Answer
a) $\sqrt{2x}=-x$
b) $2x=x^2$
c) $0$ and $2$
d) $0$
Work Step by Step
a) We isolate the radical on one side:
$$\sqrt{2x}=-x.$$
b) Square both sides:
$$2x=x^2.$$
c) Solve the quadratic equation:
$$\begin{align*}
x^2-2x&=0\\
x(x-2)&=0\\
x=0&\text{ or }x-2=0\\
x=0&\text{ or }x=2.
\end{align*}$$
The solutions of the resulting quadratic equation are $0$ and $2$.
d) We check the solutions:
$$\begin{align*}
x_1&=0\\
\sqrt{2(0)}+0&\stackrel{?}{=}0\\
0+0&\stackrel{?}{=}0\\
0&=0\checkmark\\\\
x_2&=2\\
\sqrt{2(2)}+2&\stackrel{?}{=}0\\
2+2&\stackrel{?}{=}0\\
4&\not=0.
\end{align*}$$
The solution that satisfies the original equation is $0$.
