ) < 0. We want to know when the parabola is below the x -axis.

When we ask when is ax 2 + bx + c > 0, we are asking when is f ( x ) > 0. We want to know when the parabola is above the x -axis.

Example 9.64

How to solve a quadratic inequality graphically.

Solve x 2 − 6 x + 8 < 0 x 2 − 6 x + 8 < 0 graphically. Write the solution in interval notation.

Try It 9.127

ⓐ Solve x 2 + 2 x − 8 < 0 x 2 + 2 x − 8 < 0 graphically and ⓑ write the solution in interval notation.

Try It 9.128

ⓐ Solve x 2 − 8 x + 12 ≥ 0 x 2 − 8 x + 12 ≥ 0 graphically and ⓑ write the solution in interval notation.

We list the steps to take to solve a quadratic inequality graphically.

Solve a quadratic inequality graphically.

• Step 1. Write the quadratic inequality in standard form.
• Step 2. Graph the function f ( x ) = a x 2 + b x + c . f ( x ) = a x 2 + b x + c .
• Step 3. Determine the solution from the graph.

In the last example, the parabola opened upward and in the next example, it opens downward. In both cases, we are looking for the part of the parabola that is below the x -axis but note how the position of the parabola affects the solution.

Example 9.65

Solve − x 2 − 8 x − 12 ≤ 0 − x 2 − 8 x − 12 ≤ 0 graphically. Write the solution in interval notation.

Try It 9.129

ⓐ Solve − x 2 − 6 x − 5 > 0 − x 2 − 6 x − 5 > 0 graphically and ⓑ write the solution in interval notation.

Try It 9.130

ⓐ Solve − x 2 + 10 x − 16 ≤ 0 − x 2 + 10 x − 16 ≤ 0 graphically and ⓑ write the solution in interval notation.

Solve Quadratic Inequalities Algebraically

The algebraic method we will use is very similar to the method we used to solve rational inequalities. We will find the critical points for the inequality, which will be the solutions to the related quadratic equation. Remember a polynomial expression can change signs only where the expression is zero.

We will use the critical points to divide the number line into intervals and then determine whether the quadratic expression willl be postive or negative in the interval. We then determine the solution for the inequality.

Example 9.66

How to solve quadratic inequalities algebraically.

Solve x 2 − x − 12 ≥ 0 x 2 − x − 12 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.131

Solve x 2 + 2 x − 8 ≥ 0 x 2 + 2 x − 8 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.132

Solve x 2 − 2 x − 15 ≤ 0 x 2 − 2 x − 15 ≤ 0 algebraically. Write the solution in interval notation.

In this example, since the expression x 2 − x − 12 x 2 − x − 12 factors nicely, we can also find the sign in each interval much like we did when we solved rational inequalities. We find the sign of each of the factors, and then the sign of the product. Our number line would like this:

The result is the same as we found using the other method.

We summarize the steps here.

Solve a quadratic inequality algebraically.

• Step 2. Determine the critical points—the solutions to the related quadratic equation.
• Step 3. Use the critical points to divide the number line into intervals.
• Step 4. Above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality.
• Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Example 9.67

Solve - x 2 + 6 x − 7 ≥ 0 - x 2 + 6 x − 7 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.133

Solve − x 2 + 2 x + 1 ≥ 0 − x 2 + 2 x + 1 ≥ 0 algebraically. Write the solution in interval notation.

Try It 9.134

Solve − x 2 + 8 x − 14 < 0 − x 2 + 8 x − 14 < 0 algebraically. Write the solution in interval notation.

The solutions of the quadratic inequalities in each of the previous examples, were either an interval or the union of two intervals. This resulted from the fact that, in each case we found two solutions to the corresponding quadratic equation ax 2 + bx + c = 0. These two solutions then gave us either the two x- intercepts for the graph or the two critical points to divide the number line into intervals.

This correlates to our previous discussion of the number and type of solutions to a quadratic equation using the discriminant.

For a quadratic equation of the form ax 2 + bx + c = 0, a ≠ 0 . a ≠ 0 .

The last row of the table shows us when the parabolas never intersect the x -axis. Using the Quadratic Formula to solve the quadratic equation, the radicand is a negative. We get two complex solutions.

