problem solving involving quadratic inequality

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?

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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 180,880 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

Community Q&A

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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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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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6.5 Solving Quadratic Inequalities

Learning objectives.

• Check solutions to quadratic inequalities with one variable.
• Understand the geometric relationship between solutions to quadratic inequalities and their graphs.
• Solve quadratic inequalities.

Solutions to Quadratic Inequalities

A quadratic inequality A mathematical statement that relates a quadratic expression as either less than or greater than another. is a mathematical statement that relates a quadratic expression as either less than or greater than another. Some examples of quadratic inequalities solved in this section follow.

A solution to a quadratic inequality is a real number that will produce a true statement when substituted for the variable.

Are −3, −2, and −1 solutions to x 2 − x − 6 ≤ 0 ?

Substitute the given value in for x and simplify.

x 2 − x − 6 ≤ 0 x 2 − x − 6 ≤ 0 x 2 − x − 6 ≤ 0 ( − 3 ) 2 − ( − 3 ) − 6 ≤ 0 ( − 2 ) 2 − ( − 2 ) − 6 ≤ 0 ( − 1 ) 2 − ( − 1 ) − 6 ≤ 0 9 + 3 − 6 ≤ 0 4 + 2 − 6 ≤ 0 1 + 1 − 6 ≤ 0 6 ≤ 0 ✗ 0 ≤ 0 ✓ − 4 ≤ 0 ✓

Answer: −2 and −1 are solutions and −3 is not.

Quadratic inequalities can have infinitely many solutions, one solution, or no solution. If there are infinitely many solutions, graph the solution set on a number line and/or express the solution using interval notation. Graphing the function defined by f ( x ) = x 2 − x − 6 found in the previous example we have

The result of evaluating for any x -value will be negative, zero, or positive.

f ( − 3 ) = 6 P o s i t i v e f ( x ) > 0 f ( − 2 ) = 0 Z e r o f ( x ) = 0 f ( − 1 ) = − 4 N e g a t i v e f ( x ) < 0

The values in the domain of a function that separate regions that produce positive or negative results are called critical numbers The values in the domain of a function that separate regions that produce positive or negative results. . In the case of a quadratic function, the critical numbers are the roots, sometimes called the zeros. For example, f ( x ) = x 2 − x − 6 = ( x + 2 ) ( x − 3 ) has roots −2 and 3. These values bound the regions where the function is positive (above the x -axis) or negative (below the x -axis).

Therefore x 2 − x − 6 ≤ 0 has solutions where − 2 ≤ x ≤ 3 , using interval notation [ − 2 , 3 ] . Furthermore, x 2 − x − 6 ≥ 0 has solutions where x ≤ − 2 or x ≥ 3 , using interval notation ( − ∞ , − 2 ] ∪ [ − 3 , ∞ ) .

Given the graph of f determine the solutions to f ( x ) > 0 :

From the graph we can see that the roots are −4 and 2. The graph of the function lies above the x -axis ( f ( x ) > 0 ) in between these roots.

Because of the strict inequality, the solution set is shaded with an open dot on each of the boundaries. This indicates that these critical numbers are not actually included in the solution set. This solution set can be expressed two ways,

{ x | − 4 < x < 2 } S e t   N o t a t i o n ( − 4 , 2 ) I n t e r v a l   N o t a t i o n

In this textbook, we will continue to present answers in interval notation.

Answer: ( − 4 , 2 )

Try this! Given the graph of f determine the solutions to f ( x ) < 0 :

Answer: ( − ∞ , − 4 ) ∪ ( 2 , ∞ )

Solving Quadratic Inequalities

Next we outline a technique used to solve quadratic inequalities without graphing the parabola. To do this we make use of a sign chart A model of a function using a number line and signs (+ or −) to indicate regions in the domain where the function is positive or negative. which models a function using a number line that represents the x -axis and signs (+ or −) to indicate where the function is positive or negative. For example,

The plus signs indicate that the function is positive on the region. The negative signs indicate that the function is negative on the region. The boundaries are the critical numbers, −2 and 3 in this case. Sign charts are useful when a detailed picture of the graph is not needed and are used extensively in higher level mathematics. The steps for solving a quadratic inequality with one variable are outlined in the following example.

Solve: − x 2 + 6 x + 7 ≥ 0 .

It is important to note that this quadratic inequality is in standard form, with zero on one side of the inequality.

Step 1 : Determine the critical numbers. For a quadratic inequality in standard form, the critical numbers are the roots. Therefore, set the function equal to zero and solve.

− x 2 + 6 x + 7 = 0 − ( x 2 − 6 x − 7 ) = 0 − ( x + 1 ) ( x − 7 ) = 0 x + 1 = 0 or x − 7 = 0 x = − 1 x = 7

The critical numbers are −1 and 7.

Step 2 : Create a sign chart. Since the critical numbers bound the regions where the function is positive or negative, we need only test a single value in each region. In this case the critical numbers partition the number line into three regions and we choose test values x = − 3 , x = 0 , and x = 10 .

Test values may vary. In fact, we need only determine the sign (+ or −) of the result when evaluating f ( x ) = − x 2 + 6 x + 7 = − ( x + 1 ) ( x − 7 ) . Here we evaluate using the factored form.

f ( − 3 ) = − ( − 3 + 1 ) ( − 3 − 7 ) = − ( − 2 ) ( − 10 ) = − N e g a t i v e f ( 0 ) = − ( 0 + 1 ) ( 0 − 7 ) = − ( 1 ) ( − 7 ) = + P o s i t i v e f ( 10 ) = − ( 10 + 1 ) ( 10 − 7 ) = − ( 11 ) ( 3 ) = − N e g a t i v e

Since the result of evaluating for −3 was negative, we place negative signs above the first region. The result of evaluating for 0 was positive, so we place positive signs above the middle region. Finally, the result of evaluating for 10 was negative, so we place negative signs above the last region, and the sign chart is complete.

Step 3 : Use the sign chart to answer the question. In this case, we are asked to determine where f ( x ) ≥ 0 , or where the function is positive or zero. From the sign chart we see this occurs when x -values are inclusively between −1 and 7.

Using interval notation, the shaded region is expressed as [ − 1 , 7 ] . The graph is not required; however, for the sake of completeness it is provided below.

Indeed the function is greater than or equal to zero, above or on the x -axis, for x -values in the specified interval.

Answer: [ − 1 , 7 ]

Solve: 2 x 2 − 7 x + 3 > 0 .

Begin by finding the critical numbers, in this case, the roots of f ( x ) = 2 x 2 − 7 x + 3 .

