Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

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 9.8

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.
Solve: $2 y^{2} + y = 15 2 y^{2} + y = 15$ .
If you missed this problem, review Example 6.45.
Solve $\frac{1}{x^{2} + 2 x - 8} > 0 \frac{1}{x^{2} + 2 x - 8} > 0$
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 ax² + 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:

\begin{matrix} a x^{2} + b x + c < 0 & a x^{2} + b x + c \leq 0 \\ a x^{2} + b x + c > 0 & a x^{2} + b x + c \geq 0 \end{matrix} \begin{matrix} a x^{2} + b x + c < 0 & a x^{2} + b x + c \leq 0 \\ a x^{2} + b x + c > 0 & a x^{2} + b x + c \geq 0 \end{matrix}

The graph of a quadratic function f(x) = ax² + bx + c = 0 is a parabola. When we ask when is ax² + 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 ax² + bx + c > 0, we are asking when is f(x) > 0. We want to know when the parabola is above the y-axis.

The first graph is an upward facing parabola, f of x, on an x y-coordinate plane. To the left of the function, f of x is greater than 0. Between the x-intercepts, f of x is less than 0. To the right of the function, f of x is greater than 0. The second graph is a downward-facing parabola, f of x, on an x y coordinate plane. To the left of the function, f of x is less than 0. Between the x-intercepts, f of x is greater than 0. To the right of the function, f of x is less than 0.

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.

Show solution

Solution

The figure is a table with 3 columns. The first column is Step 1: Write the quadratic inequality in standard form. The second column says the inequality is in standard form. The third column says x squared minus 6 times x plus 8 less than 0.

The figure is a table with 3 columns. The first column says Step 2-Graph the function f of x equals a times x squared plus b times x plus c using properties or transformations. The second column gives instructions and the third column shows the work for step 3 as follows. We will graph using properties. The function is f of x equals x squared minus 6 times x plus 8 where a equals 1, b equals negative 6, and c equals 8. Look at a in the function f of x equals x squared minus 6 times x plus 8. Since a is positive, the parabola opens upward. The equation of the axis of symmetry is the line x equals negative b divided by 2 times a, so x equals negative negative 6 divided by 2 times 1. X equals 3. The axis of symmetry is the line x equals 3. The vertex is on the axis of symmetry. Substitute x equals 3 into the function, so f of 3 equals 3 squared minus 6 times 3 plus 8. F of 3 equals negative 1, so the vertex is (3, negative 1). We find f of 0 in order to find the y-intercept, so f of 0 equals 0 squared minus 6 times 0 plus 8. F of 0 equals 8, so the y intercept is (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x equals 6. Point symmetric to y-intercept is (6, 8). We solve f of x equals 0 in order to find the x-intercepts. We can solve this quadratic equation by factoring. 0 equals x squared minus 6 times x plus 8, 0 equals the quantity x minus 2 times the quantity x minus 4, x equals 2 or x equals 4. The x-intercepts are (2, 0) and (4, 0). We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these 5 points to sketch the parabola shown that is upward-facing with the points found through this process.

The figure is a table with 3 columns. The first column says Step 3- Determine the solution from the graph. The second column gives instructions. X squared minus 6 x plus 8 less than 0. The inequality asks for the values of x which make the function less than 0. Which values of x make the parabola below the x-axis. We do not include the values 2, 4 as the inequality is strictly less than. The third column says The solution, in interval notation, is (2, 4).

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 $x^{2} - 8 x + 12 \geq 0 x^{2} - 8 x + 12 \geq 0$ graphically and ⓑ write the solution in interval notation.

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

How To

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 $− x^{2} - 8 x - 12 \leq 0 − x^{2} - 8 x - 12 \leq 0$ graphically. Write the solution in interval notation.

