Answer
$a= 4$, $a= 8 $
Work Step by Step
Given \begin{equation}
\sqrt{6 a+1}=1+\sqrt{5 a-4}.
\end{equation} First square both sides to eliminate the radical signs (may be repeated), and solve for $a$. \begin{equation}
\begin{aligned}
\sqrt{6 a+1}&=1+\sqrt{5 a-4}\\
\left(\sqrt{6 a+1}\right)^2&= \left( 1+\sqrt{5 a-4}\right)^2\\
6a+1& =1+2\sqrt{5 a-4} +5a-4\\
a+4&= 2\sqrt{5 a-4}\\
\left(a+4\right)^2&=\left(2\sqrt{5 a-4}\right)^2\\
a^2+8a+16&= 20a-16\\
a^2-12a+32&= 0.
\end{aligned}
\end{equation} Solve the quadratic equation using the quadratic formula:
\begin{equation}
\begin{aligned}
a & =\frac{-(-12) \pm \sqrt{(-12)^2-4 \cdot 1 \cdot(32)}}{2 \cdot 1} \\
& =\frac{12 \pm 4}{2} \\
\Longrightarrow a& =\frac{12 + 4}{2}\\
&= 8\\
a& =\frac{12 - 4}{2}\\
&= 4.
\end{aligned}
\end{equation} Check. \begin{equation}
\begin{aligned}
\sqrt{6 (4)+1}&\stackrel{?}{=}1+\sqrt{5 (4)-4}\\
5& =5\quad \quad \textbf{True}\\
\sqrt{6 (8)+1}&\stackrel{?}{=}1+\sqrt{5 (8)-4}\\
7& =7\quad \quad \textbf{True}.
\end{aligned}
\end{equation} The solution is $a=4 $, $a= 8 $.
