Lesson Objectives

- Demonstrate the ability to solve a radical equation
- Learn how to solve a radical inequality

## How to Solve a Radical Inequality

Solving radical inequalities can be quite tedious depending on the type of problem that you encounter. The method that is described below is not necessarily the fastest, especially if you have an easy problem. The benefit of this method is that it will work for any scenario.

Let's consider a very simple example. $$-5 < 4 \hspace{.15em}\text{true}$$ Let's square both sides: $$25 < 16 \hspace{.15em}\text{false}$$ We can see that we must be extra careful when squaring both sides of an inequality. To avoid these types of issues, we will use a different approach and work with the related equation.

Set your radicand (2x + 12) as greater than or equal to zero and solve the inequality: $$2x + 12 ≥ 0$$ $$x ≥ -6$$ Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$-5\sqrt{2x + 12}+ 2=-18$$ $$x=2$$ This solution gives us a boundary. We can use it to split the number line up into three intervals. A region that is less than -6, where no values will work due to the domain restriction. A region that is between -6 and 2, and a region that is greater than 2. For the numbers -6 and 2, we will consider them separately. Step 3) Test numbers in each interval to obtain your solution set:

We know that nothing in interval A (values less than -6) works due to the domain restriction.

For interval B, we can test 0: $$-5\sqrt{2(0) + 12}+ 2 > -18$$ $$-5\sqrt{12}+ 2 > -18$$ $$-10\sqrt{3}+ 2 > -18$$ Note: $$-10\sqrt{3}+ 2 ≈ -15.32$$ Since this statement is true, we can conclude that any value between -6 and 2 will work. For interval C, we can test 12: $$-5\sqrt{2(12) + 12}+ 2 > -18$$ $$-5\sqrt{36}+ 2 > -18$$ $$-5(6)+ 2 > -18$$ $$-30 + 2 > -18$$ $$-28 > -18$$ Since this statement is false, we can conclude that any value larger than 2 will not work. Lastly, we can consider our critical values of 2 and -6. 2 is the value that solves the related equation. It will be excluded here because we have a strict inequality. -6 will actually work as a solution to the inequality and will be included in our solution: $$-5\sqrt{2(-6) + 12}+ 2 > -18$$ $$2 > -18$$ We will report our answer as: $$-6 ≤ x < 2$$ $$[-6, 2)$$ Example 2: Solve each inequality. $$\sqrt{5x + 1}+ \sqrt{x + 6}< 7$$ Step 1) Find any domain restrictions:

Set your radicands (5x + 1) and (x + 6) each as greater than or equal to zero and solve the inequalities: $$5x + 1 ≥ 0$$ $$x ≥ -\frac{1}{5}$$ $$x + 6 ≥ 0$$ $$x ≥ -6$$ Since x ≥ -1/5 is more restrictive, we will reject any solution that is less than -1/5.

Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt{5x + 1}+ \sqrt{x + 6}=7$$ $$x=3$$ This solution gives us a boundary. We can use it to split the number line up into three intervals. A region that is less than -1/5, where no values will work due to the domain restriction. A region that is between -1/5 and 3, and a region that is greater than 3. For the numbers -1/5 and 3, we will consider them separately. Step 3) Test numbers in each interval to obtain your solution set:

We know that nothing in interval A (values less than -1/5) works due to the domain restriction.

For interval B, we can test 0: $$\sqrt{5(0) + 1}+ \sqrt{0 + 6}< 7$$ $$\sqrt{1}+ \sqrt{6}< 7$$ $$1 + \sqrt{6}< 7$$ Note: $$1 + \sqrt{6}≈ 3.45$$ Since this statement is true, we can conclude that any value between -1/5 and 3 will work. For interval C, we can test 4: $$\sqrt{5(4) + 1}+ \sqrt{4 + 6}< 7$$ $$\sqrt{21}+ \sqrt{10}< 7$$ Note: $$\sqrt{21}+ \sqrt{10}≈ 7.74$$ Since this statement is false, we can conclude that any value larger than 3 will not work. Lastly, we can consider our critical values of -1/5 and 3. 3 is the value that solves the related equation. It will be excluded here because we have a strict inequality. -1/5 will actually work as a solution to the inequality and will be included in our solution: $$\sqrt{5\left(-\frac{1}{5}\right) + 1}+ \sqrt{-\frac{1}{5}+ 6}< 7$$ $$\sqrt{-\frac{1}{5}+ 6}< 7$$ $$\sqrt{\frac{29}{5}}< 7$$ $$\frac{\sqrt{145}}{5}< 7$$ Note: $$\frac{\sqrt{145}}{5}≈ 2.41$$ We will report our answer as: $$-\frac{1}{5}≤ x < 3$$ $$\left[-\frac{1}{5}, 3\right)$$ Example 3: Solve each inequality. $$\sqrt[3]{22x - 7}≤ 5$$ Note: Cubing both sides is probably a better option for this scenario, however, we can still solve this problem using the method shown in this tutorial.

