Question
Solve. $\frac{\sqrt{x+7}+\sqrt{x-2}}{\sqrt{x+7}-\sqrt{x-2}}=\frac{5}{1}$
✓
Answer
$\frac{(\sqrt{x+7}+\sqrt{x-2})+(\sqrt{x+7}-\sqrt{x-2})}{(\sqrt{x+7}+\sqrt{x-2})-(\sqrt{x+7}-\sqrt{x-2})}=\frac{5+1}{5-1} \quad \ldots$ (using componendo-dividendo)
$
\begin{array}{}
\therefore \quad \frac{2 \sqrt{x+7}}{2 \sqrt{x-2}}=\frac{6}{4} \\
\therefore \frac{\sqrt{x+7}}{\sqrt{x-2}}=\frac{3}{2} \\
\therefore \quad \frac{x+7}{x-2}=\frac{9}{4} \\
\therefore \quad 4 x+28=9 x-18 \\
\therefore 28+18=9 x-4 x \\
\therefore \quad 46=5 x \\
\therefore \quad \frac{46}{5}=x \\
\end{array}
$
...(squaring both sides of the equation)
$\therefore \quad x=\frac{46}{5}$ is the solution of the given equation
$
\begin{array}{}
\therefore \quad \frac{2 \sqrt{x+7}}{2 \sqrt{x-2}}=\frac{6}{4} \\
\therefore \frac{\sqrt{x+7}}{\sqrt{x-2}}=\frac{3}{2} \\
\therefore \quad \frac{x+7}{x-2}=\frac{9}{4} \\
\therefore \quad 4 x+28=9 x-18 \\
\therefore 28+18=9 x-4 x \\
\therefore \quad 46=5 x \\
\therefore \quad \frac{46}{5}=x \\
\end{array}
$
...(squaring both sides of the equation)
$\therefore \quad x=\frac{46}{5}$ is the solution of the given equation
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