Question

How do you find extraneous solutions when solving radical equations?

Answer 1

Let's start with an equation.

`x-6=sqrtx`

This can be solved easily by:

`x^2-12x+36=x`

`x^2-13x+36=0`

`(x-9)(x-4)=0`

`x=9, x=4`

but wait! we have to be careful whenever we square both sides of an equation.

Re-substitute the two answers into the original equation.

`9-6=sqrt9`
`3=3`

That's fine, one more.

`4-6 = sqrt4`
`-2 != 2`

Our extraneous answer is `x=4`

To find the extraneous answer of any equation, just find the solutions and then re-substitute the solutions into the original equation. The answer that results in an illogical equation is the extraneous answer.