In the next example, the quadratic inequality solutions will result from the solution of the quadratic equation being complex.

Example 9.68

Solve, writing any solution in interval notation:

ⓐ x 2 − 3 x + 4 > 0 x 2 − 3 x + 4 > 0 ⓑ x 2 − 3 x + 4 ≤ 0 x 2 − 3 x + 4 ≤ 0

We are to find the solution to x 2 − 3 x + 4 > 0 . x 2 − 3 x + 4 > 0 . Since for all values of x x the graph is above the x -axis, all values of x make the inequality true. In interval notation we write ( − ∞ , ∞ ) . ( − ∞ , ∞ ) .

Since the corresponding quadratic equation is the same as in part (a), the parabola will be the same. The parabola opens upward and is completely above the x -axis—no part of it is below the x -axis.

We are to find the solution to x 2 − 3 x + 4 ≤ 0 . x 2 − 3 x + 4 ≤ 0 . Since for all values of x the graph is never below the x -axis, no values of x make the inequality true. There is no solution to the inequality.

Try It 9.135

Solve and write any solution in interval notation: ⓐ − x 2 + 2 x − 4 ≤ 0 − x 2 + 2 x − 4 ≤ 0 ⓑ − x 2 + 2 x − 4 ≥ 0 − x 2 + 2 x − 4 ≥ 0

Try It 9.136

Solve and write any solution in interval notation: ⓐ x 2 + 3 x + 3 < 0 x 2 + 3 x + 3 < 0 ⓑ x 2 + 3 x + 3 > 0 x 2 + 3 x + 3 > 0

Section 9.8 Exercises

Practice makes perfect.

In the following exercises, ⓐ solve graphically and ⓑ write the solution in interval notation.

x 2 + 6 x + 5 > 0 x 2 + 6 x + 5 > 0

x 2 + 4 x − 12 < 0 x 2 + 4 x − 12 < 0

x 2 + 4 x + 3 ≤ 0 x 2 + 4 x + 3 ≤ 0

x 2 − 6 x + 8 ≥ 0 x 2 − 6 x + 8 ≥ 0

− x 2 − 3 x + 18 ≤ 0 − x 2 − 3 x + 18 ≤ 0

− x 2 + 2 x + 24 < 0 − x 2 + 2 x + 24 < 0

− x 2 + x + 12 ≥ 0 − x 2 + x + 12 ≥ 0

− x 2 + 2 x + 15 > 0 − x 2 + 2 x + 15 > 0

In the following exercises, solve each inequality algebraically and write any solution in interval notation.

x 2 + 3 x − 4 ≥ 0 x 2 + 3 x − 4 ≥ 0

x 2 + x − 6 ≤ 0 x 2 + x − 6 ≤ 0

x 2 − 7 x + 10 < 0 x 2 − 7 x + 10 < 0

x 2 − 4 x + 3 > 0 x 2 − 4 x + 3 > 0

x 2 + 8 x > − 15 x 2 + 8 x > − 15

x 2 + 8 x < − 12 x 2 + 8 x < − 12

x 2 − 4 x + 2 ≤ 0 x 2 − 4 x + 2 ≤ 0

− x 2 + 8 x − 11 < 0 − x 2 + 8 x − 11 < 0

x 2 − 10 x > − 19 x 2 − 10 x > − 19

x 2 + 6 x < − 3 x 2 + 6 x < − 3

−6 x 2 + 19 x − 10 ≥ 0 −6 x 2 + 19 x − 10 ≥ 0

−3 x 2 − 4 x + 4 ≤ 0 −3 x 2 − 4 x + 4 ≤ 0

−2 x 2 + 7 x + 4 ≥ 0 −2 x 2 + 7 x + 4 ≥ 0

2 x 2 + 5 x − 12 > 0 2 x 2 + 5 x − 12 > 0

x 2 + 3 x + 5 > 0 x 2 + 3 x + 5 > 0

x 2 − 3 x + 6 ≤ 0 x 2 − 3 x + 6 ≤ 0

− x 2 + x − 7 > 0 − x 2 + x − 7 > 0

− x 2 − 4 x − 5 < 0 − x 2 − 4 x − 5 < 0

−2 x 2 + 8 x − 10 < 0 −2 x 2 + 8 x − 10 < 0

− x 2 + 2 x − 7 ≥ 0 − x 2 + 2 x − 7 ≥ 0

Writing Exercises

Explain critical points and how they are used to solve quadratic inequalities algebraically.

