Lesson Objectives

- Demonstrate an understanding of radicals
- Learn about the squaring property of equality
- Learn how to solve a radical equation

## How to Solve Radical Equations

A radical equation is an equation where there is a variable in the radicand. As an example: $$\sqrt{(x - 2)}=9$$ To solve a radical equation, we need more than just the addition and multiplication properties of equality. We will need to introduce a new property of equality known as the "squaring property of equality".

x = 7 or x = -7

x = -7 is an extraneous solution. It will not work in the original equation: $$x=7$$ $$-7 ≠ 7$$

Example 1: Solve each equation. $$\sqrt{(x - 3)}=2$$ Step 1) Isolate the radical:

In this case, the radical is already isolated on the left side of the equation.

Step 2) Square both sides: $$\left(\sqrt{(x - 3)}\right)^2=(2)^2$$ $$x - 3=4$$ Step 3) Combine any like terms:

In this case, each side is simplified.

Step 4) Solve the equation: $$x - 3=4$$ $$x=4 + 3$$ $$x=7$$ Step 5) Check all solutions in the original equation: $$\sqrt{(x - 3)}=2$$ Plug in a 7 for x: $$\sqrt{7 - 3}=2$$ $$\sqrt{4}=2$$ $$2=2$$ Our solution x = 7 is correct.

Example 2: Solve each equation. $$\sqrt{(21 - 2x)}=\sqrt{(3x - 29)}$$ Step 1) Isolate the radical:

In this case, each radical is isolated on one side of the equation.

Step 2) Square both sides: $$\left(\sqrt{(21 - 2x)}\right)^2=\left(\sqrt{(3x - 29)}\right)^2$$ $$\left(\sqrt{(21 - 2x)}\right)^2=$$$$\left(\sqrt{(3x - 29)}\right)^2$$ $$21 - 2x=3x - 29$$ Step 3) Combine any like terms:

In this case, each side is simplified.

Step 4) Solve the equation: $$21 - 2x=3x - 29$$ $$-2x - 3x=-29 - 21$$ $$-5x=-50$$ $$\frac{-5}{-5}x=\frac{-50}{-5}$$ $$\require{cancel}\frac{\cancel{-5}}{\cancel{-5}}x=\frac{10\cancel{-50}}{\cancel{-5}}$$ $$x=10$$ Step 5) Check all solutions in the original equation: $$\sqrt{(21 - 2x)}=\sqrt{(3x - 29)}$$ $$\sqrt{(21 - 2(10))}=\sqrt{(3(10) - 29)}$$ $$\sqrt{(21 - 2(10))}=$$$$\sqrt{(3(10) - 29)}$$ $$\sqrt{(21 - 20)}=\sqrt{(30 - 29)}$$ $$\sqrt{1}=\sqrt{1}$$ $$1=1$$ Our solution x = 10 is correct.

Example 3: Solve each equation. $$\sqrt{(2x - 3)}+ 3=x$$ Step 1) Isolate the radical:

We will subtract 3 away from each side of the equation. $$\sqrt{(2x - 3)}=x - 3$$ Step 2) Square both sides: $$\left(\sqrt{(2x - 3)}\right)^2=(x - 3)^2$$ $$2x - 3=x^2 - 6x + 9$$ Step 3) Combine any like terms:

In this case, each side is simplified.

Step 4) Solve the equation: $$2x - 3=x^2 - 6x + 9$$ $$x^2 - 8x + 12=0$$ We can solve this using factoring and our zero-product property: $$(x - 6)(x - 2)=0$$ $$x - 6=0$$ $$x=6$$ $$x - 2=0$$ $$x=2$$ x = 6 or x = 2

Step 5) Check all solutions in the original equation:

Let's start with x = 6: $$\sqrt{(2x - 3)}+ 3=x$$ $$\sqrt{(2(6) - 3)}+ 3=(6)$$ $$\sqrt{(12 - 3)}+ 3=6$$ $$\sqrt{9}+ 3=6$$ $$3 + 3=6$$ $$6=6$$ x = 6 is a valid solution. Let's check x = 2: $$\sqrt{(2x - 3)}+ 3=x$$ $$\sqrt{(2(2) - 3)}+ 3=(2)$$ $$\sqrt{4 - 3}+ 3=2$$ $$\sqrt{1}+ 3=2$$ $$1 + 3=2$$ $$4=2 \hspace{.5em}\text{false}$$ Since we obtained a false statement, we can say that x = 2 is an extraneous solution. It does not work in our original equation. The only valid solution here is x = 6.

### Squaring Property of Equality

When both sides of an equation are squared, all solutions to the original equation are among the solutions to the squared equation. The keyword here is "among"! When we use the squaring property of equality, we may end up with extraneous solutions (solutions that do not work in the original equation). As a rule, we must check all solutions to an equation when the squaring property of equality is used. Let's look at a simple example: $$x=7$$ Here x is just 7, if we square both sides: $$x^2=49$$ This new equation has two solutions:x = 7 or x = -7

x = -7 is an extraneous solution. It will not work in the original equation: $$x=7$$ $$-7 ≠ 7$$

### Solving a Radical Equation

- Isolate the radical
- If more than one radical exists, isolate one of the radicals

- Square both sides
- Combine any like terms
- When radicals remain, go back to the first step
- Solve the equation
- Check all solutions in the original equation

Example 1: Solve each equation. $$\sqrt{(x - 3)}=2$$ Step 1) Isolate the radical:

In this case, the radical is already isolated on the left side of the equation.