2 x 2 − 7 x + 3 = 0 ( 2 x − 1 ) ( x − 3 ) = 0 2 x − 1 = 0 or x − 3 = 0 2 x = 1 x = 3 x = 1 2

The critical numbers are 1 2 and 3. Because of the strict inequality > we will use open dots.

Next choose a test value in each region and determine the sign after evaluating f ( x ) = 2 x 2 − 7 x + 3 = ( 2 x − 1 ) ( x − 3 ) . Here we choose test values −1, 2, and 5.

f ( − 1 ) = [ 2 ( − 1 ) − 1 ] ( − 1 − 3 ) = ( − ) ( − ) = + f ( 2 ) = [ 2 ( 2 ) − 1 ] ( 2 − 3 ) = ( + ) ( − ) = − f ( 5 ) = [ 2 ( 5 ) − 1 ] ( 5 − 3 ) = ( + ) ( + ) = +

And we can complete the sign chart.

The question asks us to find the x -values that produce positive results (greater than zero). Therefore, shade in the regions with a + over them. This is the solution set.

Answer: ( − ∞ , 1 2 ) ∪ ( 3 , ∞ )

Sometimes the quadratic function does not factor. In this case we can make use of the quadratic formula.

Solve: x 2 − 2 x − 11 ≤ 0 .

Find the critical numbers.

x 2 − 2 x − 11 = 0

Identify a , b , and c for use in the quadratic formula. Here a = 1 , b = − 2 , and c = − 11 . Substitute the appropriate values into the quadratic formula and then simplify.

x = − b ± b 2 − 4 a c 2 a = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( − 11 ) 2 ( 1 ) = 2 ± 48 2 = 2 ± 4 3 2 = 1 ± 2 3

Therefore the critical numbers are 1 − 2 3 ≈ − 2.5 and 1 + 2 3 ≈ 4.5 . Use a closed dot on the number to indicate that these values will be included in the solution set.

Here we will use test values −5, 0, and 7.

f ( − 5 ) = ( − 5 ) 2 − 2 ( − 5 ) − 11 = 25 + 10 − 11 = + f ( 0 ) = ( 0 ) 2 − 2 ( 0 ) − 11 = 0 + 0 − 11 = − f ( 7 ) = ( 7 ) 2 − 2 ( 7 ) − 11 = 49 − 14 − 11 = +

After completing the sign chart shade in the values where the function is negative as indicated by the question ( f ( x ) ≤ 0 ).

Answer: [ 1 − 2 3 , 1 + 2 3 ]

Try this! Solve: 9 − x 2 > 0 .

Answer: ( − 3 , 3 )

It may be the case that there are no critical numbers.

Solve: x 2 − 2 x + 3 > 0 .

To find the critical numbers solve,

x 2 − 2 x + 3 = 0

Substitute a = 1 , b = − 2 , and c = 3 into the quadratic formula and then simplify.

x = − b ± b 2 − 4 a c 2 a = − ( − 2 ) ± ( − 2 ) 2 − 4 ( 1 ) ( 3 ) 2 ( 1 ) = 2 ± − 8 2 = 2 ± 2 i 2 2 = 1 + i 2

Because the solutions are not real, we conclude there are no real roots; hence there are no critical numbers. When this is the case, the graph has no x -intercepts and is completely above or below the x -axis. We can test any value to create a sign chart. Here we choose x = 0 .

f ( 0 ) = ( 0 ) 2 − 2 ( 0 ) + 3 = +

Because the test value produced a positive result the sign chart looks as follows:

We are looking for the values where f ( x ) > 0 ; the sign chart implies that any real number for x will satisfy this condition.

Answer: ( − ∞ , ∞ )

The function in the previous example is graphed below.

We can see that it has no x -intercepts and is always above the x -axis (positive). If the question was to solve x 2 − 2 x + 3 < 0 , then the answer would have been no solution. The function is never negative.

Try this! Solve: 9 x 2 − 12 x + 4 ≤ 0 .

Answer: One solution, 2 3 .

Find the domain: f ( x ) = x 2 − 4 .

Recall that the argument of a square root function must be nonnegative. Therefore, the domain consists of all real numbers for x such that x 2 − 4 is greater than or equal to zero.

x 2 − 4 ≥ 0

It should be clear that x 2 − 4 = 0 has two solutions x = ± 2 ; these are the critical values. Choose test values in each interval and evaluate f ( x ) = x 2 − 4 .

f ( − 3 ) = ( − 3 ) 2 − 4 = 9 − 4 = + f ( 0 ) = ( 0 ) 2 − 4 = 0 − 4 = − f ( 3 ) = ( 3 ) 2 − 4 = 9 − 4 = +

Shade in the x -values that produce positive results.

Answer: Domain: ( − ∞ , − 2 ] ∪ [ 2 , ∞ )

Key Takeaways

• Quadratic inequalities can have infinitely many solutions, one solution or no solution.
• We can solve quadratic inequalities graphically by first rewriting the inequality in standard form, with zero on one side. Graph the quadratic function and determine where it is above or below the x -axis. If the inequality involves “less than,” then determine the x -values where the function is below the x -axis. If the inequality involves “greater than,” then determine the x -values where the function is above the x -axis.
• We can streamline the process of solving quadratic inequalities by making use of a sign chart. A sign chart gives us a visual reference that indicates where the function is above the x -axis using positive signs or below the x -axis using negative signs. Shade in the appropriate x -values depending on the original inequality.
• To make a sign chart, use the function and test values in each region bounded by the roots. We are only concerned if the function is positive or negative and thus a complete calculation is not necessary.

Topic Exercises

Part a: solutions to quadratic inequalities.

Determine whether or not the given value is a solution.

x 2 − x + 1 < 0 ; x = − 1

x 2 + x − 1 > 0 ; x = − 2

4 x 2 − 12 x + 9 ≤ 0 ; x = 3 2

5 x 2 − 8 x − 4 < 0 ; x = − 2 5

3 x 2 − x − 2 ≥ 0 ; x = 0

4 x 2 − x + 3 ≤ 0 ; x = − 1

2 − 4 x − x 2 < 0 ; x = 1 2

5 − 2 x − x 2 > 0 ; x = 0

− x 2 − x − 9 < 0 ; x = − 3

− x 2 + x − 6 ≥ 0 ; x = 6

Given the graph of f determine the solution set.

f ( x ) ≤ 0 ;

f ( x ) ≥ 0 ;

f ( x ) > 0 ;

f ( x ) < 0 ;

Use the transformations to graph the following and then determine the solution set.

x 2 − 1 > 0

x 2 + 2 > 0

( x − 1 ) 2 > 0

( x + 2 ) 2 ≤ 0

( x + 2 ) 2 − 1 ≤ 0

( x + 3 ) 2 − 4 > 0

− x 2 + 4 ≥ 0

− ( x + 2 ) 2 > 0

− ( x + 3 ) 2 + 1 < 0

− ( x − 4 ) 2 + 9 > 0

Part B: Solving Quadratic Inequalities

Use a sign chart to solve and graph the solution set. Present answers using interval notation.

