solve the equation.\n$sqrt{2x}=x - 4$\n$x = ?$

Explanation:

Step1: Square both sides

$(\sqrt{2x})^2=(x - 4)^2$ $2x=x^{2}-8x + 16$

Step2: Rearrange to quadratic form

$x^{2}-8x-2x + 16=0$ $x^{2}-10x + 16=0$

Step3: Factor the quadratic

$(x - 2)(x - 8)=0$

Step4: Solve for x

$x-2=0$ gives $x = 2$; $x - 8=0$ gives $x=8$

Step5: Check solutions

For $x = 2$, $\sqrt{2\times2}=2$, $2-4=-2$, $2\neq-2$, so $x = 2$ is an extraneous solution. For $x = 8$, $\sqrt{2\times8}=\sqrt{16}=4$, $8 - 4=4$, so $x = 8$ is valid.

Answer:

$8$