Show solution

Solution

The quadratic inequality in standard form.		$- x^{2} - 8 x - 12 \leq 0 - x^{2} - 8 x - 12 \leq 0$
Graph the function $f (x) = − x^{2} - 8 x - 12 f (x) = − x^{2} - 8 x - 12$ .		The parabola opens downward.
Find the line of symmetry.		$x = - \frac{b}{2 a}$ $x = - \frac{- 8}{2 (−1)}$ $x = −4$
Find the vertex.		$f (x) = − x^{2} - 8 x - 12$ $f (−4) = − {(−4)}^{2} - 8 (−4) - 12 f (−4) = − {(−4)}^{2} - 8 (−4) - 12$ $f (−4) = −16 + 32 - 12 f (−4) = −16 + 32 - 12$ $f (−4) = 4 f (−4) = 4$ Vertex $(−4, 4) (−4, 4)$
Find the x-intercepts. Let $f (x) = 0 f (x) = 0$ .		$f (x) = − x^{2} - 8 x - 12$ $0 = − x^{2} - 8 x - 12$
Factor. Use the Zero Product Property.		$0 = −1 (x + 6) (x + 2)$ $x = −6 x = −2$
Graph the parabola.		x-intercepts $(−6, 0), (−2, 0) (−6, 0), (−2, 0)$
Determine the solution from the graph. We include the x-intercepts as the inequality is “less than or equal to.”		$(− \infty, − 6] \cup [− 2, \infty) (− \infty,$

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 $− x^{2} + 10 x - 16 \leq 0 − x^{2} + 10 x - 16 \leq 0$ 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 $x^{2} - x - 12 \geq 0 x^{2} - x - 12 \geq 0$ algebraically. Write the solution in interval notation.

Show solution

Solution

This figure is a table giving the instructions for solving x squared minus x minus 12 greater than or equal to 0 algebraically. It consists of 3 columns where the instructions are given in the first column, the explanation in the second, and the work in the third. Step 1 is to write the quadratic inequality in standard form. The quadratic inequality in already in standard form, so x squared minus x minus 12 greater than or equal to 0.

Step 3 is to use the critical points to divide the number line into intervals. Use negative 3 and 4 to divide the number line into intervals. A number line is shown that includes from left to right the values of negative 3, 0, and 4, with dotted lines on negative 3 and 4.

Step 4 says above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality. X equals negative 5, x equals 0, and x equals 5 are chosen to test. The expression negative x squared minus x minus 12 is given with negative 5 squared minus negative 5 minus 12 underneath, which gives 18. The expression negative x squared minus x minus 12 is given with 0 squared minus 0 minus 12 underneath, which gives 12. The expression negative x squared minus x minus 12 is given with 5 squared minus 5 minus 12 underneath, which gives 8.

For Step 5, determine the intervals where the inequality is correct. Write the solution in interval notation. x squared minus x minus 12 greater than or equal to 0 is shown. The inequality is positive in the first and last intervals and equals 0 at the points negative 4, 3 . The solution, in interval notation, is (negative infinity, negative 3] in union with [4, infinity).

Try It 9.131

Solve $x^{2} + 2 x - 8 \geq 0 x^{2} + 2 x - 8 \geq 0$ algebraically. Write the solution in interval notation.

Try It 9.132

Solve $x^{2} - 2 x - 15 \leq 0 x^{2} - 2 x - 15 \leq 0$ 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 figure shows the expression x squared minus x minus 12 factored to the quantity of x plus 3 times the quantity of x minus 4. The image shows a number line showing dotted lines on negative 3 and 4. It shows the signs of the quantity x plus 3 to be negative, positive, positive, and the signs of the quantity x minus 4 to be negative, negative, positive. Under the number line, it shows the quantity x plus 3 times the quantity x minus 4 with the signs positive, negative, positive.

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

We summarize the steps here.

How To

Solve a quadratic inequality algebraically.

Step 1. Write the quadratic inequality in standard form.
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 $x^{2} + 6 x - 7 \geq 0 x^{2} + 6 x - 7 \geq 0$ algebraically. Write the solution in interval notation.

