Answer
$x= 20$
Work Step by Step
Given
\begin{equation}
\sqrt{x+5}+\sqrt{3 x+4}=13.
\end{equation} Start by squaring both sides to eliminate the radical sign on the left hand side. Rearrange the equation and then square both sides to eliminate the radical sign on the right hand side and then solve for $x$. \begin{equation}
\begin{aligned}
\sqrt{x+5}+\sqrt{3 x+4}&=13\\
\left(\sqrt{x+5}\right)^{2} & =\left(13-\sqrt{3 x+4}\right)^{2} \\
x+5& =3 x-26 \sqrt{3 x+4}+173\\
-(2x+168) & = -26 \sqrt{3 x+4} \\
x+84 & = 13\sqrt{3 x+4} \\
\left(x+84\right)^{2}& =\left(13\sqrt{3 x+4}\right)^{2}\\
x^2+168 x+7056&= 507 x+676\\
x^2-339 x+6380&= 0.
\end{aligned}
\end{equation} Solve the quadratic equation: \begin{equation}
\begin{aligned}
a & =1, b=-339 , c=6380 \\
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
x & =\frac{339 \pm \sqrt{(-399)^2-4 \cdot 1\cdot(6380)}}{2 \cdot 1} \\
& =\frac{339 \pm 299}{2 }.
\end{aligned}
\end{equation} This gives: \begin{equation}
x=319,\quad x=20.
\end{equation} Check \begin{equation}
\begin{aligned}
\sqrt{319+5}+\sqrt{3 \cdot 319+4}& \stackrel{?}{=} 13\\
49 & =13 \quad \text {False } \\
\sqrt{20+5}+\sqrt{3 \cdot 20+4}& \stackrel{?}{=} 13\\
13 & =13 \quad \text {True}.
\end{aligned}
\end{equation} The solution is $x= 20$.
