Lesson Objectives
- Demonstrate an understanding of radicals
- Learn how to solve an equation with radicals
- Learn how to identify extraneous solutions
How to Solve Equations with Radicals
When we solve equations with radicals, we need a new rule. When both sides of an equation are raised to the same power, all solutions of the original equation are solutions to the new equation. We must be careful here, each solution of the new equation is not necessarily a solution to the original. When working with radical equations, we must always check for extraneous solutions or solutions that do not work in the original equation. To see this in action, let's think about a very simple example.
x = 7
Let's suppose we square both sides of the above equation:
x2 = 49
This new equation has two solutions, x can be 7 or (-7).
(7)2 = 49
(-7)2 = 49
-7 does not work as a solution in our original equation.
-7 = 7 (false)
This is an extraneous solution and occurred from the squaring operation. When we square a number, a positive remains positive, and a negative becomes positive. It stands to reason, when we square both sides of an equation, there will be a loss of information. In other words, we don't know if x was 7 or x was -7 since 49 is the result of either being squared. This information loss is why we must always check our solutions in the original equation after squaring both sides.
Example 1: Solve each equation. $$\require{color}\sqrt{2x - 7}=1$$ Step 1) Isolate one of the radicals.
In this case, our radical is already isolated on the left side of the equation. $$\sqrt{2x - 7}=1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation: $$\left(\sqrt{2x - 7}\right)^2=1^2$$ $$2x - 7=1$$ Step 3) Solve the equation. $$2x - 7=1$$ $$2x=8$$ $$x=4$$ Step 4) Check all solutions in the original equation. $$\sqrt{2x - 7}=1$$ $$\sqrt{2(4) - 7}=1$$ $$\sqrt{8 - 7}=1$$ $$\sqrt{1}=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Example 2: Solve each equation. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ Step 1) Isolate one of the radicals.
We will isolate the leftmost radical: $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation. $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ $$\left(\sqrt{4x + 1}\right)^2=\left(\sqrt{2x + 4}+ 1\right)^2$$ $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ In this case, we still have a radical involved. We will repeat our first two steps.
Step 1) Isolate one of the radicals.
Let's isolate our radical on the right side of the equation: $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ $$2x - 4=2\sqrt{2x + 4}$$ 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. $$\left(2x - 4\right)^2=\left(2\sqrt{2x + 4}\right)^2$$ $$4x^2 - 16x + 16=4(2x + 4)$$ $$4x^2 - 16x + 16=8x + 16$$ $$4x^2 - 24x=0$$ Step 3) Solve the equation. $$4x^2 - 24x=0$$ $$4x(x - 6)=0$$ $$4x=0$$ $$x=0$$ $$x - 6=0$$ $$x=6$$ Step 4) Check all solutions in the original equation.
We have two proposed solutions, x = 6 and x = 0. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(0) + 1}- \sqrt{2(0) + 4}=1$$ $$\sqrt{1}- \sqrt{4}=1$$ $$1 - 2=1$$ $$-1=1 \hspace{.2em}\color{red}{✖}$$ $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(6) + 1}- \sqrt{2(6) + 4}=1$$ $$\sqrt{24 + 1}- \sqrt{12 + 4}=1$$ $$\sqrt{25}- \sqrt{16}=1$$ $$5 - 4=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Our solution of x = 0 does not work in the original equation, therefore, this is an extraneous solution.
Our only valid solution is x = 6.
Example 3: 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. $$\sqrt[3]{2x - 11}- \sqrt[3]{5x + 1}=0$$ $$\sqrt[3]{2(-4) - 11}- \sqrt[3]{5(-4) + 1}=0$$ $$\sqrt[3]{-19}- \sqrt[3]{-19}=0$$ $$0=0 \hspace{.2em}\color{green}{✔}$$
x = 7
Let's suppose we square both sides of the above equation:
x2 = 49
This new equation has two solutions, x can be 7 or (-7).
(7)2 = 49
(-7)2 = 49
-7 does not work as a solution in our original equation.
-7 = 7 (false)
This is an extraneous solution and occurred from the squaring operation. When we square a number, a positive remains positive, and a negative becomes positive. It stands to reason, when we square both sides of an equation, there will be a loss of information. In other words, we don't know if x was 7 or x was -7 since 49 is the result of either being squared. This information loss is why we must always check our solutions in the original equation after squaring both sides.
