# Solving Radical Equations

*How to solve equations with square roots, cube roots, etc.*

## Radical Equations

A Radical Equation is an equation with a square root or cube root, etc. |

## Solving Radical Equations

We can get rid of a square root by squaring (or cube roots by cubing, etc).

Warning: this can sometimes create "solutions" which don't actually work when we put them into the original equation. So we need to Check!

Follow these steps:

- isolate the square root on one side of the equation
- square both sides of the equation

Then continue with our solution!

### Example: solve √(2x+9) − 5 = 0

Now it should be easier to solve!

Check: √(2·8+9) − 5 = √(25) − 5 = 5 − 5 = 0

That one worked perfectly.

## More Than One Square Root

What if there are two or more square roots? Easy! Just repeat the process for each one.

It will **take longer** (lots more steps) ... but nothing too hard.

### Example: solve √(2x−5) − √(x−1) = 1

^{2}

We have removed one square root.

Now do the "square root" thing again:

^{2}

We have now successfully removed both square roots.

Let us continue on with the solution.

^{2}− 10x + 25)/4

It is a Quadratic Equation! So let us put it in standard form.

^{2}− 10x + 25

^{2}+ 10x − 25 = 0

^{2}+ 14x − 29 = 0

^{2}− 14x + 29 = 0

Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:

2.53 and 11.47 (to 2 decimal places)

Let us check the solutions:

2.53: √(2×2.53−5) − √(2.53−1) ≈ **−1** Oops! Should be plus 1.

11.47: √(2×11.47−5) − √(11.47−1) ≈ **1** Yes that one works.

There is **really only one solution**:

Answer: 11.47 (to 2 decimal places)

See? This method **can** sometimes produce solutions that don't really work!

The root that seemed to work, but wasn't right when we checked it, is called an **"Extraneous Root"**

So checking is important.