# 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. (Cube roots by cubing, etc)

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

Follow these simple 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

- isolate the square root: √(2x+9) = 5
- square both sides: 2x+9 = 25

Now it should be easier to solve!

Move 9 to right: 2x = 25 − 9 = 16

Divide by 2: x = 16/2 = 8

Answer: x = 8

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

- isolate one of the square roots: √(2x−5) = 1 + √(x−1)
- square both sides: 2x−5 = (1 + √(x−1))
^{2}

We have removed one square root.

Expand right hand side: 2x−5 = 1 + 2√(x−1) + (x−1)

Simplify: 2x−5 = 2√(x−1) + x

Subtract x from both sides: x−5 = 2√(x−1)

Now do the "square root" thing again:

- isolate the square root: √(x−1) = (x−5)/2
- square both sides: x−1 = ((x−5)/2)
^{2}

We have now successfully removed both square roots.

Let us continue on with the solution.

Expand right hand side: x−1 = (x^{2} − 10x + 25)/4

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

Multiply by 4 to remove division: 4x−4 = x^{2} − 10x + 25

Bring all to left: 4x − 4 − x^{2} + 10x − 25 = 0

Combine like terms: −x^{2} + 14x − 29 = 0

Swap all signs: x^{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.