Solve x 2 + 2 x ≥ 8 x 2 + 2 x ≥ 8 both graphically and algebraically. Which method do you prefer, and why?

Describe the steps needed to solve a quadratic inequality graphically.

Describe the steps needed to solve a quadratic inequality algebraically.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

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/intermediate-algebra-2e/pages/1-introduction

• Authors: Lynn Marecek, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/9-8-solve-quadratic-inequalities

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Quadratic Inequalities

Just like we solved and graphed Linear Inequalities , we can do the same with Quadratic Inequalities. There are also more types of inequalities that can be found here in the  Solving Inequalities section .

Graphing Quadratic Inequality Functions

We learned how to graph inequalities with two variables way back in the Coordinate System and Graphing Lines section. We can do the same with quadratics and the shading is pretty much the same: when we have (positive) “$ y<$ ”, we always shade in under the line that we draw, and when we have (positive) “$ y>$ ”, we always shade above the line that we draw. We can also see these on a graphing calculator.

Solving Quadratic Inequalities

You will also have to know how to solve quadratic inequalities, which make things a little messy. Examples of Quadratic Inequalities can be as “simple” as $ {{x}^{2}}>4$  or $ 4{{x}^{2}}\le 28x$ , or as complicated as $ 2{{x}^{2}}-7x\le -3$   or $ \displaystyle {{x}^{2}}+5x-9<0$ . We’ll use these as examples below.

Remember if you have a negative coefficient of $ \boldsymbol{{{x}^{2}}}$, you can move everything to the other side to make it positive – but be careful of the inequality signs! There are three main methods used to solve Quadratic Inequalities.  The Sign Pattern or Sign Chart Method is the most preferred, but I’ll cover a couple of methods here first.

Solving Using Graphing

You can solve quadratic inequalities by graphing the two sides of an inequality and seeing what the $ x$-intervals are for where one graph lies either below ($ <$) or above ($ >$) the other one.

Solving Algebraically, including Completing the Square

When solving algebraically, we can take the square root of each side, but we need to pay attention to the inequality sign. We need to break the equation into two equations like we did in the Solving Absolute Value Inequalities section (one with a plus, one with a minus), but the equation with the minus must have an inequality sign change. The reason we’re breaking up the inequality into two equations is a square root may be either positive or negative .

After solving both inequalities, put the solutions together in interval notation. Remember that with >, we have an “or” and with < we have an “and”.

Sign Chart (Sign Pattern) Method – the Easiest Method!

The sign chart or sign pattern method is used quite frequently in Algebra, including here in Solving Polynomial Inequalities from the Graphing and Finding Roots of Polynomial Functions section , and here in Rational Inequalities from the  Graphing Rational Functions, including Asymptotes section . It looks difficult at first, but really isn’t too bad at all!

A sign chart or sign pattern is a number line that is separated into partitions, or intervals or regions, with boundary points, that are called “ critical values “. You obtain the critical values by setting the quadratic to 0 and solving for $ x$ (the roots).

The idea of sign charts is to pick any point in between the critical values , and see if the whole quadratic is positive or negative , and then use that information to get the solution to the inequality.

It’s a good idea to put open or closed circles  on the  critical values  to remind ourselves if we have   inclusive  points (inequalities with equal signs, such as  $ \le $ and $ \ge $) or  exclusive  points (inequalities without equal signs, or  factors in the denominators ).

Real World Quadratic Inequality Example

Learn these rules, and practice, practice, practice!

Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the  Mathway  site, where you can register for the  full version  (steps included) of the software.  You can even get math worksheets.

You can also go to the  Mathway  site here , where you can register, or just use the software for free without the detailed solutions.  There is even a Mathway App for your mobile device.  Enjoy!

On to Quadratic Applications – you are ready!

• Mathematicians
• Math Lessons
• Square Roots
• Math Calculators
• Quadratic Inequalities – Explanation & Examples

Solving Quadratic Inequalities – Explanation & Examples

A quadratic inequality is an equation of second degree that uses an inequality sign instead of an equal sign.

