Answer
The solution is $x=8$
Work Step by Step
$\sqrt{2x}-x+4=0$
Leave the square root alone on the left side of the equation:
$\sqrt{2x}=x-4$
Square both sides:
$(\sqrt{2x})^{2}=(x-4)^{2}$
$2x=x^{2}-8x+16$
Take $2x$ to the right side and simplify:
$0=x^{2}-8x+16-2x$
$0=x^{2}-10x+16$
Rearrange:
$x^{2}-10x+16=0$
Solve by factoring:
$(x-2)(x-8)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-2=0$
$x=2$
$x-8=0$
$x=8$
Check the solutions found by plugging them into the original equation:
$x=2$
$\sqrt{2(2)}-2+4=0$
$\sqrt{4}-2+4=0$
$2-2+4=0$
$4\ne0$ False
$x=8$
$\sqrt{2(8)}-8+4=0$
$\sqrt{16}-8+4=0$
$4-8+4=0$
$0=0$ True
The solution is $x=8$
