Question

If $x$ is a solution of the equation, $\sqrt{2x+1}-\sqrt{2x-1}=1,(x\ge \frac{1}{2}),$ then $\sqrt{{4x}^2-1}$ is equal to

• A. $2$
• B. $\frac{3}{4}$
• C. $2\sqrt{2}$
• D. $\frac{4}{7}$

Answer 1

The correct option is B $\frac{3}{4}$

Given ,$\sqrt{2x+1}-\sqrt{2x-1}=1$

$\Rightarrow \sqrt{2x+1}=1+\sqrt{2x-1}$

squaring both sides, we get

$\Rightarrow 2x+1=1+2x-1+2\sqrt{2x-1}$

$\Rightarrow 0=-1+2\sqrt{2x-1}$

$\Rightarrow 2\sqrt{2x-1}=1.$

squaring both sides,we get.

$4(\sqrt{2x-1}{)}^2=1$

$\Rightarrow 2x-1=\frac{1}{4}$

$\Rightarrow 2x=\frac{1}{4}+1$

$\Rightarrow 2x=\frac{5}{4}$

$\Rightarrow x=\frac{5}{8}$

$\therefore \sqrt{x^2-1}=\sqrt{4\times (\frac{5}{8}{)}^2-1}=\sqrt{4\times \frac{25}{64}-1}$

$=\sqrt{\frac{100-64}{64}}$

$=\sqrt{\frac{36}{64}}$

$=\frac{6}{8}$

$=\frac{3}{4}$