The solutions to quadratic inequality always give the two roots. The nature of the roots may differ and can be determined by discriminant (b 2 – 4ac).

The general forms of the quadratic inequalities are:

ax 2 + bx + c < 0

ax 2 + bx + c ≤ 0

ax 2 + bx + c > 0

ax 2 + bx + c ≥ 0

Examples of quadratic inequalities are:

x 2 – 6x – 16 ≤ 0, 2x 2  – 11x + 12 > 0, x 2  + 4 > 0, x 2  – 3x + 2 ≤ 0 etc.

How to Solve Quadratic Inequalities?

Examples of quadratic inequalities are: x 2 – 6x – 16 ≤ 0, 2x 2  – 11x + 12 > 0, x 2  + 4 > 0, x 2  – 3x + 2 ≤ 0 etc.

Solving a quadratic inequality in Algebra is similar to solving a quadratic equation. The only exception is that, with quadratic equations, you equate the expressions to zero, but with inequalities, you’re interested in knowing what’s on either side of the zero i.e. negatives and positives.

Quadratic equations can be solved by either the factorization method or by use of the quadratic formula . Before we can learn how to solve quadratic inequalities, let’s recall how quadratic equations are solved by handling a few examples.

How Quadratic Equations are Solved by Factorization Method?

Since we know we can similarly solve quadratic inequalities as quadratic equations, it is useful to understand how to factorize the given equation or inequality.

Let’s see a few examples here.

• 6x 2 – 7x + 2 = 0

⟹ 6x 2  – 4x – 3x + 2 = 0

Factorize the expression;

⟹ 2x (3x – 2) – 1(3x – 2) = 0

⟹ (3x – 2) (2x – 1) = 0

⟹ 3x – 2 = 0 or 2x – 1 = 0

⟹ 3x = 2 or 2x = 1

⟹ x = 2/3 or x = 1/2

Therefore, x = 2/3, ½

• Solve 3x 2 – 6x + 4x – 8 = 0

Factorize the expression on the left-hand side.

⟹ 3x 2  – 6x + 4x – 8 = 0

⟹ 3x (x – 2) + 4(x – 2) = 0

⟹ (x – 2) (3x + 4) = 0

⟹ x – 2 = 0 or 3x + 4 = 0

⟹ x = 2 or x = -4/3

Therefore, the roots of the quadratic equation are, x = 2, -4/3.

• Solve 2(x 2 + 1) = 5x

2x 2 + 2 = 5x

⟹ 2x 2  – 5x + 2 = 0

⟹ 2x 2  – 4x – x + 2 = 0

⟹ 2x (x – 2) – 1(x – 2) = 0

⟹ (x – 2) (2x – 1) = 0

⟹ x – 2 = 0 or 2x – 1 = 0

⟹ x = 2 or x = 1/2

Therefore, the solutions are x = 2, 1/2.

• (2x – 3) 2 = 25

Expand and factorize the expression.

(2x – 3) 2  = 25

⟹ 4x 2  – 12x + 9 – 25 = 0

⟹ 4x 2  – 12x – 16 = 0

⟹ x 2  – 3x – 4 = 0

⟹ (x – 4) (x + 1) = 0

⟹ x = 4 or x = -1

• Solve x 2 + (4 – 3y) x – 12y = 0

Expand the equation;

x 2  + 4x – 3xy – 12y = 0

⟹ x (x + 4) – 3y (x + 4) = 0

x + 4) (x – 3y) = 0

⟹ x + 4 = 0 or x – 3y = 0

⟹ x = -4 or x = 3y

Thus, x = -4 or x = 3y

To solve a quadratic inequality, we also apply the same method as illustrated in the procedure below:

• Write the quadratic inequality in standard form: ax 2  + bx + c where a, b and are coefficients and a ≠ 0
• Determine the roots of the inequality.
• Write the solution in inequality notation or interval notation.

Solve the inequality x 2  – 4x > –3

First, make one side one side of the inequality zero by adding both sides by 3.

x 2  – 4x > –3 ⟹ x 2  – 4x + 3 > 0

Factor the left side of the inequality.

x 2  – 4x + 3 > 0 ⟹ (x – 3) (x – 1) > 0

Solve for all the zeroes for the inequality;

For, (x – 1) > 0 ⟹ x > 1 and for, (x – 3) > 0 ⟹ x>3

Since y is positive, we therefore choose the values of x which the curve will be above the x-axis. x < 1 or x > 3

Solve the inequality x 2  – x > 12.

