Lesson Objectives
• Learn how to solve radical equations with higher-level roots

## How to Solve Radical Equations with Higher-Level Roots

Let's begin by restating our procedure for solving an equation with radicals. We can then apply this procedure to equations with higher-level roots.

### Solving Equations with Radicals

• Isolate one of the radicals
• Raise both sides of the equation to a power equal to the index of the radical
• Repeat the previous two steps if necessary
• Solve the equation
• Check all solutions in the original equation

### Solving Radical Equations with Higher-Level Roots

Example 1: Solve each equation $$\sqrt[3]{2x - 11}- \sqrt[3]{5x + 1}=0$$ Step 1) Isolate one of the radicals.
In this case, we will move the rightmost radical to the right side of the equation. This will isolate both radicals. $$\sqrt[3]{2x - 11}=\sqrt[3]{5x + 1}$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a cube root, this means we want to cube both sides of the equation. $$\left(\sqrt[3]{2x - 11}\right)^3=\left(\sqrt[3]{5x + 1}\right)^3$$ $$2x - 11=5x + 1$$ Step 3) Solve the equation. $$2x - 11=5x + 1$$ $$-3x=12$$ $$x=-4$$ Step 4) Check all solutions in the original equation. $$\sqrt[3]{2x - 11}- \sqrt[3]{5x + 1}=0$$ $$\sqrt[3]{2(-4) - 11}- \sqrt[3]{5(-4) + 1}=0$$ $$\sqrt[3]{-19}- \sqrt[3]{-19}=0$$ $$0=0 \hspace{.2em}\color{green}{✔}$$

#### Skills Check:

Example #1

Solve each equation. $$\sqrt[3]{2x + 3}=5$$

A
$$x=-\frac{1}{8}, 7$$
B
$$x=-25, 13$$
C
$$x=-5$$
D
$$x=61$$
E
$$x=-1$$

Example #2

Solve each equation. $$\sqrt{2 \sqrt{x - 3}}=\sqrt{4 - x}$$

A
$$x=-13, 17$$
B
$$x=-\frac{1}{8}, 7$$
C
$$x=-3, 7$$
D
$$x=1 - 5 \sqrt{3}$$
E
$$x=6 - 2 \sqrt{2}$$

Example #3

Solve each equation. $$\sqrt[4]{2x - 9}=0$$

A
$$x=-\frac{1}{2}$$
B
$$x=\frac{9}{2}$$
C
$$x=-3, 7$$
D
$$x=-2, 1$$
E
$$x=-\frac{1}{4}, 9$$

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