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

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Answer 1 · 2016-05-27T16:15:15

First it should be $x>=0$ and $x-4>=0$ hence $x>=4$

Now square the relation to get

$2x=(x-4)^2=>2x=x^2-8x+16=> x^2-10x+16=0=>(x-2)*(x-8)=0$

The only acceptable solution is $x=8$

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