Step 1) Find any domain restrictions:

Since we have a cube root, which is an odd index, we don't have to worry about this step. We can take the cube root of a negative number in the real number system.

Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt[3]{22x - 7}=5$$ $$x=6$$ This solution gives us a boundary. We can use it to split the number line up into two intervals. A region that is less than 6, and a region that is greater than 6. We will consider 6 separately. Step 3) Test numbers in each interval to obtain your solution set:

For interval A, we can test 0: $$\sqrt[3]{22(0) - 7}≤ 5$$ $$\sqrt[3]{-7}≤ 5$$ Note: $$\sqrt[3]{-7}≈ -1.91$$ Since this statement is true, we can conclude that any value that is less than 6 will work. For interval B, we can test 7: $$\sqrt[3]{22(7) - 7}≤ 5$$ $$\sqrt[3]{147}≤ 5$$ Note: $$\sqrt[3]{147}≈ 5.28$$ Since this statement is false, we can conclude that any value larger than 6 will not work. Lastly, we can consider our critical value of 6. 6 is the value that solves the related equation. It will be included here because we have a non-strict inequality. $$\sqrt[3]{22(6) - 7}≤ 5$$ $$\sqrt[3]{125}≤ 5$$ $$5 ≤ 5$$ We will report our answer as: $$x ≤ 6$$ $$(-∞, 6]$$

Let's consider a very simple example. $$-5 < 4 \hspace{.15em}\text{true}$$ Let's square both sides: $$25 < 16 \hspace{.15em}\text{false}$$ We can see that we must be extra careful when squaring both sides of an inequality. To avoid these types of issues, we will use a different approach and work with the related equation.

### Solving a Radical Inequality:

- Find any domain restrictions
- Unless we are told otherwise, we will assume we are working with real numbers
- We can't have a negative radicand with an even index

- Solve the related equation
- Replace the inequality symbol with an equality symbol

- Split the number line up into intervals based on the critical values
- Domain restriction
- Solution(s) of the related equation

- Test numbers in each interval to obtain your solution set
- Consider the critical values separately

Set your radicand (2x + 12) as greater than or equal to zero and solve the inequality: $$2x + 12 ≥ 0$$ $$x ≥ -6$$ Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$-5\sqrt{2x + 12}+ 2=-18$$ $$x=2$$ This solution gives us a boundary. We can use it to split the number line up into three intervals. A region that is less than -6, where no values will work due to the domain restriction. A region that is between -6 and 2, and a region that is greater than 2. For the numbers -6 and 2, we will consider them separately. Step 3) Test numbers in each interval to obtain your solution set:

We know that nothing in interval A (values less than -6) works due to the domain restriction.

For interval B, we can test 0: $$-5\sqrt{2(0) + 12}+ 2 > -18$$ $$-5\sqrt{12}+ 2 > -18$$ $$-10\sqrt{3}+ 2 > -18$$ Note: $$-10\sqrt{3}+ 2 ≈ -15.32$$ Since this statement is true, we can conclude that any value between -6 and 2 will work. For interval C, we can test 12: $$-5\sqrt{2(12) + 12}+ 2 > -18$$ $$-5\sqrt{36}+ 2 > -18$$ $$-5(6)+ 2 > -18$$ $$-30 + 2 > -18$$ $$-28 > -18$$ Since this statement is false, we can conclude that any value larger than 2 will not work. Lastly, we can consider our critical values of 2 and -6. 2 is the value that solves the related equation. It will be excluded here because we have a strict inequality. -6 will actually work as a solution to the inequality and will be included in our solution: $$-5\sqrt{2(-6) + 12}+ 2 > -18$$ $$2 > -18$$ We will report our answer as: $$-6 ≤ x < 2$$ $$[-6, 2)$$ Example 2: Solve each inequality. $$\sqrt{5x + 1}+ \sqrt{x + 6}< 7$$ Step 1) Find any domain restrictions:

Set your radicands (5x + 1) and (x + 6) each as greater than or equal to zero and solve the inequalities: $$5x + 1 ≥ 0$$ $$x ≥ -\frac{1}{5}$$ $$x + 6 ≥ 0$$ $$x ≥ -6$$ Since x ≥ -1/5 is more restrictive, we will reject any solution that is less than -1/5.

Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt{5x + 1}+ \sqrt{x + 6}=7$$ $$x=3$$ This solution gives us a boundary. We can use it to split the number line up into three intervals. A region that is less than -1/5, where no values will work due to the domain restriction. A region that is between -1/5 and 3, and a region that is greater than 3. For the numbers -1/5 and 3, we will consider them separately. Step 3) Test numbers in each interval to obtain your solution set:

We know that nothing in interval A (values less than -1/5) works due to the domain restriction.

For interval B, we can test 0: $$\sqrt{5(0) + 1}+ \sqrt{0 + 6}< 7$$ $$\sqrt{1}+ \sqrt{6}< 7$$ $$1 + \sqrt{6}< 7$$ Note: $$1 + \sqrt{6}≈ 3.45$$ Since this statement is true, we can conclude that any value between -1/5 and 3 will work. For interval C, we can test 4: $$\sqrt{5(4) + 1}+ \sqrt{4 + 6}< 7$$ $$\sqrt{21}+ \sqrt{10}< 7$$ Note: $$\sqrt{21}+ \sqrt{10}≈ 7.74$$ Since this statement is false, we can conclude that any value larger than 3 will not work. Lastly, we can consider our critical values of -1/5 and 3. 3 is the value that solves the related equation. It will be excluded here because we have a strict inequality. -1/5 will actually work as a solution to the inequality and will be included in our solution: $$\sqrt{5\left(-\frac{1}{5}\right) + 1}+ \sqrt{-\frac{1}{5}+ 6}< 7$$ $$\sqrt{-\frac{1}{5}+ 6}< 7$$ $$\sqrt{\frac{29}{5}}< 7$$ $$\frac{\sqrt{145}}{5}< 7$$ Note: $$\frac{\sqrt{145}}{5}≈ 2.41$$ We will report our answer as: $$-\frac{1}{5}≤ x < 3$$ $$\left[-\frac{1}{5}, 3\right)$$ Example 3: Solve each inequality. $$\sqrt[3]{22x - 7}≤ 5$$ Note: Cubing both sides is probably a better option for this scenario, however, we can still solve this problem using the method shown in this tutorial.

Step 1) Find any domain restrictions:

Since we have a cube root, which is an odd index, we don't have to worry about this step. We can take the cube root of a negative number in the real number system.

Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt[3]{22x - 7}=5$$ $$x=6$$ This solution gives us a boundary. We can use it to split the number line up into two intervals. A region that is less than 6, and a region that is greater than 6. We will consider 6 separately. Step 3) Test numbers in each interval to obtain your solution set:

For interval A, we can test 0: $$\sqrt[3]{22(0) - 7}≤ 5$$ $$\sqrt[3]{-7}≤ 5$$ Note: $$\sqrt[3]{-7}≈ -1.91$$ Since this statement is true, we can conclude that any value that is less than 6 will work. For interval B, we can test 7: $$\sqrt[3]{22(7) - 7}≤ 5$$ $$\sqrt[3]{147}≤ 5$$ Note: $$\sqrt[3]{147}≈ 5.28$$ Since this statement is false, we can conclude that any value larger than 6 will not work. Lastly, we can consider our critical value of 6. 6 is the value that solves the related equation. It will be included here because we have a non-strict inequality. $$\sqrt[3]{22(6) - 7}≤ 5$$ $$\sqrt[3]{125}≤ 5$$ $$5 ≤ 5$$ We will report our answer as: $$x ≤ 6$$ $$(-∞, 6]$$

#### Skills Check:

Example #1

Solve each inequality. $$\sqrt{2x - 4}+ 9 ≤ 11$$

Please choose the best answer.

A

$$x ≥ 4$$

B

$$2 < x ≤ 4$$

C

$$2 ≤ x ≤ 4$$

D

$$x ≤ 2$$

E

$$x ≤ 2 \hspace{.1em}\text{or}\hspace{.1em}x ≥ 4$$

Example #2

Solve each inequality. $$\sqrt{x + 9}> \sqrt{3 - 2x}$$

Please choose the best answer.

A

$$-\frac{3}{2}≤ x < 2$$

B

$$x > \frac{3}{2}$$

C

$$x ≤ -2$$

D

$$-2 < x ≤ \frac{3}{2}$$

E

$$-2 ≤ x < \frac{3}{2}$$

Example #3

Solve each inequality. $$5 + \sqrt{3x - 8}≥ 7$$

Please choose the best answer.

A

$$x ≤ \frac{8}{3}$$

B

$$x ≥ \frac{8}{3}$$

C

$$-2 ≤ x ≤ 4$$

D

$$x ≥ 4$$

E

$$x ≤ 4$$

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