# How do you solve #sqrt(x-3) = x - 5# and find any extraneous solutions?

##### 1 Answer

Aug 29, 2016

#### Answer:

Both solutions are valid.

#### Explanation:

The value under the root may not be negative.

Square both sides of the equation.

A common error is to square each of the terms on the right side separately, instead of as a square of a binomial.

Find factors of 28 which add to give 11. (

The signs will be the same, both negative.

Putting each factor equal to 0 gives:

Both solutions are valid.