How do you solve $sqrt(8-2x) = x-4$ ?

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Answer 1 · 2017-10-02T20:26:52

$x=2$
$x=4$

$sqrt(8-2x)= x-4$

Square both sides to remove radical:

$8-2x= (x-4)^2$

Expand right hand side:

$8-2x= x^2-8x+16$

Collect like terms on one side of the equation:

$x^2 -6x +8=0$

Factor $x^2 -6x +8=0$ :

$(x -2)(x-4)=0=> x= 2$ and $x=4$

How do you solve sqrt(8-2x) = x-4sqrt(8-2x) = x-4?