Lesson Objectives
• Demonstrate an understanding of inequalities and interval notation
• Learn how to solve a quadratic inequality

## How to Solve Quadratic Inequalities

In this lesson, we will learn how to solve quadratic inequalities.
• Replace the inequality symbol with an equality symbol and solve the equation
• The solutions will give us the boundary points or endpoints
• These endpoints separate the solution regions from the non-solution regions
• Use the endpoints to set up intervals on the number line
• Substitute a test number from each interval into the original inequality
• If the test number makes the inequality true, the region that includes that test number is in the solution set
• If a test number makes the inequality false, the region that includes that test number is not in the solution set
• The endpoints are included for a non-strict inequality and excluded for a strict inequality
Let's take a look at an example.
Example 1: Solve each inequality $$x^2 - 5x - 6 > 0$$ Step 1) We will change this inequality into an equality and solve for x. $$x^2 - 5x - 6=0$$ $$(x - 6)(x + 1)=0$$ $$x=-1, 6$$ Step 2) Use the solutions to the equation in step 1 as the endpoints. These endpoints will allow us to set up intervals on the number line.
On the horizontal number line, we can set up three intervals: We have split the number line up into three intervals. One interval contains any number less than -1 and is labeled with the letter "A". The next interval is between -1 and 6. This interval is labeled with the letter "B". Lastly, we have an interval that consists of any number that is greater than 6. This interval is labeled with the letter "C".
Step 3) Substitute a test number from each interval into the original inequality.
Let's begin with interval A, we can choose any value that is less than -1. Let's choose -2, we will plug this in for x in the original inequality. $$(-2)^2 - 5(-2) - 6 > 0$$ $$4 + 10 - 6 > 0$$ $$\require{color}8 > 0 \hspace{.25em}\color{green}{✔}$$ Since the test number of (-2) produces a true statement, we know values that are less than -1 will satisfy the inequality. Let's now choose a number in interval B. Since 0 is easy to work with, we will plug this in for x in the original inequality. $$(0)^2 - 5(0) - 6 > 0$$ $$-6 > 0 \hspace{.25em}\color{red}{✖}$$ Since the test number of 0 produces a false statement, we know values that are between -1 and 6 will not satisfy the inequality. Let's now choose a number in interval C. Let's choose 7, we will plug this in for x in the original inequality. $$(7)^2 - 5(7) - 6 > 0$$ $$49 - 35 - 6 > 0$$ $$8 > 0 \hspace{.25em}\color{green}{✔}$$ Since the test number of 7 produces a true statement, we know values that are greater than 6 will satisfy the inequality. Now we can put together a final answer. We know the endpoints are excluded since we have a strict inequality. We can state our answer as: $$x < -1 \hspace{.2em}or\hspace{.2em}x > 6$$ Our solution in interval notation: $$(-\infty, -1) ∪ (6,∞)$$ We can also graph the interval on the number line:

#### Skills Check:

Example #1

Solve each inequality. $$12x^2 + x - 30 ≤ 11x^2$$

A
$$-6 ≤ x ≤ 5$$
B
$$-4 ≤ x ≤ \frac{1}{2}$$
C
$$-7 ≤ x ≤ 1$$
D
$$x ≤ -2 \hspace{.1em}\text{or}\hspace{.1em}x ≥ 5$$
E
$$x ≤ -1 \hspace{.1em}\text{or}\hspace{.1em}x ≥ 4$$

Example #2

Solve each inequality. $$2x^2 - 2x + 1 > x^2$$

A
$$x < -1 \hspace{.1em}\text{or}\hspace{.1em}x > 1$$
B
$$x < 1 \hspace{.1em}\text{or}\hspace{.1em}x > 1$$
C
$$x > -1$$
D
$$x < -1$$
E
$$-1 < x < 1$$

Example #3

Solve each inequality. $$-x^2 - 3x - 35 > 9x$$

A
$$-9 < x < \frac{2}{3}$$
B
$$x < -9$$
C
$$-7 < x < -5$$
D
$$x > -9$$
E
$$x < 5 \hspace{.1em}\text{or}\hspace{.1em}x > 7$$