x 2 − x − 12 > 0

x 2 − 10 x + 16 > 0

x 2 + 2 x − 24 < 0

x 2 + 15 x + 54 < 0

x 2 − 23 x − 24 ≤ 0

x 2 − 12 x + 20 ≤ 0

2 x 2 − 11 x − 6 ≥ 0

3 x 2 + 17 x − 6 ≥ 0

8 x 2 − 18 x − 5 < 0

10 x 2 + 17 x + 6 > 0

9 x 2 + 30 x + 25 ≤ 0

16 x 2 − 40 x + 25 ≤ 0

4 x 2 − 4 x + 1 > 0

9 x 2 + 12 x + 4 > 0

− x 2 − x + 30 ≥ 0

− x 2 − 6 x + 27 ≤ 0

x 2 − 64 < 0

x 2 − 81 ≥ 0

4 x 2 − 9 ≥ 0

16 x 2 − 25 < 0

25 − 4 x 2 ≥ 0

1 − 49 x 2 < 0

x 2 − 8 > 0

x 2 − 75 ≤ 0

2 x 2 + 1 > 0

4 x 2 + 3 < 0

x − x 2 > 0

3 x − x 2 ≤ 0

x 2 − x + 1 < 0

x 2 + x − 1 > 0

4 x 2 − 12 x + 9 ≤ 0

5 x 2 − 8 x − 4 < 0

3 x 2 − x − 2 ≥ 0

4 x 2 − x + 3 ≤ 0

2 − 4 x − x 2 < 0

5 − 2 x − x 2 > 0

− x 2 − x − 9 < 0

− x 2 + x − 6 ≥ 0

− 2 x 2 + 4 x − 1 ≥ 0

− 3 x 2 − x + 1 ≤ 0

Find the domain of the function.

f ( x ) = x 2 − 25

f ( x ) = x 2 + 3 x

g ( x ) = 3 x 2 − x − 2

g ( x ) = 12 x 2 − 9 x − 3

h ( x ) = 16 − x 2

h ( x ) = 3 − 2 x − x 2

f ( x ) = x 2 + 10

f ( x ) = 9 + x 2

A robotics manufacturing company has determined that its weekly profit in thousands of dollars is modeled by P ( n ) = − n 2 + 30 n − 200 where n represents the number of units it produces and sells. How many units must the company produce and sell to maintain profitability. (Hint: Profitability occurs when profit is greater than zero.)

The height in feet of a projectile shot straight into the air is given by h ( t ) = − 16 t 2 + 400 t where t represents the time in seconds after it is fired. In what time intervals is the projectile under 1,000 feet? Round to the nearest tenth of a second.

Part C: Discussion Board

Does the sign chart for any given quadratic function always alternate? Explain and illustrate your answer with some examples.

Research and discuss other methods for solving a quadratic inequality.

Explain the difference between a quadratic equation and a quadratic inequality. How can we identify and solve each? What is the geometric interpretation of each?

[ − 4 , 2 ]

[ − 1 , 3 ]

( − ∞ , ∞ )

( − ∞ , 4 ) ∪ ( 8 , ∞ )

( − ∞ , − 1 ) ∪ ( 1 , ∞ )

( − ∞ , 1 ) ∪ ( 1 , ∞ )

[ − 3 , − 1 ]

[ − 2 , 2 ]

( − ∞ , − 4 ) ∪ ( − 2 , ∞ )

( − ∞ , − 3 ) ∪ ( 4 , ∞ )

( − 6 , 4 )

[ − 1 , 24 ]

• ( − ∞ , − 1 2 ] ∪ [ 6 , ∞ )
• ( − 1 4 , 5 2 )
• ( − ∞ , 1 2 ) ∪ ( 1 2 , ∞ )

[ − 6 , 5 ]

( − 8 , 8 )

• ( − ∞ , − 3 2 ] ∪ [ 3 2 , ∞ )
• [ − 5 2 , 5 2 ]

( − ∞ , − 2 2 ) ∪ ( 2 2 , ∞ )

• ( − ∞ , − 2 3 ] ∪ [ 1 , ∞ )

( − ∞ , − 2 − 6 ) ∪ ( − 2 + 6 , ∞ )

• [ 2 − 2 2 , 2 + 2 2 ]
• ( − ∞ , − 5 ] ∪ [ 5 , ∞ )

[ − 4 , 4 ]

The company must produce and sell more than 10 units and fewer than 20 units each week.

Answer may vary

Quadratic Inequalities

Instructions: Use this calculator to solve quadratic inequalities, showing all the steps. Please type the inequality you want to solve in the box below.

More about Quadratic Inequalities

This quadratic inequality calculate will provide you with solutions to inequalities showing all the steps. For example, the inequality you provide can something like 'x^2 - 1/2 > 0', and in general, quadratic inequalities do not get too hard to solve.

Once you provide a valid inequality involving quadratic expression , you can click on "Calculate" to get all the steps of the calculation shown to you, including a graph of the inequality solutions.

Quadratic inequalities are one of the few types (along with linear inequalities ) that you are guaranteed to be able to find an exact solution, and this process involves solving a quadratic equation.

What is the definition of this type of inequality

We haven't stated it explicitly, but it comes sort of clear from its name: a quadratic inequality is a specific type of inequality in which all the terms involved in it are polynomials of at most degree 2. In this context, one example is

which is quadratic by the fact that both sides of the inequality are polynomials of order at most 2. Now, if you had:

then the inequality is not quadratic any longer, because of the term \(x^3\) on the right hand side. For these inequalities we have a clear roadmap to find the solution.

Steps for Solving Inequalities that are Quadratic

• Step 1: Make sure that you have a quadratic inequality, as the method used in this case is valid only for this type of inequality
• Step 2: As with most inequalities, pass everything to the left side of the inequality, and solve the associated equation
• Step 3: If the associated quadratic equation does not have real roots, we then know that either the whole real line is a solution, or there is no solution. So you test any point and see if it solves the inequality, and if it does, the solution is the whole real line (-∞, ∞), otherwise, the solution is empty.
• Step 4: If the associated quadratic equation has only one real solution, it means that associated quadratic graph touches tangentially the x-axis. So depending on the inequality sign, you may have only the touching point being the solution, or everything except the touching point being the solution, or the whole real line (-∞, ∞), for which you need to test the touching point, and a point outside of that (to the left and right of the point)
• Step 4: If the associated quadratic equation has two different real solutions, you check intervals defined by these roots to determine which portions of the real line will be a part of the solution

One you analyze the pieces, if needed, you join them by using the "union" operator, which is used to piece intervals together.

