How to Solve Quadratic Inequalities

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A quadratic inequality has two roots, or x-intercepts because it has an $x^2$ term. You get a parabola when you plot this inequality on a coordinate plane. Finding the values of $x$ that make the inequality true is what it means to solve an inequality. By drawing the inequality, you can show these solutions in algebra or on a number line or coordinate plane.

Follow these steps to figure out how to solve a quadratic inequality:

Step1:
Solve the inequality as if it were an equation.
Step2:
The boundary points for the solution to the inequality are the real answers to the equation.
Step3:
If the original inequality did not include equality, make the boundary points open circles. If it did, make the boundary points solid circles.
Step4:
Choose points from each of the areas that the boundary points make. Change these "test points" in the original inequality.
Step5:
If a test point meets the original inequality, then the area that contains that test point is part of the solution.
Step6:
Show the solution both as a picture and as a solution set.

Example:

Solve $(x \ + \ 1) \ (2x \ - \ 4) \ < \ 0$ .

Figure out $(x \ + \ 1) \ (2x \ - \ 4) \ = \ 0$ . Using the zero-product property,

$(x \ + \ 1) \ = \ 0$ or $(2x \ - \ 4) \ = \ 0$ $⇒ \ x \ = \ -1 \ , \ x \ = \ 2$

Make the points of the border. Here, the boundary points are open circles because the original inequality doesn't include equality.

Points of the border:

Quadratic Inequality

There are now three different areas:

Quadratic Inequality2

Choose test points from these three areas: $x \ = \ -3 \ , \ x \ = \ 0 \ , \ x \ = \ 3$

Check to see if the test points meet the first inequality.

$x \ = \ -3 \ ⇒ \ (-3 \ + \ 1) \ (2(-3) \ - \ 4) \ < \ 0 \ ⇒ \ (-2)(-10) \ < \ 0 \ ⇒ \ 20 \ < \ 0$
$x \ = \ 0 \ ⇒ \ (0 \ + \ 1) \ (2(0) \ - \ 4) \ < \ 0 \ ⇒ \ (1)(-4) \ < \ 0 \ ⇒ \ -4 \ < \ 0$
$x \ = \ 3 \ ⇒ \ (3 \ + \ 1) \ (2(3) \ - \ 4) \ < \ 0 \ ⇒ \ (4)(2) \ < \ 0 \ ⇒ \ 8 \ < \ 0$

Since $x \ = \ 0$ solves the original problem, the area $-1 \ < \ x \ < \ 2$ is part of the answer. Since $x \ = \ -3$ doesn't solve the original problem, the area $x \ < \ -1$ isn't part of the answer. Since $x \ = \ 3$ doesn't solves the original problem, the area $x \ > \ 2$ isn't part of the answer. Now show the solution in a graphic form and a solution set.

The graphic form:

Quadratic Inequality3

The form of the solution set: $\{x \ | \ -1 \ < \ x \ < \ 2\}$

Free printable Worksheets

Exercises for Solving Quadratic Inequalities

1) Solve: $(2x \ + \ 2)(3x \ - \ 6) \ > \ 0$

2) Solve: $(2x \ + \ 6)(3x \ - \ 12) \ < \ 0$

3) Solve: $(-2x \ + \ 6)(4x \ + \ 4) \ < \ 0$

4) Solve: $2x^2 \ - \ 10x \ + \ 8 \ < \ 0$

5) Solve: $3x^2 \ + \ 12x \ + \ 9 \ > \ 0$

6) Solve: $x^2 \ - \ 1 \ > \ 0$

7) Solve: $x^2 \ - \ 4 \ < \ 0$

8) Solve: $(2x \ + \ 2)(3x \ - \ 6) \ < \ 0$

9) Solve: $x^2 \ - \ x \ - \ 12 \ > \ 0$

10) Solve: $(-x \ - \ 1)(x \ + \ 3) \ > \ 0$

Answers

1) Solve: $(2x \ + \ 2)(3x \ - \ 6) \ > \ 0$

$\color{red}{x_1 \ = \ -1, \ x_2 \ = \ 2}$
Quadratic Inequalities
$\color{red}{x \ = \ -2, \ x \ = \ 0, x \ = \ 3}$
So, $\color{red}{⇒ \ x \ < \ -1, \ x \ > \ 2}$ is the solution.
Solution set: $\color{red}{\{x \ | \ x \ < \ -1, \ x \ > \ 2\}}$

