If x is a solution of the equation, sqrt | vedclass

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Question

$\sqrt{2 x+1}-\sqrt{2 x-1}=1..........(1)$

$\Rightarrow \quad 2 x+1+2 x-1-2 \sqrt{4 x^{2}-1}=1$

$\Rightarrow 4 x-1=2 \sqrt{4 x^{2}-1}$

$\Righ

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

$\sqrt{2 x+1}-\sqrt{2 x-1}=1..........(1)$

$\Rightarrow \quad 2 x+1+2 x-1-2 \sqrt{4 x^{2}-1}=1$

$\Rightarrow 4 x-1=2 \sqrt{4 x^{2}-1}$

$\Rightarrow 16 x^{2}-8 x+1=16 x^{2}-4$

$\Rightarrow 8 x=5$

$\Rightarrow \quad x=\frac{5}{8}$ which satisfies equation $(1)$

So, $\sqrt{4 x^{2}-1}=\frac{3}{4}$

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

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