How do you draw a quadratic inequality?

Graphing inequalities provides a great way of visually understanding how the solution looks like. In terms of the procedure, you need to know whether you are working with a one variable inequality, or you have more variables.

If you have an inequality such as

you have only one variable, and then the solution will be a subset of the real line. On the other hand, if you had something like

then you have only two variables x and y, and then the solution of the inequality wiil be a subset of the x-y plane.

Importance of Quadratic Expressions

Quadratic expressions involved in equations and inequalities play a fundamental role in Mathematics. Perhaps, quadratic is the most broadly used type of structure after linear.

When dealing with Calculus and Algebra you will find countless applications of quadratics in problems of maximization and minimization, integration, and a lot more. Digging a bit you will find applications of quadratic expressions all over different scientific displines

Example: Quadratic Inequalities

Solve this quadratic inequality: \(\frac{1}{3}x^2 + \frac{5}{4}x - \frac{5}{6} < 0\)

We need to first solve the following auxiliary quadratic equation \(\displaystyle \frac{1}{3}x^2+\frac{5}{4}x-\frac{5}{6}=0\).

Applying the Quadratic Formula

The quadratic equation is:

In this case, we have:

Plugging these values into the formula for the roots we get:

so then, we find that:

Critical Points

The list of critical points found organized in ascending order is: \(-\frac{1}{8}\sqrt{385}-\frac{15}{8}\), \(\frac{1}{8}\sqrt{385}-\frac{15}{8}\).

Then, we need to analyze the following critical intervals:

• For the interval \(\left(-\infty, -\frac{1}{8}\sqrt{385}-\frac{15}{8}\right)\): The left-hand side is positive, so then \(\left(-\infty, -\frac{1}{8}\sqrt{385}-\frac{15}{8}\right)\) is not a part of the solution.

• For the interval \(\left(-\frac{1}{8}\sqrt{385}-\frac{15}{8}, \frac{1}{8}\sqrt{385}-\frac{15}{8}\right)\): The left-hand side is negative, which implies that \(\left(-\frac{1}{8}\sqrt{385}-\frac{15}{8}, \frac{1}{8}\sqrt{385}-\frac{15}{8}\right)\) is a part of the solution.

• For the interval \(\left(\frac{1}{8}\sqrt{385}-\frac{15}{8}, \infty\right)\): The left-hand side is positive, which means that \(\left(\frac{1}{8}\sqrt{385}-\frac{15}{8}, \infty\right)\) is not a part of the solution.

Based on the inequality provided, and analyzing the critical points, we find that the solution to the inequality is: \(-\frac{1}{8}\sqrt{385}-\frac{15}{8}< x \le \frac{1}{8}\sqrt{385}-\frac{15}{8}\).

Using interval notation, the solution is written as:

Example: More Quadratic Inequalities

Solve: \(x^2 < 4\)

The given inequality is:

which derives to the quadratic equation \(\displaystyle x^2-4=0\).

Quadratic Formula

For a quadratic equation of the form \(a x^2 + bx + c = 0\), the roots are computed using the following quadratic formula :

In this case, we have that the equation we need to solve is \(\displaystyle x^2-4 = 0\), which implies that corresponding coefficients are:

First, we will compute the discriminant to assess the nature of the roots. The discriminating is computed as:

Since in this case we get the discriminant is \(\Delta = \displaystyle 16 > 0\), which is positive, we know that the equation has two different real roots.

Now, plugging these values into the formula for the roots we get:

In this case, the quadratic equation \( \displaystyle x^2-4 = 0 \), has two real roots, so then:

so then the original polynomial is factored as \(\displaystyle p(x) = x^2-4 = \left(x+2\right)\left(x-2\right) \), which completes the factorization.

Analyzing Critical Points

The list of critical points found organized in ascending order is: \(-2\), \(2\).

Based on this, we need to analyze the following intervals:

• For the interval \(\left(-\infty, -2\right)\): The left-hand side is positive, so then \(\left(-\infty, -2\right)\) is not a part of the solution.

• For the interval \(\left(-2, 2\right)\): The left-hand side is negative, so then \(\left(-2, 2\right)\) is part of the solution.

• For the interval \(\left(2, \infty\right)\): The left-hand side is positive, which means that \(\left(2, \infty\right)\) is not a part of the solution.

Solution to the Inequality

Based on the inequality provided, and analyzing the critical points, we find that the solution to the inequality is: \(-2< x \le 2\).

Other useful inequality calculators

The simplest inequality type you will be able to solve is linear inequalities . After that, you have quadratic inequalities, in terms of ease.

Then, you have polynomial inequalities in general, with degree higher than 2. Those should be straightforward, but they still could be formidable tasks in terms of the amount of laborious work required to solve them.

Having an inequality calculator for inequalities in general can really come in handy, as it will address the different types of inequalities that are amenable to the search of exact solutions.

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• x^{2}+2x-12<0
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• How do you solve quadratic inequalities?
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• What is quadratic inequality?
• A quadratic inequality is a second degree equation with an inequality sign
• What is the rule for quadratic inequalities?
• The rule for inequalities, when multiplying or divising by a negative number, flip the inequality sign
• What is the difference between quadratic equation and quadratic inequalities?
• The difference between quadratic equations and quadratic inequalities is that a quadratic equation is an equation that equates a quadratic expression to a constant or to zero, whereas a quadratic inequality is an inequality that involves a quadratic expression.

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• High School Math Solutions – Inequalities Calculator, Quadratic Inequalities We’ve learned how to solve linear inequalities. Now, it’s time to learn how to solve quadratic inequalities. Solving...

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Solving Inequality Word Questions

(You might like to read Introduction to Inequalities and Solving Inequalities first.)

In Algebra we have "inequality" questions like:

Sam and Alex play in the same soccer team. Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals. What are the possible number of goals Alex scored?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

• Read the whole thing first
• Do a sketch if needed
• Assign letters for the values
• Find or work out formulas

We should also write down what is actually being asked for , so we know where we are going and when we have arrived!

The best way to learn this is by example, so let's try our first example:

Assign Letters:

• the number of goals Alex scored: A
• the number of goals Sam scored: S

We know that Alex scored 3 more goals than Sam did, so: A = S + 3

And we know that together they scored less than 9 goals: S + A < 9

We are being asked for how many goals Alex might have scored: A

Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals.

Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals .

• When S = 0, then A = 3 and S + A = 3, and 3 < 9 is correct
• When S = 1, then A = 4 and S + A = 5, and 5 < 9 is correct
• When S = 2, then A = 5 and S + A = 7, and 7 < 9 is correct
• (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)

Lots More Examples!

