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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