Show solution

Solution

Write the quadratic inequality in standard form.	$- x^{2} + 6 x - 7 \geq 0 - x^{2} + 6 x - 7 \geq 0$
Multiply both sides of the inequality by $−1 −1$ . Remember to reverse the inequality sign.	$x^{2} - 6 x + 7 \leq 0$
Determine the critical points by solving the related quadratic equation.	$x^{2} - 6 x + 7 = 0$
Write the Quadratic Formula.	$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Then substitute in the values of $a, b, c a, b, c$ .	$x = \frac{- (−6) \pm \sqrt{{(−6)}^{2} - 4 \cdot 1 \cdot (7)}}{2 \cdot 1}$
Simplify.	$x = \frac{6 \pm \sqrt{8}}{2}$
Simplify the radical.	$x = \frac{6 \pm 2 \sqrt{2}}{2}$
Remove the common factor, 2.	$x = \frac{2 (3 \pm \sqrt{2})}{2}$ $x = 3 \pm \sqrt{2}$ $x = 3 + \sqrt{2} x = 3 - \sqrt{2}$ $x \approx 1.6 x \approx 4.4$
Use the critical points to divide the number line into intervals. Test numbers from each interval in the original inequality.
Determine the intervals where the inequality is correct. Write the solution in interval notation.	$− x^{2} + 6 x - 7 \geq 0 − x^{2} + 6 x - 7 \geq 0$ in the middle interval $[3 - \sqrt{2}, 3 + \sqrt{2}] [3 - \sqrt{2},]$

Try It 9.133

Solve $− x^{2} + 2 x + 1 \geq 0 − x^{2} + 2 x + 1 \geq 0$ 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 ax² + 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 ax² + bx + c = 0, $a \neq 0 . a \neq 0 .$

The figure is a table with 3 columns. Column 1 is labeled discriminant, column 2 is Number/Type of solution, and column 3 is Typical Graph. Reading across the columns, if b squared minus 4 times a times c is greater than 0, there will be 2 real solutions because there are 2 x-intercepts on the graph. The image of a typical graph an upward or downward parabola with 2 x-intercepts. If the discriminant b squared minus 4 times a times c is equals to 0, then there is 1 real solution because there is 1 x-intercept on the graph. The image of the typical graph is an upward- or downward-facing parabola that has a vertex on the x-axis instead of crossing through it. If the discriminant b squared minus 4 times a times c is less than 0, there are 2 complex solutions because there is no x-intercept. The image of the typical graph shows an upward- or downward-facing parabola that does not cross the x-axis.

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:

ⓐ $x^{2} - 3 x + 4 > 0 x^{2} - 3 x + 4 > 0$ ⓑ $x^{2} - 3 x + 4 \leq 0 x^{2} - 3 x + 4 \leq 0$

Show solution

Solution

ⓐ

Write the quadratic inequality in standard form.	$- x^{2} - 3 x + 4 > 0$
Determine the critical points by solving the related quadratic equation.	$x^{2} - 3 x + 4 = 0$
Write the Quadratic Formula.	$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
Then substitute in the values of $a, b, c a, b, c$ .	$x = \frac{- (− 3) \pm \sqrt{{(− 3)}^{2} - 4 \cdot 1 \cdot (4)}}{2 \cdot 1} x = \frac{- (− 3) \pm \sqrt{{(− 3)}^{2} - 4 \cdot 1 \cdot (4)}}{2 \cdot 1}$
Simplify.	$x = \frac{3 \pm \sqrt{- 7}}{2} x = \frac{3 \pm \sqrt{- 7}}{2}$
Simplify the radicand.	$x = \frac{3 \pm \sqrt{7} i}{2} x = \frac{3 \pm \sqrt{7} i}{2}$
The complex solutions tell us the parabola does not intercept the x-axis. Also, the parabola opens upward. This tells us that the parabola is completely above the x-axis.	Complex solutions

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 $(− \infty, \infty) . (− \infty, \infty) .$

ⓑ
$\begin{matrix} Write the quadratic inequality in standard form. & x^{2} - 3 x + 4 \leq 0 \\ \begin{matrix} Determine the critical points by solving \\ the related quadratic equation \end{matrix} & x^{2} - 3 x + 4 = 0 \end{matrix} \begin{matrix} Write the quadratic inequality in standard form. & x^{2} - 3 x + 4 \leq 0 \\ \begin{matrix} Determine the critical points by solving \\ the related quadratic equation \end{matrix} & x^{2} - 3 x + 4 = 0 \end{matrix}$

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 $x^{2} - 3 x + 4 \leq 0 . x^{2} - 3 x + 4 \leq 0 .$ 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:
ⓐ $− x^{2} + 2 x - 4 \leq 0 − x^{2} + 2 x - 4 \leq 0$ ⓑ $− x^{2} + 2 x - 4 \geq 0 − x^{2} + 2 x - 4 \geq 0$

Try It 9.136

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

Practice Makes Perfect

Solve Quadratic Inequalities Graphically

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

363.

