# How do you solve sqrt(2x)=x-4 and find any extraneous solutions?

First it should be $x \ge 0$ and $x - 4 \ge 0$ hence $x \ge 4$
2x=(x-4)^2=>2x=x^2-8x+16=> x^2-10x+16=0=>(x-2)*(x-8)=0
The only acceptable solution is $x = 8$