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
Example 1: Solve each equation. $$\require{color}\sqrt{2x - 7}=1$$ Step 1) Isolate one of the radicals.
In this case, our radical is already isolated on the left side of the equation. $$\sqrt{2x - 7}=1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation: $$\left(\sqrt{2x - 7}\right)^2=1^2$$ $$2x - 7=1$$ Step 3) Solve the equation. $$2x - 7=1$$ $$2x=8$$ $$x=4$$ Step 4) Check all solutions in the original equation. $$\sqrt{2x - 7}=1$$ $$\sqrt{2(4) - 7}=1$$ $$\sqrt{8 - 7}=1$$ $$\sqrt{1}=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Example 2: Solve each equation. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ Step 1) Isolate one of the radicals.
We will isolate the leftmost radical: $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ Step 2) Raise both sides of the equation to a power equal to the index of the radical.
In this case, we have a square root, which means we want to square both sides of the equation. $$\sqrt{4x + 1}=\sqrt{2x + 4}+ 1$$ $$\left(\sqrt{4x + 1}\right)^2=\left(\sqrt{2x + 4}+ 1\right)^2$$ $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ In this case, we still have a radical involved. We will repeat our first two steps.
Step 1) Isolate one of the radicals.
Let's isolate our radical on the right side of the equation: $$4x + 1=2x + 5 + 2\sqrt{2x + 4}$$ $$2x - 4=2\sqrt{2x + 4}$$ 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. $$\left(2x - 4\right)^2=\left(2\sqrt{2x + 4}\right)^2$$ $$4x^2 - 16x + 16=4(2x + 4)$$ $$4x^2 - 16x + 16=8x + 16$$ $$4x^2 - 24x=0$$ Step 3) Solve the equation. $$4x^2 - 24x=0$$ $$4x(x - 6)=0$$ $$4x=0$$ $$x=0$$ $$x - 6=0$$ $$x=6$$ Step 4) Check all solutions in the original equation.
We have two proposed solutions, x = 6 and x = 0. $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(0) + 1}- \sqrt{2(0) + 4}=1$$ $$\sqrt{1}- \sqrt{4}=1$$ $$1 - 2=1$$ $$-1=1 \hspace{.2em}\color{red}{✖}$$ $$\sqrt{4x + 1}- \sqrt{2x + 4}=1$$ $$\sqrt{4(6) + 1}- \sqrt{2(6) + 4}=1$$ $$\sqrt{24 + 1}- \sqrt{12 + 4}=1$$ $$\sqrt{25}- \sqrt{16}=1$$ $$5 - 4=1$$ $$1=1 \hspace{.2em}\color{green}{✔}$$ Our solution of x = 0 does not work in the original equation, therefore, this is an extraneous solution.
Our only valid solution is x = 6.
Example 3: 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. $$\sqrt[3]{2x - 11}- \sqrt[3]{5x + 1}=0$$ $$\sqrt[3]{2(-4) - 11}- \sqrt[3]{5(-4) + 1}=0$$ $$\sqrt[3]{-19}- \sqrt[3]{-19}=0$$ $$0=0 \hspace{.2em}\color{green}{✔}$$
Skills Check:
Example #1
Solve each equation. $$\sqrt{2x - 5}=1 - \sqrt{2x - 2}$$
Please choose the best answer.
A
$$x=8$$
B
$$\text{No Solution}$$
C
$$x=-7, 8$$
D
$$x=5, 8$$
E
$$x=-1, 8$$
Example #2
Solve each equation. $$\sqrt{2x - 10}=2 + \sqrt{7 - x}$$
Please choose the best answer.
A
$$x=7$$
B
$$x=-7, 7$$
C
$$\text{No Solution}$$
D
$$x=-3, 7$$
E
$$x=-10, 7$$
Example #3
Solve each equation. $$\sqrt[3]{9x - 7}=5$$
Please choose the best answer.
A
$$\text{No Solution}$$
B
$$x=-\frac{44}{3}$$
C
$$x=\frac{12}{5}, 19$$
D
$$x=\frac{44}{3}$$
E
$$x=-\frac{12}{5}, 9$$
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