To write the inequality in standard form, subtract both sides of the inequality by 12.

x 2  – x > 12 ⟹ x 2  – x – 12 > 0.

Factorize the quadratic inequality to get to;

( x  – 4) ( x  + 3) > 0

For, (x + 3) > 0 ⟹ x > -3

For x – 4 > 0 ⟹ x > 4

The values x < –3 or x > 4 are therefore the solution of this quadratic inequality.

Solve 2x 2  < 9x + 5

Write the inequality in standard form by making one side of the inequality zero.

2x 2  < 9x + 5 ⟹ 2x 2  – 9x – 5 < 0

Factor the left side of the quadratic inequality.

2x 2  – 9x – 5 < 0 ⟹ (2x + 1) (x – 5) < 0

Solve for all the zeroes for the inequality

For, (x – 5) < 0 ⟹ x < 5 and for (2x + 1) < 0 ⟹ x < -1/2

Since y is negative for the equation 2x 2  – 9x – 5 < 0, we therefore choose the values of x which the curve will be below the x axis.

Therefore, the solution is   -1/2 < x < 5

Solve – x 2 + 4 < 0.

Since the inequality is already in standard form, we therefore factor the expression.

-x 2 + 4 < 0 ⟹ (x + 2) (x – 2) < 0

For, (x + 2) < 0 ⟹ x < -2 and for, (x – 2) < 0 ⟹ x < 2

The y for –x 2 + 4 < 0 is negative; therefore, we choose the values of x in which the curve will below the x- axis: –2 < x > 2

Solve 2x 2  + x − 15 ≤ 0.

Factor the quadratic equation.

2x 2  + x − 15 = 0

2x 2  + 6x – 5x− 15 = 0

2x (x + 3) – 5(x + 3) = 0

(2x – 5) (x + 3) = 0

For, 2x – 5 = 0 ⟹ x= 5/2 and for, x + 3= 0 ⟹ x = -3

Since the y for 2x 2  + x − 15 ≤ 0 is negative, the we choose the values of x in which the curve will be below the x axis. Therefore, x ≤ -3 or x ≥5/2 is the solution.

Solve – x 2  + 3x − 2 ≥ 0

Multiply the quadratic equation by -1 and remember to change the sign.

x 2  – 3x + 2 = 0

x 2  – 1x – 2x + 2 = 0

x (x – 1) – 2(x – 1) = 0

(x – 2) (x – 1) = 0

For, x – 2 = 0 ⟹ x = 2 and for, x – 1= 0 ⟹x=1

Therefore, the solution to the quadratic inequality is 1 ≤ x ≤ 2

Solve x 2 − 3x + 2 > 0

Factorize the expression to get;

x 2 − 3x + 2 > 0 ⟹ (x − 2) (x − 1) > 0

Now solve for the roots of the inequality as;

(x − 2) > 0 ⟹ x > 2

(x − 1) > 0 ⟹x > 1

The curve for x 2 − 3x + 2 > 0 has positive y, therefore which choose the values of x in which the curve will be above the x-axis. The solution is hence, x < 1 or x > 2.

Solve −2x 2 + 5x + 12 ≥ 0

Multiply the entire expression by -1 and change the inequality sign

−2x 2 + 5x + 12 ≥ 0 ⟹2x 2 − 5x − 12 ≤ 0

(2x + 3) (x − 4) ≤ 0.

Solve the roots;

(2x + 3) ≤ 0 ⟹ x ≤ -3/2.

(x − 4) ≤ 0 ⟹ x ≤ 4.

By applying the rule; (x – a) (x – b) ≥ 0, then a ≤ x ≤ b, we can comfortably write the solutions of this quadratic inequality as:

-3/2 ≤ x ≤ 4.

x 2  − x − 6 < 0

Factorize x 2  − x − 6 to get;

(x + 2) (x − 3) < 0

Find the roots of the equation as;

(x + 2) (x − 3) = 0

x = −2 or x = +3 Because y is negative for x 2  − x − 6 < 0, then we choose an interval in which the curve will be below the x axis. Therefore, -2 < x < 3 is the solution.

