Question

How do you solve `-x = sqrt(2x+15)` and find any extraneous solutions?

Answer 1

Solve for x then substitute the values in to verify that they are not extraneous solutions; x=5 is extraneous.

Explanation:

First, we solve for x:
`(-x)^2=2x+15`
`x^2-2x-15=0`
`(x-5)(x+3)=0`
`x=-3, 5`

If `x=-3`
`-(-3)=sqrt(2(-3)+15)`
`3=sqrt(-6+15)=sqrt(9)=3`
therefore `x=-3` is not extraneous

If `x=5`
`-5=sqrt(2(5)+15)`
Since the square root of a number cannot be negative, `x=5` is an extraneous solution.