Answer
$n= 5 $
Work Step by Step
Given \begin{equation}
\sqrt{2 n+6}+\sqrt{n-4}=5.
\end{equation} This is a radical equation. The solution steps are shown below.
\begin{equation}
\begin{aligned}
\sqrt{2 n+6}+\sqrt{n-4}&=5\\
\sqrt{2 n+6} &= 5-\sqrt{n-4}\\
\left( \sqrt{2 n+6} \right)^2 & = \left(5-\sqrt{n-4} \right)^2\\
2n+6&= 25-10\sqrt{n-4}+n-4\\
n-15 &= -10\sqrt{n-4}\\
\left( n-15 \right)^2&= \left( -10\sqrt{n-4} \right)^2\\
n^2-30n+225& = 100n-400\\
n^2-130n+625&= 0
\end{aligned}
\end{equation} We use the quadratic formula: \begin{equation}
\begin{aligned}
a& =1,\ b=-130,\ c=625 \\
n & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
n & =\frac{-(-130) \pm \sqrt{(-130)^2-4 \cdot 1 \cdot(625)}}{2 \cdot 1} \\
& =\frac{130 \pm 120}{2} \\
\Longrightarrow n& =\frac{130 + 120}{2}\\
&=125\\
n& =\frac{130 - 120}{2}\\
&=5.
\end{aligned}
\end{equation} Check \begin{equation}
\begin{aligned}
\sqrt{2 \cdot (125)+6}+\sqrt{125-4}&\stackrel{?}{=} 5\\
27& =5\quad \quad \textbf{False}\\
\sqrt{2 \cdot (5)+6}+\sqrt{5-4}&\stackrel{?}{=} 5\\
5& =5\quad \quad \textbf{True}\\
\end{aligned}
\end{equation} The solution is $n= 5 $.