$x^{2} + 6 x + 5 > 0 x^{2} + 6 x + 5 > 0$

364.

$x^{2} + 4 x - 12 < 0 x^{2} + 4 x - 12 < 0$

365.

$x^{2} + 4 x + 3 \leq 0 x^{2} + 4 x + 3 \leq 0$

366.

$x^{2} - 6 x + 8 \geq 0 x^{2} - 6 x + 8 \geq 0$

367.

$− x^{2} - 3 x + 18 \leq 0 − x^{2} - 3 x + 18 \leq 0$

368.

$− x^{2} + 2 x + 24 < 0 − x^{2} + 2 x + 24 < 0$

369.

$− x^{2} + x + 12 \geq 0 − x^{2} + x + 12 \geq 0$

370.

$− x^{2} + 2 x + 15 > 0 − x^{2} + 2 x + 15 > 0$

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

371.

$x^{2} + 3 x - 4 \geq 0 x^{2} + 3 x - 4 \geq 0$

372.

$x^{2} + x - 6 \leq 0 x^{2} + x - 6 \leq 0$

373.

$x^{2} - 7 x + 10 < 0 x^{2} - 7 x + 10 < 0$

374.

$x^{2} - 4 x + 3 > 0 x^{2} - 4 x + 3 > 0$

375.

$x^{2} + 8 x > - 15 x^{2} + 8 x > - 15$

376.

$x^{2} + 8 x < - 12 x^{2} + 8 x < - 12$

377.

$x^{2} - 4 x + 2 \leq 0 x^{2} - 4 x + 2 \leq 0$

378.

$− x^{2} + 8 x - 11 < 0 − x^{2} + 8 x - 11 < 0$

379.

$x^{2} - 10 x > - 19 x^{2} - 10 x > - 19$

380.

$x^{2} + 6 x < - 3 x^{2} + 6 x < - 3$

381.

$−6 x^{2} + 19 x - 10 \geq 0 −6 x^{2} + 19 x - 10 \geq 0$

382.

$−3 x^{2} - 4 x + 4 \leq 0 −3 x^{2} - 4 x + 4 \leq 0$

383.

$−2 x^{2} + 7 x + 4 \geq 0 −2 x^{2} + 7 x + 4 \geq 0$

384.

$2 x^{2} + 5 x - 12 > 0 2 x^{2} + 5 x - 12 > 0$

385.

$x^{2} + 3 x + 5 > 0 x^{2} + 3 x + 5 > 0$

386.

$x^{2} - 3 x + 6 \leq 0 x^{2} - 3 x + 6 \leq 0$

387.

$− x^{2} + x - 7 > 0 − x^{2} + x - 7 > 0$

388.

$− x^{2} - 4 x - 5 < 0 − x^{2} - 4 x - 5 < 0$

389.

$−2 x^{2} + 8 x - 10 < 0 −2 x^{2} + 8 x - 10 < 0$

390.

$− x^{2} + 2 x - 7 \geq 0 − x^{2} + 2 x - 7 \geq 0$

Writing Exercises

391.

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

392.

Solve $x^{2} + 2 x \geq 8 x^{2} + 2 x \geq 8$ both graphically and algebraically. Which method do you prefer, and why?

393.

Describe the steps needed to solve a quadratic inequality graphically.

394.

Describe the steps needed to solve a quadratic inequality algebraically.

Self Check

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

This figure is a list to assess your understanding of the concepts presented in this section. It has 4 columns labeled I can…, Confidently, With some help, and No-I don’t get it! Below I can…, there is solve quadratic inequalities graphically and solve quadratic inequalities algebraically. The other columns are left blank for you to check you understanding.

ⓑ 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?