Answer
See below
Work Step by Step
Given: $\sqrt 2x=x-4$
Square both sides: $(x-4)^2=(\sqrt 2x)^2\\x^2-8x+16=2x\\x^2-10x+16=0$
Solve for x: $x=8 \lor x=2$
Check: $8-4=\sqrt 2(8)=4$
$2-4=-2\ne\sqrt 2(2)=2$
Hence, the solution is $x=8$.
$x=2$ is an extraneous solution.
