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. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $2$
• D. $2\sqrt{2}$

Answer 1

Answer: (A)

$\frac{3}{4}$

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

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

squaring both sides, we get

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

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

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

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

Squaring again on both sides, we get

$1=4(2x-1)$

$1=8x-4$

$8x=1+4$

$8x=5$

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

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

$=\sqrt{\frac{25}{16}-1}$

$=\sqrt{\frac{25-16}{16}}$

$=\sqrt{\frac{9}{16}}=\frac{3}{4}$