2) Solve: $(2x \ + \ 6)(3x \ - \ 12) \ < \ 0$

$\color{red}{x_1 \ = \ -3, \ x_2 \ = \ 4}$
Quadratic Inequalities2
$\color{red}{x \ = \ -4, \ x \ = \ 0, x \ = \ 5}$
So, $\color{red}{⇒ \ -3 \ < \ x \ < \ 4}$ is the solution.
Solution set: $\color{red}{\{x \ | \ -3 \ < \ x \ < \ 4\}}$

3) Solve: $(-2x \ + \ 6)(4x \ + \ 4) \ < \ 0$

$\color{red}{x_1 \ = \ 3, \ x_2 \ = \ -1}$
Quadratic Inequalities3

$\color{red}{x \ = \ -2, \ x \ = \ 0, x \ = \ 4}$
So, $\color{red}{⇒ \ x \ < \ -1, \ x \ > \ 3}$ is the solution.
Solution set: $\color{red}{\{x \ | \ x \ < \ -1, \ x \ > \ 3\}}$

4) Solve: $2x^2 \ - \ 10x \ + \ 8 \ < \ 0$

$\color{red}{x_1 \ = \ 1, \ x_2 \ = \ 4}$
Quadratic Inequalities4 $\color{red}{x \ = \ 0, \ x \ = \ 2, x \ = \ 5}$
So, $\color{red}{⇒ \ 1 \ < \ x \ < \ 4}$ is the solution.
Solution set: $\color{red}{\{x \ | \ 1 \ < \ x \ < \ 4\}}$

5) Solve: $3x^2 \ + \ 12x \ + \ 9 \ > \ 0$

$\color{red}{x_1 \ = \ -1, \ x_2 \ = \ -3}$
Quadratic Inequalities5
$\color{red}{x \ = \ -4, \ x \ = \ -2, x \ = \ 0}$
So, $\color{red}{⇒ \ x \ < \ -3, \ x \ > \ 4}$ is the solution.
Solution set: $\color{red}{\{x \ | \ x \ < \ -3, \ x \ > \ 4\}}$

6) Solve: $x^2 \ - \ 1 \ > \ 0$

$\color{red}{x_1 \ = \ -1, \ x_2 \ = \ 1}$
Quadratic_Inequalities6 $\color{red}{x \ = \ -2, \ x \ = \ 0, x \ = \ 2}$
So, $\color{red}{⇒ \ x \ < \ -1, \ x \ > \ 1}$ is the solution.
Solution set: $\color{red}{\{x \ | \ x \ < \ -1, \ x \ > \ 1\}}$

7) Solve: $x^2 \ - \ 4 \ < \ 0$

$\color{red}{x_1 \ = \ -2, \ x_2 \ = \ 2}$
Quadratic Inequalities7 $\color{red}{x \ = \ -3, \ x \ = \ 0, x \ = \ 3}$
So, $\color{red}{⇒ \ -2 \ < \ x \ < \ 2}$ is the solution.
Solution set: $\color{red}{\{x \ | \ -2 \ < \ x \ < \ 2\}}$

8) Solve: $(2x \ + \ 2)(3x \ - \ 6) \ < \ 0$

$\color{red}{x_1 \ = \ -1, \ x_2 \ = \ 2}$
Quadratic Inequalities
$\color{red}{x \ = \ -2, \ x \ = \ 0, x \ = \ 3}$
So, $\color{red}{⇒ \ -1 \ < \ x \ < \ 2}$ is the solution.
Solution set: $\color{red}{\{x \ | \ -1 \ < \ x \ < \ 2\}}$

9) Solve: $x^2 \ - \ x \ - \ 12 \ > \ 0$

$\color{red}{x_1 \ = \ -3, \ x_2 \ = \ 4}$
Quadratic Inequalities2
$\color{red}{x \ = \ -4, \ x \ = \ 0, x \ = \ 5}$
So, $\color{red}{⇒ \ x \ < \ -3, \ x \ > \ 4}$ is the solution.
Solution set: $\color{red}{\{x \ | \ x \ < \ -3, \ x \ > \ 4\}}$

10) Solve: $(-x \ - \ 1)(x \ + \ 3) \ > \ 0$

$\color{red}{x_1 \ = \ -1, \ x_2 \ = \ -3}$
Quadratic Inequalities5
$\color{red}{x \ = \ -4, \ x \ = \ -2, x \ = \ 0}$
So, $\color{red}{⇒ \ -3 \ < \ x \ < \ -1}$ is the solution.
Solution set: $\color{red}{\{x \ | \ -3 \ < \ x \ < \ -1\}}$

How to Solve Quadratic Inequalities