# Week 5: Solving Radical Equations

This week in Pre Calculus 11, I learned how to solve radical equations. It was a little bit difficult for me at some points, because I didn’t understand squaring. Today I will teach you how to deal with these kinds of expressions.

Steps:

1. Solving
2. Restrictions
3. Checking

Example

$\sqrt{x}$ = 6

SOLVING

• remove the radical sign by squaring both sides of the equation.

$(\sqrt{x})^2$ = $6^2$

• Simplify

x = 36

Restrictions

Note: a number under a  radical sign can not be negative because a number times itself is positive weather it be -x or x. The only exception is if it has a odd power.

In the example above x needs to be a positive number so its restriction would be x > 0 or x = 0.

Check It Out

• $\sqrt{36}$ = 6
• 6 = 6

This is how to solve a basic radical equation. Now I will move on to a more difficult equation to maximize your understanding.

Example 2

1 $-3\sqrt{2x}$ = -3 $-2\sqrt{2x}$

Note: When solving eq, what you do to one side you have to do to the other.

SOLVING

1. Place the constants on side of the equation
• 1 + 3 $-3\sqrt{2x}$ = -3 + 3 $-2\sqrt{2x}$
•  4 $-3\sqrt{2x}$ = $-2\sqrt{2x}$

Note: The radicans are the same so they can be combined together. Place them on one side of the equation.

• 4 $-3\sqrt{2x}$ = $-2\sqrt{2x}$
• 4 $-3\sqrt{2x}$ + $3\sqrt{2x}$ = $-2\sqrt{2x}$ + $3\sqrt{2x}$
• 4 = $1\sqrt{2x}$

3. Square both sides to remove the radical sign.

• $4^2$$(\sqrt{2x})^2$
• $4^2$ = 2x
• 16 = 2x

4. Isolate x

• $\frac{16}{2}$$\frac{2x}{2}$
• $\frac{16}{2}$ = x
• 8 = x

RESTRICTIONS

x > 0   or   x = 0

Check It Out

• 1 – $3\sqrt{2(8)}$ = -3 – $2\sqrt{2(8)}$
• 1 – $3\sqrt{16}$ = -3 – $2\sqrt{16}$
• 1 – $3\cdot4$ = -3 – $2\cdot4$
• 1 – 12 = -3 – 8
• -11 = -11

This is how you solve radical equations.

Comment below if this helped you or if you have any questions. 🙂