Example: Of 8 pups, there are more girls than boys. How many girl pups could there be?

• the number of girls: g
• the number of boys: b

We know that there are 8 pups, so: g + b = 8, which can be rearranged to

We also know there are more girls than boys, so:

We are being asked for the number of girl pups: g

So there could be 5, 6, 7 or 8 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, and the question isn't clear on that point (sometimes questions are like that).

• When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
• When g = 7, then b = 1 and g > b is correct
• When g = 6, then b = 2 and g > b is correct
• When g = 5, then b = 3 and g > b is correct
• (But if g = 4, then b = 4 and g > b is incorrect)

A speedy example:

Example: Joe enters a race where he has to cycle and run. He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed. Joe completes the race in less than 2½ hours, what can we say about his average speeds?

• Average running speed: s
• So average cycling speed: 2s
• Speed = Distance Time
• Which can be rearranged to: Time = Distance Speed

We are being asked for his average speeds: s and 2s

The race is divided into two parts:

• Distance = 25 km
• Average speed = 2s km/h
• So Time = Distance Average Speed = 25 2s hours
• Distance = 20 km
• Average speed = s km/h
• So Time = Distance Average Speed = 20 s hours

Joe completes the race in less than 2½ hours

• The total time < 2½
• 25 2s + 20 s < 2½

So his average speed running is greater than 13 km/h and his average speed cycling is greater than 26 km/h

In this example we get to use two inequalities at once:

Example: The velocity v m/s of a ball thrown directly up in the air is given by v = 20 − 10t , where t is the time in seconds. At what times will the velocity be between 10 m/s and 15 m/s?

• velocity in m/s: v
• the time in seconds: t
• v = 20 − 10t

We are being asked for the time t when v is between 5 and 15 m/s:

So the velocity is between 10 m/s and 15 m/s between 0.5 and 1 second after.

And a reasonably hard example to finish with:

Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area. The perimeter of the room is 16 m. What could the width and length of the room be?

Make a sketch: we don't know the size of the tables, only their area, they may fit perfectly or not!

• the length of the room: L
• the width of the room: W

The formula for the perimeter is 2(W + L) , and we know it is 16 m

• 2(W + L) = 16
• L = 8 − W

We also know the area of a rectangle is the width times the length: Area = W × L

And the area must be greater than or equal to 7:

• W × L ≥ 7

We are being asked for the possible values of W and L

Let's solve:

So the width must be between 1 m and 7 m (inclusive) and the length is 8−width .

• Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m 2 (fits exactly 7 tables)
• Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m 2 (7 won't fit)
• Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m 2 (7 fit easily)
• Likewise for W around 7 m

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2.4: Inequalities with Absolute Value and Quadratic Functions

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• Page ID 80767

• Carl Stitz & Jeff Zeager
• Lakeland Community College & Lorain County Community College

In this section, not only do we develop techniques for solving various classes of inequalities analytically, we also look at them graphically. The first example motivates the core ideas.

Example 2.4.1

Let \(f(x) = 2x-1\) and \(g(x)=5\).

• Solve \(f(x) = g(x)\).
• Solve \(f(x) < g(x)\).
• Solve \(f(x) > g(x)\).
• Graph \(y=f(x)\) and \(y=g(x)\) on the same set of axes and interpret your solutions to parts 1 through 3 above.
• To solve \(f(x) = g(x)\), we replace \(f(x)\) with \(2x-1\) and \(g(x)\) with \(5\) to get \(2x-1 = 5\). Solving for \(x\), we get \(x=3\).
• The inequality \(f(x) < g(x)\) is equivalent to \(2x-1 < 5\). Solving gives \(x < 3\) or \((-\infty, 3)\).
• To find where \(f(x) > g(x)\), we solve \(2x-1 > 5\). We get \(x > 3\), or \((3, \infty)\).

To see the connection between the graph and the Algebra, we recall the Fundamental Graphing Principle for Functions in Section 1.6 : the point \((a,b)\) is on the graph of \(f\) if and only if \(f(a)=b\). In other words, a generic point on the graph of \(y=f(x)\) is \((x,f(x))\), and a generic point on the graph of \(y=g(x)\) is \((x,g(x))\). When we seek solutions to \(f(x)=g(x)\), we are looking for \(x\) values whose \(y\) values on the graphs of \(f\) and \(g\) are the same. In part 1, we found \(x=3\) is the solution to \(f(x)=g(x)\). Sure enough, \(f(3) = 5\) and \(g(3) = 5\) so that the point \((3,5)\) is on both graphs. In other words, the graphs of \(f\) and \(g\) intersect at \((3,5)\). In part 2, we set \(f(x) < g(x)\) and solved to find \(x < 3\). For \(x < 3\), the point \((x,f(x))\) is below \((x,g(x))\) since the \(y\) values on the graph of \(f\) are less than the \(y\) values on the graph of \(g\) there. Analogously, in part 3, we solved \(f(x) > g(x)\) and found \(x > 3\). For \(x > 3\), note that the graph of \(f\) is above the graph of \(g\), since the \(y\) values on the graph of \(f\) are greater than the \(y\) values on the graph of \(g\) for those values of \(x\).

The preceding example demonstrates the following, which is a consequence of the Fundamental Graphing Principle for Functions.

Graphical Interpretation of Equations and Inequalities

Suppose \(f\) and \(g\) are functions.

• The solutions to \(f(x)=g(x)\) are the \(x\) values where the graphs of \(y=f(x)\) and \(y=g(x)\) intersect.
• The solution to \(f(x) < g(x)\) is the set of \(x\) values where the graph of \(y=f(x)\) is below the graph of \(y=g(x)\).
• The solution to \(f(x) > g(x)\) is the set of \(x\) values where the graph of \(y=f(x)\) above the graph of \(y=g(x)\).

The next example turns the tables and furnishes the graphs of two functions and asks for solutions to equations and inequalities.

Example 2.4.2

The graphs of \(f\) and \(g\) are below. (The graph of \(y=g(x)\) is bolded.) Use these graphs to answer the following questions.

• Solve \(f(x) \geq g(x)\).
• To solve \(f(x)=g(x)\), we look for where the graphs of \(f\) and \(g\) intersect. These appear to be at the points \((-1,2)\) and \((1,2)\), so our solutions to \(f(x) = g(x)\) are \(x = -1\) and \(x=1\).
• To solve \(f(x) < g(x)\), we look for where the graph of \(f\) is below the graph of \(g\). This appears to happen for the \(x\) values less than \(-1\) and greater than \(1\). Our solution is \((-\infty, -1) \cup (1,\infty)\).
• To solve \(f(x) \geq g(x)\), we look for solutions to \(f(x)=g(x)\) as well as \(f(x) > g(x)\). We solved the former equation and found \(x = \pm 1\). To solve \(f(x) > g(x)\), we look for where the graph of \(f\) is above the graph of \(g\). This appears to happen between \(x=-1\) and \(x=1\), on the interval \((-1,1)\). Hence, our solution to \(f(x) \geq g(x)\) is \([-1,1]\).

