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

Question

3079 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-07T09:41:26

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

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.

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