Practice Questions

Previous lesson  |  main page | next lesson.

Quadratic Inequalities

A quadratic inequality or inequality of degree two is a second-degree polynomial inequality in one variable. The standard form of a quadratic inequality presents the same organization of terms as a quadratic equation , but instead of the equal sign, it presents the signs \(>\), \(<\), \(\geq\), \(\leq\).

A quadratic inequality is of the form:

\[ax^2 + bx + c \gt 0 \quad \text{where} \quad a \neq 0\]

• The coefficients \(a\), \(b\), and \(c\) are constants and \(x\) represents the variable.
• \(a\) is the coefficient of the quadratic term \(x^2\), \(b\) the coefficient of the linear term \(x\) and \(c\) the constant term.
• When \(a\) is equal to zero, the equation degenerates into a linear equation \(bx + c = 0\), or a constant equation, depending on the values of \(b\) and \(c\).

Resolution method

The first step in solving a quadratic inequality is to find solutions to the corresponding second-degree equation. Then, based on the sign of the inequality, the range of values that satisfy it can be determined. Two widely used methods exist to obtain the solutions of the quadratic equation: the quadratic formula and the factorization method .

The quadratic formula provides solutions to any quadratic equation in a mechanical way, while the factorization method involves factoring the quadratic equation to obtain its solutions. Both methods are practical and widely used for finding equation solutions.

Given an inequality in standard form \(ax^2 +bx +c \gt 0\) or \(ax^2 +bx +c \lt 0\), let’s translate it into the associated equation by setting the polynomial equal to zero. Thus, we obtain a second-degree equation in the form: \[ax^2 +bx +c = 0\]

Following the solution of a second-degree equation, the subsequent step involves determining the range of values that satisfy the inequality. This range must be determined to ensure that the inequality is valid for all values within the specified range. This process is crucial as it allows for the accurate analysis and interpretation of solutions.

Solutions when \(\Delta \geq 0\)

Quadratic inequalities, unlike equations, typically yield a solution set consisting of a range of values, determined based on the sign of the inequality. Given \(\Delta \geq 0\), the discriminant of the quadratic formula , and \(x_1, x_2\) the solutions of the second-degree equation associated with the inequality, we have the following cases:

When the inequality is of the form \(ax^2+bx+c \geq 0 \) or \(ax^2+bx+c \gt 0 \), and the solutions of the associated equation are of the form \(x_1 \neq x_2\), assuming for simplicity’s sake that \(x_1 \lt x_2\), we have:

• \(x_1 \leq x \lor x \geq x_2 \; \) for \( \; ax^2+bx+c \geq 0 \)
• \(x_1 \lt x \lor x \gt x_2 \; \) for \( \; ax^2+bx+c \gt 0 \)

When the inequality is of the form \(ax^2 +bx +c \leq 0 \) or \(ax^2 +bx +c \lt 0 \), and the solutions of the associated equation are of the form \(x_1 \neq x_2\), assuming for simplicity’s sake that \(x_1 \lt x_2\), we have:

• \(x_1 \leq x \leq x_2 \;\) for \( \; ax^2 +bx +c \leq 0 \)
• \(x_1 \lt x \lt x_2 \;\) for \( \; ax^2 +bx +c \lt 0 \)

When the inequality is of the form \(ax^2+bx+c \geq 0 \) or \(ax^2+bx+c \gt 0 \), and the solution of the associated equation is of the form \(x_1 = x_2\), we have:

• \(\forall x \; \) for \( \;ax^2+bx+c \geq 0 \)
• \(\forall x \neq x_1 \; \) for \( \; ax^2+bx+c \gt 0 \),

When the inequality is of the form \(ax^2+bx+c \leq 0 \) or \(ax^2+bx+c \lt 0 \), and the solution of the associated equation is of the form \(x_1 = x_2\), we have:

• \(\not\exists \, x \; \) for \( \; ax^2+bx+c \leq 0 \)
• \(x=x_1 \; \) for \( \; ax^2+bx+c \lt 0 \)

Solutions when \(\Delta \lt 0\)

Given \(\Delta \lt 0 \), the discriminant of the quadratic formula , and \(x_1, x_2\) the solutions of the second-degree equation associated with the inequality, we have the following cases:

When the inequality is of the form \(ax^2+bx+c \geq 0 \) or \(ax^2+bx+c \gt 0 \) we have for both cases: \(\forall x\)

When the inequality is of the form \(ax^2+bx+c \leq 0 \) or \(ax^2+bx+c \lt 0 \) we have for both cases: \(\not\exists \; x\)

Solve the quadratic inequality: \[2x^2 +5x – 3 \gt 0 \]

As illustrated above, the first fundamental step is to move to the associated quadratic equation by setting the polynomial equal to zero:

\[2x^2+5x-3 = 0 \]

Now we use the quadratic formula to find the solutions of the equation.

\[ \begin{align} x_{1,2} &= \frac{-5 \pm \sqrt{5^2 -4\cdot(2)\cdot(-3)}}{2\cdot 2} \\[1em] &= \frac{-5 \pm \sqrt{25 +24}}{4} \\[1em] &= \frac{-5 \pm 7}{4} \\[1em] \end{align}\]

We obtain: \[x_1 = \frac{2}{4} \to \frac{1}{2} \] \[x_2 = – \frac{12}{4} \to -3 \]

Now determine the range of solutions to the inequality. We can use the graphical method to determine it visually. The inequality is of the form \(ax^2 +bx +c \gt 0\). We have:

The solution of the inequality is: \[-3 \lt x \lor x \gt 2 \quad \text{or} \quad x \in (-\infty, -3] \cup [2, +\infty)\]

If the inequality had been of the form \(2x^2 +5x – 3 \lt 0\) we would have had:

In this case, the range of values to be considered is \(-3 \lt x \lt 2\) and the solution to the equation is \(x \in [3,2]\).

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MATH 1111 - College Algebra: 3.2 Quadratic Functions and Quadratic Inequalities

• 1.1 Sets and Set Operations
• 1.2 Linear Equations and Inequalities
• 1.3 Systems of Linear Equations
• 1.4 Polynomials; Operations with Polynomials
• 1.5 Factoring Polynomials
• 1.6 Quadratic Equations
• 1.7 Rational Expressions and Equations
• 1.8 Complex Numbers
• 2.1 Cartesian Coordinates/Relations
• 2.2 Intro to Functions
• 2.3 Operations with Functions
• 2.4 Graph of Functions
• 3.1 Linear Functions

3.2 Quadratic Functions and Quadratic Inequalities

• 4.1 Finding Zeros of Polynomial Functions
• 4.2 Graphing Polynomial Functions
• 4.3 Rational Functions
• 4.4 Rational Inequalities
• 5.1 Composition of Functions
• 5.2 Inverse Functions
• 5.3 Introduction to Exponential and Logarithmic Functions

At the end of this section students will be able to:

• Determine the general and standard form of a quadratic function
• Graph a quadratic function using its vertex and x-and y-intercepts
• Graph a quadratic function using transformation techniques
• Solve applications using quadratic functions
• Solve quadratic inequalities

Required Reading

2.3 Quadratic Functions

Stitz-Zeager College Algebra  - pages  188-199

2.4 Inequalities with Absolute Value and Quadratic Functions    

Stitz-Zeager College Algebra  - pages  208-217

Practice Exercises

Quadratic Functions

Stitz-Zeager College Algebra  - pages 200-201

Answers to practice exercises can be found on pages 203-207.

Quadratic Inequalities

Stitz-Zeager College Algebra  - pages 220

Answers to practice exercises can be found on page 222.

Supplemental Resources

Graphs of Quadratic Functions  (tutorial):  West Texas A&M University Virtual Math Lab (College Algebra Tutorial 34)

Quadratic Inequalities (tutorial):  West Texas A&M University Virtual Math Lab (College Algebra Tutorial 23A)

Find the Standard Form (Vertex Form) and the Vertex of a Quadratic Function:

Graphing a Parabola Using its Vertex and x-intercepts:

Graphing Quadratic Functions Using Transformations Techniques:

Solving Quadratic Inequalities: 

Solving Applications of Quadratic Functions: 

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• Next: Module 4 >>
• Last Updated: Apr 2, 2024 2:52 PM
• URL: https://libguides.gcsu.edu/math1111