We now turn our attention to solving inequalities involving the absolute value. We have the following theorem from Intermediate Algebra to help us.

Theorem 2.4. Inequalities Involving the Absolute Value

Let \(c\) be a real number.

• For \(c > 0\), \(|x| < c\) is equivalent to \(-c<x<c\).
• For \(c > 0\), \(|x| \leq c\) is equivalent to \(-c \leq x \leq c\).
• For \(c \leq 0\), \(|x| < c\) has no solution, and for \(c < 0\), \(|x| \leq c\) has no solution.
• For \(c \geq 0\), \(|x| > c\) is equivalent to \(x<-c\) or \(x>c\).
• For \(c \geq 0\), \(|x| \geq c\) is equivalent to \(x \leq -c\) or \(x \geq c\).
• For \(c < 0\), \(|x| > c\) and \(|x| \geq c\) are true for all real numbers.

As with Theorem 2.1 in Section 2.2 , we could argue Theorem 2.4 using cases. However, in light of what we have developed in this section, we can understand these statements graphically. For instance, if \(c > 0\), the graph of \(y=c\) is a horizontal line which lies above the \(x\)-axis through \((0,c)\). To solve \(|x| < c\), we are looking for the \(x\) values where the graph of \(y=|x|\) is below the graph of \(y=c\). We know that the graphs intersect when \(|x|=c\), which, from Section 2.2 , we know happens when \(x=c\) or \(x=-c\). Graphing, we get

We see that the graph of \(y=|x|\) is below \(y=c\) for \(x\) between \(-c\) and \(c\), and hence we get \(|x| < c\) is equivalent to \(-c < x < c\). The other properties in Theorem 2.4 can be shown similarly.

Example 2.4.3

Solve the following inequalities analytically; check your answers graphically.

• \(|x-1| \geq 3\)
• \(4 - 3|2x+1| > -2\)
• \(2 < |x-1| \leq 5\)
• \(|x+1|\geq \dfrac{x+4}{2}\)

We see that the graph of \(y=|x-1|\) is above the horizontal line \(y=3\) for \(x < -2\) and \(x > 4\) hence this is where \(|x-1| > 3\). The two graphs intersect when \(x=-2\) and \(x=4\), so we have graphical confirmation of our analytic solution.

We now turn our attention to quadratic inequalities. In the last example of Section 2.3 , we needed to determine the solution to \(x^2 - x -6 < 0\). We will now re-visit this problem using some of the techniques developed in this section not only to reinforce our solution in Section 2.3 , but to also help formulate a general analytic procedure for solving all quadratic inequalities. If we consider \(f(x) = x^2-x-6\) and \(g(x)=0\), then solving \(x^2 - x -6 < 0\) corresponds graphically to finding the values of \(x\) for which the graph of \(y=f(x)=x^2-x-6\) (the parabola) is below the graph of \(y=g(x)=0\) (the \(x\)-axis). We’ve provided the graph again for reference.

We can see that the graph of \(f\) does dip below the \(x\)-axis between its two \(x\)-intercepts. The zeros of \(f\) are \(x=-2\) and \(x=3\) in this case and they divide the domain (the \(x\)-axis) into three intervals: \((-\infty, -2)\), \((-2,3)\) and \((3, \infty)\). For every number in \((-\infty, -2)\), the graph of \(f\) is above the \(x\)-axis; in other words, \(f(x) > 0\) for all \(x\) in \((-\infty, -2)\). Similarly, \(f(x) < 0\) for all \(x\) in \((-2,3)\), and \(f(x) > 0\) for all \(x\) in \((3, \infty)\). We can schematically represent this with the sign diagram below.

Here, the \((+)\) above a portion of the number line indicates \(f(x) > 0\) for those values of \(x\); the \((-)\) indicates \(f(x) < 0\) there. The numbers labeled on the number line are the zeros of \(f\), so we place \(0\) above them. We see at once that the solution to \(f(x) < 0\) is \((-2,3)\).

Our next goal is to establish a procedure by which we can generate the sign diagram without graphing the function. An important property 2 of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphically, this means that a parabola can’t be above the \(x\)-axis at one point and below the \(x\)-axis at another point without crossing the \(x\)-axis. This allows us to determine the sign of all of the function values on a given interval by testing the function at just one value in the interval. This gives us the following.

Steps for Solving a Quadratic Inequality

• Rewrite the inequality, if necessary, as a quadratic function \(f(x)\) on one side of the inequality and \(0\) on the other.
• Find the zeros of \(f\) and place them on the number line with the number \(0\) above them.
• Choose a real number, called a test value , in each of the intervals determined in step 2.
• Determine the sign of \(f(x)\) for each test value in step 3, and write that sign above the corresponding interval.
• Choose the intervals which correspond to the correct sign to solve the inequality.

Example 2.4.4

Solve the following inequalities analytically using sign diagrams. Verify your answer graphically.

• \(2x^2 \leq 3-x\)
• \(x^2 - 2x > 1\)
• \(x^2+1 \leq 2x\)
• \(2x-x^2 \geq |x-1|-1\)

To verify our solution graphically, we refer to the original inequality, \(2x^2 \leq 3-x\). We let \(g(x) = 2x^2\) and \(h(x)=3-x\). We are looking for the \(x\) values where the graph of \(g\) is below that of \(h\) (the solution to \(g(x) < h(x)\)) as well as the points of intersection (the solutions to \(g(x)=h(x)\)). The graphs of \(g\) and \(h\) are given on the right with the sign chart on the left.

Combining these into one sign diagram, we have that our solution is \([0,2]\). Graphically, to check \(2x-x^2 \geq |x-1|-1\), we set \(h(x) = 2x-x^2\) and \(i(x) = |x-1|-1\) and look for the \(x\) values where the graph of \(h\) is above the the graph of \(i\) (the solution of \(h(x) > i(x)\)) as well as the \(x\)-coordinates of the intersection points of both graphs (where \(h(x)=i(x)\)). The combined sign chart is given on the left and the graphs are on the right.

One of the classic applications of inequalities is the notion of tolerances. 5 Recall that for real numbers \(x\) and \(c\), the quantity \(|x-c|\) may be interpreted as the distance from \(x\) to \(c\). Solving inequalities of the form \(|x-c| \leq d\) for \(d \geq 0\) can then be interpreted as finding all numbers \(x\) which lie within \(d\) units of \(c\). We can think of the number \(d\) as a ‘tolerance’ and our solutions \(x\) as being within an accepted tolerance of \(c\). We use this principle in the next example.

