Answer
$x=12 $
Work Step by Step
Given\begin{equation}
\sqrt{x+4}-\sqrt{3 x}=-2.
\end{equation} Let's take a look at the student's solution and see where she went wrong and make the necessary corrections.
\begin{equation}
\begin{aligned}
(\sqrt{x+4}-\sqrt{3 x})^2 & =(-2)^2 \\
x+4-3 x & =4 \\
-2 x+4 & =4 \\
-2 x & =0 \\
x & =0.
\end{aligned}
\end{equation} The first mistake is that the two radical terms remained on the left hand side while squaring both sides. This move will only complicate the problem without eliminating the radical terms. The squaring of the radicals was poorly done because\begin{equation}
(\sqrt{x+4}-\sqrt{3 x})^2\neq x+4-3 x .
\end{equation} Here is how the radical equation should be solved.
\begin{equation}
\begin{aligned}
\sqrt{x+4}-\sqrt{3 x} & =-2 \\
\sqrt{x+4} & =\sqrt{3 x}-2 \\
\left( \sqrt{x+4} \right)^2 & =\left(\sqrt{3 x}-2 \right)^2\\
x+4 & =3x-4\sqrt{3 x}+4 \\
x+4-3x-4& =-4\sqrt{3 x} \\
-2x & =-4\sqrt{3 x}\\
x& = 2\sqrt{3 x}\\
\left(x\right)^2&= \left( 2\sqrt{3 x} \right)\\
x^2&= 4\cdot 3x= 12x\\
x^2-12x&= 0\\
x(x-12)&= 0\\
\implies x= 0\\
x&= 12.
\end{aligned}
\end{equation} Check. \begin{equation}
\begin{aligned}
\sqrt{0+4}-\sqrt{3 \cdot 0}& \stackrel{?}{=}-2 \\
2& =-2\quad \textbf{False}\\
\sqrt{12+4}-\sqrt{3 \cdot 12}& \stackrel{?}{=}-2 \\
-2& =-2\quad \textbf{True}
\end{aligned}
\end{equation} The solution is $x=12 $.
