Lesson Objectives

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

## How to Solve Radical Inequalities with Variables Outside of the Radical

As we discussed in the last lesson, 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. Here, we will look at a more challenging type of problem, one where variables occur outside of the radical.

Set your radicand (x - 3) as greater than or equal to zero and solve the inequality: $$x - 3 ≥ 0$$ $$x ≥ 3$$ Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt{x - 3}=2x - 12$$ $$x=7$$ 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 3, where no values will work due to the domain restriction. A region that is between 3 and 7, and a region that is greater than 7. For the numbers 3 and 7, 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 3) works due to the domain restriction. For interval B we can test 4: $$\sqrt{4 - 3}< 2(4) - 12$$ $$1 < -4$$ This statement is false, so we know that numbers less than or equal to 7 won't work as a solution. Let's try a number from region C, where values are larger than 7. Let's try a value of 12. $$\sqrt{12 - 3}< 2(12) - 12$$ $$3 < 12$$ This statement is true, so we know that numbers that are greater than 7 will work as a solution. 7 itself is excluded here since we have a non-strict inequality. We will report our answer as: $$x > 7$$ $$(7, \infty)$$

### Solving a Radical Inequality:

- Find any domain restrictions
- Replace the inequality symbol with an equality symbol
- Solve the resulting equation
- The solution gives us the critical values or boundaries

- Split the number line up into intervals based on the critical values and domain restrictions
- Test numbers in each interval to obtain your solution set

Set your radicand (x - 3) as greater than or equal to zero and solve the inequality: $$x - 3 ≥ 0$$ $$x ≥ 3$$ Step 2) Replace the inequality symbol with an equality symbol and solve the resulting equation: $$\sqrt{x - 3}=2x - 12$$ $$x=7$$ 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 3, where no values will work due to the domain restriction. A region that is between 3 and 7, and a region that is greater than 7. For the numbers 3 and 7, 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 3) works due to the domain restriction. For interval B we can test 4: $$\sqrt{4 - 3}< 2(4) - 12$$ $$1 < -4$$ This statement is false, so we know that numbers less than or equal to 7 won't work as a solution. Let's try a number from region C, where values are larger than 7. Let's try a value of 12. $$\sqrt{12 - 3}< 2(12) - 12$$ $$3 < 12$$ This statement is true, so we know that numbers that are greater than 7 will work as a solution. 7 itself is excluded here since we have a non-strict inequality. We will report our answer as: $$x > 7$$ $$(7, \infty)$$

#### Skills Check:

Example #1

Solve each inequality. $$\sqrt{3x - 5}≥ 4x - 10$$

Please choose the best answer.

A

$$x ≥ \frac{5}{3}$$

B

$$x ≤ 3$$

C

$$\frac{5}{3}≤ x ≤ 3$$

D

$$x ≤ -3$$

E

$$x ≤ \frac{5}{3}\hspace{.1em}\text{or}\hspace{.1em}x ≥ 3$$

Example #2

Solve each inequality. $$\sqrt[3]{5x + 2}> x - 2$$

Please choose the best answer.

A

$$x > 5$$

B

$$x < 5$$

C

$$-\frac{5}{2}< x < 5$$

D

$$-\frac{5}{2}< x < 2$$

E

$$x ≥ 2$$

Example #3

Solve each inequality. $$\sqrt{x^2 +x-2}≥ x + 3$$

Please choose the best answer.

A

$$x ≤ -\frac{11}{5}$$

B

$$x ≥ -\frac{3}{5}$$

C

$$-\frac{3}{5}≤ x ≤ \frac{11}{5}$$

D

$$x ≤ -\frac{3}{5}$$

E

$$x ≥ -\frac{11}{5}$$

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