Example 2.4.5

The area \(A\) (in square inches) of a square piece of particle board which measures \(x\) inches on each side is \(A(x) = x^2\). Suppose a manufacturer needs to produce a \(24\) inch by \(24\) inch square piece of particle board as part of a home office desk kit. How close does the side of the piece of particle board need to be cut to \(24\) inches to guarantee that the area of the piece is within a tolerance of \(0.25\) square inches of the target area of \(576\) square inches?

Mathematically, we express the desire for the area \(A(x)\) to be within \(0.25\) square inches of \(576\) as \(|A - 576| \leq 0.25\). Since \(A(x) = x^2\), we get \(|x^2 - 576| \leq 0.25\), which is equivalent to \(-0.25 \leq x^2 - 576 \leq 0.25\). One way to proceed at this point is to solve the two inequalities \(-0.25 \leq x^2 - 576\) and \(x^2 - 576 \leq 0.25\) individually using sign diagrams and then taking the intersection of the solution sets. While this way will (eventually) lead to the correct answer, we take this opportunity to showcase the increasing property of the square root: if \(0 \leq a \leq b\), then \(\sqrt{a} \leq \sqrt{b}\). To use this property, we proceed as follows

\[\begin{array}{rclr} -0.25 \leq & x^2 - 576 & \leq 0.25 & \\ 575.75 \leq & x^2 & \leq 576.25 & \text{(add $576$ across the inequalities.)} \\ \sqrt{575.75} \leq & \sqrt{x^2} & \leq \sqrt{576.25} & \text{(take square roots.)} \\ \sqrt{575.75} \leq & |x| & \leq \sqrt{576.25} & \text{($\sqrt{x^2} = |x|$)} \\ \end{array}\]

By Theorem 2.4 , we find the solution to \(\sqrt{575.75} \leq |x|\) to be \(\left(-\infty, -\sqrt{575.75} \, \right] \cup \left[\sqrt{575.75}, \infty \right)\) and the solution to \(|x| \leq \sqrt{576.25}\) to be \(\left[-\sqrt{576.25}, \sqrt{576.25} \, \right]\). To solve \(\sqrt{575.75} \leq |x| \leq \sqrt{576.25}\), we intersect these two sets to get \([-\sqrt{576.25}, -\sqrt{575.75}] \cup [\sqrt{575.75},\sqrt{576.25}]\). Since \(x\) represents a length, we discard the negative answers and get \([\sqrt{575.75},\sqrt{576.25}]\). This means that the side of the piece of particle board must be cut between \(\sqrt{575.75} \approx 23.995\) and \(\sqrt{576.25} \approx 24.005\) inches, a tolerance of (approximately) \(0.005\) inches of the target length of \(24\) inches.

Our last example in the section demonstrates how inequalities can be used to describe regions in the plane, as we saw earlier in Section 1.2 .

Example 2.4.6

Sketch the following relations.

• \(R = \{ (x,y) : y > |x| \}\)
• \(S = \{ (x,y) : y \leq 2-x^2 \}\)
• \(T = \{ (x,y) : |x| < y \leq 2-x^2 \}\)
• The relation \(R\) consists of all points \((x,y)\) whose \(y\)-coordinate is greater than \(|x|\). If we graph \(y=|x|\), then we want all of the points in the plane above the points on the graph. Dotting the graph of \(y=|x|\) as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region below on the left.

2.4.1. Exercises

In Exercises 1-32, solve the inequality. Write your answer using interval notation.

• \(|3x - 5| \leq 4\)
• \(|7x + 2| > 10\)
• \(|2x+1| - 5 < 0\)
• \(|2-x| - 4 \geq -3\)
• \(|3x+5| + 2 < 1\)
• \(2|7-x| +4 > 1\)
• \(2 \leq |4-x| < 7\)
• \(1 < |2x - 9| \leq 3\)
• \(|x + 3| \geq |6x + 9|\)
• \(|x-3| - |2x+1| < 0\)
• \(|1-2x| \geq x + 5\)
• \(x + 5 < |x+5|\)
• \(x \geq |x+1|\)
• \(|2x + 1| \leq 6-x\)
• \(x + |2x-3| < 2\)
• \(|3-x| \geq x-5\)
• \(x^{2} + 2x - 3 \geq 0\)
• \(16x^2+8x+1 > 0\)
• \(x^2+9 < 6x\)
• \(9x^2 + 16 \geq 24x\)
• \(x^2+4 \leq 4x\)
• \(x^{2} + 1 < 0\)
• \(3x^{2} \leq 11x + 4\)
• \(x > x^{2}\)
• \(2x^2-4x-1 > 0\)
• \(5x+4 \leq 3x^2\)
• \(2 \leq |x^{2} - 9| < 9\)
• \(x^{2} \leq |4x - 3|\)
• \(x^{2} + x + 1 \geq 0\)
• \(x^2 \geq |x|\)
• \(x |x+5| \geq -6\)
• \(x |x-3| < 2\)
• The profit, in dollars, made by selling \(x\) bottles of \(100 \%\) All-Natural Certified Free-Trade Organic Sasquatch Tonic is given by \(P(x) = -x^2+25x-100\), for \(0 \leq x \leq 35\). How many bottles of tonic must be sold to make at least \(\$50\) in profit?
• Suppose \(C(x) = x^2-10x+27\), \(x \geq 0\) represents the costs, in hundreds of dollars, to produce \(x\) thousand pens. Find the number of pens which can be produced for no more than \(\$1100\).
• The temperature \(T\), in degrees Fahrenheit, \(t\) hours after 6 AM is given by \(T(t) = -\frac{1}{2} t^2 + 8t+32\), for \(0 \leq t \leq 12\). When is it warmer than \(42^{\circ}\) Fahrenheit?
• The height \(h\) in feet of a model rocket above the ground \(t\) seconds after lift-off is given by \(h(t) = -5t^2+100t\), for \(0 \leq t \leq 20\). When is the rocket at least \(250\) feet off the ground? Round your answer to two decimal places.
• If a slingshot is used to shoot a marble straight up into the air from 2 meters above the ground with an initial velocity of 30 meters per second, for what values of time \(t\) will the marble be over 35 meters above the ground? (Refer to Exercise 25 in Section 2.3 for assistance if needed.) Round your answers to two decimal places.
• What temperature values in degrees Celsius are equivalent to the temperature range \(50^{\circ}F\) to \(95^{\circ}F\)? (Refer to Exercise 35 in Section 2.1 for assistance if needed.)

