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

## How to Solve Quadratic & Rational Inequalities

In this lesson, we will learn how to solve quadratic and rational inequalities. Let's begin with quadratic inequalities.

### Quadratic Inequalities

A quadratic inequality is of the form: $$ax^2 + bx + c > 0$$ Where a ≠ 0, and our ">" can be replaced with any inequality symbol.
To solve a quadratic inequality:
• 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: Let's look at another example.
Example 2: Solve each inequality. $$-3x^{2}- 7x + 6 > 0$$ Step 1) We will change this inequality into an equality and solve for x. $$-3x^{2}- 7x + 6=0$$ Let's use the quadratic formula: $$a=-3, b=-7, c=6$$ $$x=\frac{-b \pm \sqrt{b^{2}- 4ac}}{2a}$$ $$x=\frac{-(-7) \pm \sqrt{(-7)^{2}- 4(-3)(6)}}{2(-3)}$$ $$x=\frac{7 \pm 11}{-6}$$ Let's split this into two solutions: $$x=\frac{7 + 11}{-6}=-3$$ $$\text{or}$$ $$x=\frac{7 - 11}{-6}=\frac{2}{3}$$ 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 -3 and is labeled with the letter "A". The next interval is between -3 and 2/3. This interval is labeled with the letter "B". Lastly, we have an interval that consists of any number that is greater than 2/3. 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 -3. Let's choose -4, we will plug this in for x in the original inequality. $$-3(-4)^{2}- 7(-4) + 6 > 0$$ $$-3(16) + 28 + 6 > 0$$ $$-48 + 34 > 0$$ $$-14 > 0 \hspace{.25em}\color{red}{✖}$$ Since the test number of -4 produces a false statement, we know values that are less than -3 will not satisfy the inequality. Let's now choose a number in interval B. Let's choose 0, we will plug this in for x in the original inequality. $$-3(0)^{2}- 7(0) + 6 > 0$$ $$6 > 0 \hspace{.25em}\color{green}{✔}$$ Since the test number of 0 produces a true statement, we know values that are between -3 and 2/3 will satisfy the inequality. Let's now choose a number in interval C. Let's choose 1, we will plug this in for x in the original inequality. $$-3(1)^{2}- 7(1) + 6 > 0$$ $$-3 - 7 + 6 > 0$$ $$-10 + 6 > 0$$ $$-4 > 0 \hspace{.25em}\color{red}{✖}$$ Since the test number of -1 produces a false statement, we know values that are greater than 2/3 will not 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: $$-3 < x < \frac{2}{3}$$ Our solution in interval notation: $$\left(-3, \frac{2}{3}\right)$$ We can also graph the interval on the number line:

### Rational Inequalities

We will also encounter inequalities that involve rational expressions. These inequalities are known as "rational inequalities". To solve a rational inequality:
• Write the inequality so that 0 is on one side and a single fraction is on the other
• Find the numbers that make the numerator or denominator equal to 0
• Use the numbers from the previous step to set up intervals on the number line
• Test a number from each interval to determine if that interval satisfies the inequality
• The endpoints are included for a non-strict inequality and excluded for a strict inequality
• We exclude any number that makes the denominator zero
Let's look at an example.
Example 3: Solve each inequality. $$\frac{-1}{x - 3}> 1$$ Step 1) Write the inequality so that 0 is on one side and a single fraction is on the other. $$\frac{-1}{x - 3}- 1 > 0$$ $$\frac{-1}{x - 3}- \frac{x - 3}{x - 3}> 0$$ $$\frac{-1 - (x - 3)}{x - 3}> 0$$ $$\frac{-1 - x + 3}{x - 3}> 0$$ $$\frac{-x + 2}{x - 3}> 0$$ Step 2) Find the numbers that make the numerator or denominator equal to 0.
Let's set the numerator and denominator equal to zero and solve the equations. $$-x + 2=0$$ $$x=2$$ $$x - 3=0$$ $$x=3$$ At this point, we know that x = 3 can't be in the solution set. This is because it makes the denominator 0 and division by zero is undefined.
Step 3) Use the numbers from the previous step 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 into three intervals. One interval contains any number less than 2 and is labeled with the letter "A". The next interval is between 2 and 3. This interval is labeled with the letter "B". Lastly, we have an interval that consists of any number that is greater than 3. This interval is labeled with the letter "C".
Step 4) Test a number from each interval to determine if that interval satisfies the inequality.
Let's begin with interval A, we will test the number 0 in the original inequality. $$\frac{-1}{0 - 3}> 1$$ $$\frac{-1}{-3}> 1$$ $$\frac{1}{3}> 1 \hspace{.25em}\color{red}{✖}$$ Since the test of the number 0 produces a false statement, we know values that are less than 2 will not satisfy the inequality. Let's choose a number from interval B. We can use 5/2 since it is between 2 and 3. $$\large{\frac{-1}{\frac{5}{2}- 3}}> 1$$ $$\large{\frac{-1}{\frac{5}{2}- \frac{6}{2}}}> 1$$ $$\large{\frac{-1}{-\frac{1}{2}}}> 1$$ $$2 > 1 \hspace{.25em}\color{green}{✔}$$ Since the test of the number 5/2 produces a true statement, we know values that are between 2 and 3 will satisfy the inequality. Let's choose a number from interval C. We can use 4 since it is larger than 3. $$\frac{-1}{4 - 3}> 1$$ $$\frac{-1}{1}> 1$$ $$-1 > 1 \hspace{.25em}\color{red}{✖}$$ Since the test of the number 4 produces a false statement, we know values that are greater than 4 will not satisfy the inequality. We also know the endpoints are excluded since 3 creates a denominator of zero and we have a strict inequality. We can state our answer as: $$2 < x < 3$$ Our solution in interval notation: $$(2,3)$$ We can also graph the interval on the number line: Let's look at another example.
Example 4: Solve each inequality. $$\frac{-5x - 6}{4x + 4}≤ -1$$ Step 1) Write the inequality so that 0 is on one side and a single fraction is on the other. $$\frac{-5x - 6}{4x + 4}+ 1 ≤ 0$$ $$\frac{-5x - 6}{4x + 4}+ \frac{4x + 4}{4x + 4}≤ 0$$ $$\frac{-5x - 6 + 4x + 4}{4x + 4}≤ 0$$ $$\frac{-x - 2}{4x + 4}≤ 0$$ Step 2) Find the numbers that make the numerator or denominator equal to 0. Let's set the numerator and denominator equal to zero and solve the equations. $$-x - 2=0$$ $$x=-2$$ $$4x + 4=0$$ $$4x=-4$$ $$x=-1$$ At this point, we know that x = -1 can't be in the solution set. This is because it makes the denominator 0 and division by zero is undefined.
Step 3) Use the numbers from the previous step 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 into three intervals. One interval contains any number less than -2 and is labeled with the letter "A". The next interval is between -2 and -1. This interval is labeled with the letter "B". Lastly, we have an interval that consists of any number that is greater than -1. This interval is labeled with the letter "C".
Step 4) Test a number from each interval to determine if that interval satisfies the inequality. Let's begin with interval A, we will test the number -4 in the original inequality: $$\frac{-5(-4) - 6}{4(-4) + 4}≤ -1$$ $$\frac{14}{-12}≤ -1$$ $$-\frac{7}{6}≤ -1 \hspace{.25em}\color{green}{✔}$$ Since the test of the number -4 produces a true statement, we know values that are less than -2 will satisfy the inequality. Let's choose a number from interval B. We can use -3/2 since it is between -2 and -1. $$\frac{-5\left(-\frac{3}{2}\right) - 6}{4\left(-\frac{3}{2}\right) + 4}≤ -1$$ $$\frac{\frac{15}{2}- 6}{-6 + 4}≤ -1$$ $$\frac{\frac{15}{2}- 6}{-2}≤ -1$$ Multiply the numerator and denominator of the complex fraction by 2, the LCD of all denominators: $$\frac{15 - 12}{-4}≤ -1$$ $$-\frac{3}{4}≤ -1 \hspace{.25em}\color{red}{✖}$$ Since the test of the number -3/2 produces a false statement, we know values that are between -2 and -1 will not satisfy the inequality. Let's choose a number from interval C. We can use 0 since it is larger than -1. $$\frac{-5(0) - 6}{4(0) + 4}≤ -1$$ $$\frac{-6}{4}≤ -1$$ $$-\frac{3}{2}≤ -1 \hspace{.25em}\color{green}{✔}$$ Since the test of the number 0 produces a true statement, we know values that are greater than -1 will satisfy the inequality. We also know the endpoint of -2 is included since we have a non-strict inequality. Additionally, since -1 creates a denominator of zero, this endpoint is restricted from the answer. We can state our answer as: $$x ≤ -2$$ $$\text{or}$$ $$x > -1$$ Our solution in interval notation: $$(-∞, -2] ∪ (-1, ∞)$$ We can also graph the interval on the number line:

#### Skills Check:

Example #1

Solve each inequality. $$-x^{2}- 10x - 16 ≥ 0$$

Please choose the best answer.

A
$$[-8, -2]$$
B
$$[2, 8]$$
C
$$[8, ∞)$$
D
$$(-∞, -8) ∪ (-2, ∞)$$
E
$$[2, ∞)$$

Example #2

Solve each inequality. $$3x^{2}- 8x + 12 < 8$$

Please choose the best answer.

A
$$\left(-∞, \frac{2}{3}\right] ∪ [2, ∞)$$
B
$$\left(-∞, \frac{2}{3}\right] ∪ (2, ∞)$$
C
$$\left[\frac{2}{3}, 2\right]$$
D
$$\left(\frac{2}{3}, 2\right)$$
E
$$\left(\frac{2}{3}, 2\right) ∪ [3, ∞)$$

Example #3

Solve each inequality. $$\frac{5x + 13}{4x + 12}> 1$$

Please choose the best answer.

A
$$(-3, -1)$$
B
$$(-3, -1]$$
C
$$\left(\frac{1}{3}, 5\right)$$
D
$$(-5, -2) ∪ (3, 4)$$
E
$$(-∞, -3) ∪ (-1, ∞)$$

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