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.

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:

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:

Example 3: Solve each inequality. $$x^2 - 6x + 9 < 0$$ Step 1) Here, we will immediately notice that the trinomial on the left is a perfect square trinomial. This means we can factor the left side into a binomial squared: $$(x - 3)^2 < 0$$ Step 2) A simple analysis of the problem will lead to a result of no solution. It does not matter what is plugged in for x, the squaring operation will always produce either 0 or a positive number. Therefore, there is no solution for this problem.

No Solution

Example 4: Solve each inequality. $$x^2 - 10x + 25 ≤ 0$$ Step 1) Here, we will immediately notice that the trinomial on the left is a perfect square trinomial. This means we can factor the left side into a binomial squared: $$(x - 5)^2 ≤ 0$$ Step 2) In this case, we have a non-strict inequality. Although our left side will never be less than zero, it could be equal to zero. $$(x - 5)^2=0$$ Square Root Property: $$x - 5=0$$ $$x=5$$ This tells us that when x is 5, our inequality will be true. This is due to the fact that an x-value of 5 makes the left side 0.

Example 5: Solve each inequality. $$x^2 - 14x + 49 ≥ 0$$ Step 1) Here, we will immediately notice that the trinomial on the left is a perfect square trinomial. This means we can factor the left side into a binomial squared: $$(x - 7)^2 ≥ 0$$ Step 2) In this case, we have a non-strict inequality. Our left side will always be either 0 or some positive value, therefore, it is true for all real numbers.

All Real Numbers

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

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:

### Quadratic Inequality with a Perfect Square Trinomial

In some cases, we will have a quadratic inequality with a perfect square trinomial. This type of trinomial can be factored into a binomial squared. Recall that the squaring operation produces a result that is non-negative. Let's look at a few examples.Example 3: Solve each inequality. $$x^2 - 6x + 9 < 0$$ Step 1) Here, we will immediately notice that the trinomial on the left is a perfect square trinomial. This means we can factor the left side into a binomial squared: $$(x - 3)^2 < 0$$ Step 2) A simple analysis of the problem will lead to a result of no solution. It does not matter what is plugged in for x, the squaring operation will always produce either 0 or a positive number. Therefore, there is no solution for this problem.

No Solution

Example 4: Solve each inequality. $$x^2 - 10x + 25 ≤ 0$$ Step 1) Here, we will immediately notice that the trinomial on the left is a perfect square trinomial. This means we can factor the left side into a binomial squared: $$(x - 5)^2 ≤ 0$$ Step 2) In this case, we have a non-strict inequality. Although our left side will never be less than zero, it could be equal to zero. $$(x - 5)^2=0$$ Square Root Property: $$x - 5=0$$ $$x=5$$ This tells us that when x is 5, our inequality will be true. This is due to the fact that an x-value of 5 makes the left side 0.

Example 5: Solve each inequality. $$x^2 - 14x + 49 ≥ 0$$ Step 1) Here, we will immediately notice that the trinomial on the left is a perfect square trinomial. This means we can factor the left side into a binomial squared: $$(x - 7)^2 ≥ 0$$ Step 2) In this case, we have a non-strict inequality. Our left side will always be either 0 or some positive value, therefore, it is true for all real numbers.

All Real Numbers

#### Skills Check:

Example #1

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

Please choose the best answer.

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. $$2x^2 - 2x + 1 > x^2$$

Please choose the best answer.

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. $$-x^2 - 3x - 35 > 9x$$

Please choose the best answer.

A

$$-9 < x < \frac{2}{3}$$

B

$$x < -9$$

C

$$-7 < x < -5$$

D

$$x > -9$$

E

$$x < 5 \hspace{.1em}\text{or}\hspace{.1em}x > 7$$

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