In Exercises 39-42, write and solve an inequality involving absolute values for the given statement.

• Find all real numbers \(x\) so that \(x\) is within \(4\) units of \(2\).
• Find all real numbers \(x\) so that \(3x\) is within \(2\) units of \(-1\).
• Find all real numbers \(x\) so that \(x^2\) is within \(1\) unit of \(3\).
• Find all real numbers \(x\) so that \(x^2\) is at least \(7\) units away from \(4\).
• The surface area \(S\) of a cube with edge length \(x\) is given by \(S(x) = 6x^{2}\) for \(x > 0\). Suppose the cubes your company manufactures are supposed to have a surface area of exactly 42 square centimeters, but the machines you own are old and cannot always make a cube with the precise surface area desired. Write an inequality using absolute value that says the surface area of a given cube is no more than 3 square centimeters away (high or low) from the target of 42 square centimeters. Solve the inequality and write your answer using interval notation.
• Suppose \(f\) is a function, \(L\) is a real number and \(\varepsilon\) is a positive number. Discuss with your classmates what the inequality \(|f(x) - L| < \varepsilon\) means algebraically and graphically. 6

In Exercises 45 - 50, sketch the graph of the relation.

• \(R = \left\{ (x,y) : y \leq x-1 \right\}\)
• \(R = \left\{ (x,y) : y > x^2 +1 \right\}\)
• \(R = \left\{ (x,y) : -1 < y \leq 2x + 1 \right\}\)
• \(R = \left\{ (x,y) : x^2 \leq y < x+2 \right\}\)
• \(R = \left\{ (x,y) : |x|-4 < y < 2-x \right\}\)
• \(R = \left\{ (x,y) : x^{2} < y \leq |4x - 3| \right\}\) [sketchregionineqlast]
• Prove the second, third and fourth parts of Theorem 2.4 .

2.4.2 Answers

• \(\left[\frac{1}{3}, 3\right]\)
• \(\left(-\infty, -\frac{12}{7} \right) \cup \left(\frac{8}{7}, \infty\right)\)
• \((-\infty,1] \cup [3,\infty)\)
• No solution
• \((-\infty, \infty)\)
• \((-3,2] \cup [6,11)\)
• \([3, 4) \cup (5, 6]\)
• \(\left[-\frac{12}{7}, -\frac{6}{5}\right]\)
• \((-\infty, -4) \cup \left( \frac{2}{3}, \infty\right)\)
• \(\left(-\infty, -\frac{4}{3} \right] \cup [6, \infty)\)
• \((-\infty, -5)\)
• No Solution.
• \(\left[ -7, \frac{5}{3}\right]\)
• \(\left( 1, \frac{5}{3} \right)\)
• \((-\infty, -3] \cup [1, \infty)\)
• \(\left(-\infty, -\frac{1}{4}\right) \cup \left(-\frac{1}{4}, \infty \right)\)
• \(\left\{2 \right\}\)
• \(\left[-\frac{1}{3}, 4 \right]\)
• \(\left(-\infty, 1-\frac{\sqrt{6}}{2} \right) \cup \left(1+\frac{\sqrt{6}}{2}, \infty \right)\)
• \(\left(-\infty, \frac{5 - \sqrt{73}}{6} \right] \cup \left[\frac{5 + \sqrt{73}}{6}, \infty \right)\)
• \(\left(-3\sqrt{2}, -\sqrt{11} \right] \cup \left[-\sqrt{7}, 0 \right) \cup \left(0, \sqrt{7} \right] \cup \left[\sqrt{11}, 3\sqrt{2} \right)\)
• \(\left[-2-\sqrt{7}, -2+\sqrt{7} \right] \cup [1, 3]\)
• \((-\infty, -1] \cup \left\{ 0 \right\} \cup [1,\infty)\)
• \([-6,-3] \cup [-2, \infty)\)
• \((-\infty, 1) \cup \left(2, \frac{3+\sqrt{17}}{2}\right)\)
• \(P(x) \geq 50\) on \([10,15]\). This means anywhere between 10 and 15 bottles of tonic need to be sold to earn at least \(\$50\) in profit.
• \(C(x) \leq 11\) on \([2,8]\). This means anywhere between 2000 and 8000 pens can be produced and the cost will not exceed \(\$1100\).
• \(T(t) > 42\) on \((8-2\sqrt{11}, 8+2\sqrt{11}) \approx (1.37, 14.63)\), which corresponds to between 7:22 AM (1.37 hours after 6 AM) to 8:38 PM (14.63 hours after 6 AM.) However, since the model is valid only for \(t\), \(0 \leq t \leq 12\), we restrict our answer and find it is warmer than \(42^{\circ}\) Fahrenheit from 7:22 AM to 6 PM.
• \(h(t) \geq 250\) on \([10-5\sqrt{2}, 10+5\sqrt{2}] \approx [2.93, 17.07]\). This means the rocket is at least 250 feet off the ground between 2.93 and 17.07 seconds after lift off.
• \(s(t) = -4.9t^2 + 30t + 2\). \(s(t) > 35\) on (approximately) \((1.44, 4.68)\). This means between 1.44 and 4.68 seconds after it is launched into the air, the marble is more than 35 feet off the ground.
• From our previous work \(C(F) = \frac{5}{9}(F - 32)\) so \(50 \leq F \leq 95\) becomes \(10 \leq C \leq 35\).
• \(|x-2| \leq 4\), \([-2,6]\)
• \(|3x+1| \leq 2\), \(\left[-1, \frac{1}{3}\right]\)
• \(|x^2-3| \leq 1\), \([-2, -\sqrt{2} \,] \cup [\sqrt{2}, 2]\)
• \(|x^2 -4| \geq 7\), \((-\infty, -\sqrt{11} \,] \cup [\sqrt{11}, \infty)\)
• Solving \(|S(x) - 42| \leq 3\), and disregarding the negative solutions yields \(\left[\sqrt{\frac{13}{2}}, \sqrt{\frac{15}{2}}\right] \approx [2.550, 2.739]\). The edge length must be within \(2.550\) and \(2.739\) centimeters.

1 See Definition 1.2 in Section 1.1.1 .

2 We will give this property a name in Chapter 3 and revisit this concept then.

3 We have to choose something in each interval. If you don’t like our choices, please feel free to choose different numbers. You’ll get the same sign chart.

4 In this case, we say the line \(\ y=2 x\) is tangent to \(\ y=x^{2}+1\) at (1, 2). Finding tangent lines to arbitrary functions is a fundamental problem solved, in general, with Calculus.

5 The underlying concept of Calculus can be phrased in terms of tolerances, so this is well worth your attention

6 Understanding this type of inequality is really important in Calculus.