Step 2) Square both sides: $$\left(\sqrt{(x - 3)}\right)^2=(2)^2$$ $$x - 3=4$$ Step 3) Combine any like terms:

In this case, each side is simplified.

Step 4) Solve the equation: $$x - 3=4$$ $$x=4 + 3$$ $$x=7$$ Step 5) Check all solutions in the original equation: $$\sqrt{(x - 3)}=2$$ Plug in a 7 for x: $$\sqrt{7 - 3}=2$$ $$\sqrt{4}=2$$ $$2=2$$ Our solution x = 7 is correct.

Example 2: Solve each equation. $$\sqrt{(21 - 2x)}=\sqrt{(3x - 29)}$$ Step 1) Isolate the radical:

In this case, each radical is isolated on one side of the equation.

Step 2) Square both sides: $$\left(\sqrt{(21 - 2x)}\right)^2=\left(\sqrt{(3x - 29)}\right)^2$$ $$\left(\sqrt{(21 - 2x)}\right)^2=$$$$\left(\sqrt{(3x - 29)}\right)^2$$ $$21 - 2x=3x - 29$$ Step 3) Combine any like terms:

In this case, each side is simplified.

Step 4) Solve the equation: $$21 - 2x=3x - 29$$ $$-2x - 3x=-29 - 21$$ $$-5x=-50$$ $$\frac{-5}{-5}x=\frac{-50}{-5}$$ $$\require{cancel}\frac{\cancel{-5}}{\cancel{-5}}x=\frac{10\cancel{-50}}{\cancel{-5}}$$ $$x=10$$ Step 5) Check all solutions in the original equation: $$\sqrt{(21 - 2x)}=\sqrt{(3x - 29)}$$ $$\sqrt{(21 - 2(10))}=\sqrt{(3(10) - 29)}$$ $$\sqrt{(21 - 2(10))}=$$$$\sqrt{(3(10) - 29)}$$ $$\sqrt{(21 - 20)}=\sqrt{(30 - 29)}$$ $$\sqrt{1}=\sqrt{1}$$ $$1=1$$ Our solution x = 10 is correct.

Example 3: Solve each equation. $$\sqrt{(2x - 3)}+ 3=x$$ Step 1) Isolate the radical:

We will subtract 3 away from each side of the equation. $$\sqrt{(2x - 3)}=x - 3$$ Step 2) Square both sides: $$\left(\sqrt{(2x - 3)}\right)^2=(x - 3)^2$$ $$2x - 3=x^2 - 6x + 9$$ Step 3) Combine any like terms:

In this case, each side is simplified.

Step 4) Solve the equation: $$2x - 3=x^2 - 6x + 9$$ $$x^2 - 8x + 12=0$$ We can solve this using factoring and our zero-product property: $$(x - 6)(x - 2)=0$$ $$x - 6=0$$ $$x=6$$ $$x - 2=0$$ $$x=2$$ x = 6 or x = 2

Step 5) Check all solutions in the original equation:

Let's start with x = 6: $$\sqrt{(2x - 3)}+ 3=x$$ $$\sqrt{(2(6) - 3)}+ 3=(6)$$ $$\sqrt{(12 - 3)}+ 3=6$$ $$\sqrt{9}+ 3=6$$ $$3 + 3=6$$ $$6=6$$ x = 6 is a valid solution. Let's check x = 2: $$\sqrt{(2x - 3)}+ 3=x$$ $$\sqrt{(2(2) - 3)}+ 3=(2)$$ $$\sqrt{4 - 3}+ 3=2$$ $$\sqrt{1}+ 3=2$$ $$1 + 3=2$$ $$4=2 \hspace{.5em}\text{false}$$ Since we obtained a false statement, we can say that x = 2 is an extraneous solution. It does not work in our original equation. The only valid solution here is x = 6.

#### Skills Check:

Example #1

Solve each equation. $$\sqrt{12 - 3x}=6$$

Please choose the best answer.

A

$$x=8, 10$$

B

$$x=-1, 3$$

C

$$x=-\frac{2}{5}, 7$$

D

$$x=-8$$

E

$$x=10$$

Example #2

Solve each equation. $$8=1 + \sqrt{6x - 5}$$

Please choose the best answer.

A

$$x=9$$

B

$$x=3$$

C

$$x=1, \frac{7}{2}$$

D

$$x=8, 10$$

E

$$x=-\frac{3}{2}, \frac{1}{2}$$

Example #3

Solve each equation. $$\sqrt{14 - 2x}=\sqrt{12 - x}$$

Please choose the best answer.

A

$$x=-9, 9$$

B

$$x=2$$

C

$$x=-2, 4$$

D

$$x=-1, 3$$

E

$$x=-2$$

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