Lesson Objectives

- Learn how to solve radical equations with higher-level roots
- Learn how to solve radical equations with nested 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.

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, which 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. $$\sqrt[3]{2x - 11}- \sqrt[3]{5x + 1}=0$$ $$\sqrt[3]{2(-4) - 11}- \sqrt[3]{5(-4) + 1}=0$$ $$\sqrt[3]{-19}- \sqrt[3]{-19}=0$$ $$0=0 \hspace{.2em}\color{green}{✔}$$ Example 2: Solve each equation. $$\sqrt[4]{2x-7}=2$$ Step 1) Isolate one of the radicals.

In this case, our radical is isolated on the left side of the equation. $$\sqrt[4]{2x-7}=2$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.

In this case, we have a fourth root, which means we want to raise both sides of the equation to the 4th power. $$(\sqrt[4]{2x-7})^4=(2)^4$$ $$2x - 7=16$$ Step 3) Solve the equation. $$2x - 7=16$$ $$2x=23$$ $$x=\frac{23}{2}$$ Step 4) Check all solutions in the original equation. $$\sqrt[4]{2 \cdot\frac{23}{2}-7}=2$$ $$\sqrt[4]{23 - 7}=2$$ $$\sqrt[4]{16}=2$$ $$2=2 \hspace{.2em}\color{green}{✔}$$

In this case, one of our radicals is isolated on the right side of the equation. $$\sqrt{x}+ 1=\sqrt{7\sqrt{x}- 5}$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.

In this case, we have a square root, which means we want to square both sides of the equation. $$(\sqrt{x}+ 1)^2=\left(\sqrt{7\sqrt{x}- 5}\right)^2$$ $$x + 2\sqrt{x}+ 1=7\sqrt{x}- 5$$ $$x - 5\sqrt{x}+ 6=0$$ Since a radical remains, we will repeat steps 1 and 2:

Step 1) Isolate one of the radicals. $$x + 6=5\sqrt{x}$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.

In this case, we have a square root, which means we want to square both sides of the equation. $$(x + 6)^2=(5\sqrt{x})^2$$ $$x^2 + 12x + 36=25x$$ Step 3) Solve the equation. $$x^2 + 12x + 36=25x$$ $$x^2 - 13x + 36=0$$ $$(x - 9)(x - 4)=0$$ $$x=9, 4$$ Step 4) Check all solutions in the original equation. $$\sqrt{9}+ 1=\sqrt{7\sqrt{9}- 5}$$ $$4=4 \hspace{.2em}\color{green}{✔}$$ $$\sqrt{4}+ 1=\sqrt{7\sqrt{4}- 5}$$ $$3=3 \hspace{.2em}\color{green}{✔}$$

### 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, which 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. $$\sqrt[3]{2x - 11}- \sqrt[3]{5x + 1}=0$$ $$\sqrt[3]{2(-4) - 11}- \sqrt[3]{5(-4) + 1}=0$$ $$\sqrt[3]{-19}- \sqrt[3]{-19}=0$$ $$0=0 \hspace{.2em}\color{green}{✔}$$ Example 2: Solve each equation. $$\sqrt[4]{2x-7}=2$$ Step 1) Isolate one of the radicals.

In this case, our radical is isolated on the left side of the equation. $$\sqrt[4]{2x-7}=2$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.

In this case, we have a fourth root, which means we want to raise both sides of the equation to the 4th power. $$(\sqrt[4]{2x-7})^4=(2)^4$$ $$2x - 7=16$$ Step 3) Solve the equation. $$2x - 7=16$$ $$2x=23$$ $$x=\frac{23}{2}$$ Step 4) Check all solutions in the original equation. $$\sqrt[4]{2 \cdot\frac{23}{2}-7}=2$$ $$\sqrt[4]{23 - 7}=2$$ $$\sqrt[4]{16}=2$$ $$2=2 \hspace{.2em}\color{green}{✔}$$

### Solving Radical Equations with Nested Roots

Example 3: Solve each equation. $$\sqrt{x}+ 1=\sqrt{7\sqrt{x}- 5}$$ Step 1) Isolate one of the radicals.In this case, one of our radicals is isolated on the right side of the equation. $$\sqrt{x}+ 1=\sqrt{7\sqrt{x}- 5}$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.

In this case, we have a square root, which means we want to square both sides of the equation. $$(\sqrt{x}+ 1)^2=\left(\sqrt{7\sqrt{x}- 5}\right)^2$$ $$x + 2\sqrt{x}+ 1=7\sqrt{x}- 5$$ $$x - 5\sqrt{x}+ 6=0$$ Since a radical remains, we will repeat steps 1 and 2:

Step 1) Isolate one of the radicals. $$x + 6=5\sqrt{x}$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.

In this case, we have a square root, which means we want to square both sides of the equation. $$(x + 6)^2=(5\sqrt{x})^2$$ $$x^2 + 12x + 36=25x$$ Step 3) Solve the equation. $$x^2 + 12x + 36=25x$$ $$x^2 - 13x + 36=0$$ $$(x - 9)(x - 4)=0$$ $$x=9, 4$$ Step 4) Check all solutions in the original equation. $$\sqrt{9}+ 1=\sqrt{7\sqrt{9}- 5}$$ $$4=4 \hspace{.2em}\color{green}{✔}$$ $$\sqrt{4}+ 1=\sqrt{7\sqrt{4}- 5}$$ $$3=3 \hspace{.2em}\color{green}{✔}$$

#### Skills Check:

Example #1

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

Please choose the best answer.

A

$$x=-\frac{1}{8}, 7$$

B

$$x=-25, 13$$

C

$$x=-5$$

D

$$x=61$$

E

$$x=-1$$

Example #2

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

Please choose the best answer.

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

Please choose the best answer.

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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