Question

Solve for $x:\sqrt{ax+1}-\sqrt{ax-1}=\sqrt{ax}$

• A. $-\frac{2}{3a}$
• B. $\frac{2}{3a}$
• C. $\frac{\pm 2\sqrt{3}}{3a}$
• D. $\frac{2\sqrt{3}}{3a}$

Answer 1

Answer: (C)

$\frac{\pm 2\sqrt{3}}{3a}$

Given, $\sqrt{ax+1}-\sqrt{ax-1}=\sqrt{ax}$
Squaring on both the sides, we get
$(\sqrt{ax+1}-\sqrt{ax-1}{)}^2=(\sqrt{ax}{)}^2$
$\Rightarrow (\sqrt{ax+1}{)}^2+(\sqrt{ax-1}{)}^2-2\sqrt{ax+1}.\sqrt{ax-1}=ax$
$\Rightarrow ax+1+ax-1-2\sqrt{ax+1}.\sqrt{ax-1}=ax$
$\Rightarrow 2ax-ax=2\sqrt{ax+1}.\sqrt{ax-1}$
$\Rightarrow ax=2\sqrt{ax+1}.\sqrt{ax-1}$
Squaring on both the sides, we get
$a^2{x}^2=2^2(\sqrt{ax+1}.\sqrt{ax-1}{)}^2$
$\Rightarrow a^2{x}^2=4\left[\right(ax{)}^2-1]$
$\Rightarrow a^2{x}^2=4a^2{x}^2-4$
$\Rightarrow 3a^2{x}^2=4$
$\Rightarrow x^2=\frac{4}{3a^2}$
$\Rightarrow x=\pm \frac{2}{\sqrt{3}a}$ or
So, $x=\pm \frac{2\sqrt{3}}{3a}$