If x is a solution of the equation sqrt{2x+1} | vedclass

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(A) Given the equation: $\sqrt{2x+1} - \sqrt{2x-1} = 1$ $\dots(1)$ Squaring both sides: $(\sqrt{2x+1} - \sqrt{2x-1})^2 = 1^2$ $(2x+1) + (2x-1) - 2\sqr

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) Given the equation: $\sqrt{2x+1} - \sqrt{2x-1} = 1$ $\dots(1)$ Squaring both sides: $(\sqrt{2x+1} - \sqrt{2x-1})^2 = 1^2$ $(2x+1) + (2x-1) - 2\sqrt{(2x+1)(2x-1)} = 1$ $4x - 2\sqrt{4x^2-1} = 1$ $2\sqrt{4x^2-1} = 4x - 1$ Squaring again: $4(4x^2-1) = (4x-1)^2$ $16x^2 - 4 = 16x^2 - 8x + 1$ $8x = 5 \implies x = \frac{5}{8}$ Now,substitute $x = \frac{5}{8}$ into $\sqrt{4x^2-1}$: $\sqrt{4(\frac{5}{8})^2 - 1} = \sqrt{4(\frac{25}{64}) - 1} = \sqrt{\frac{25}{16} - 1} = \sqrt{\frac{9}{16}} = \frac{3}{4}$

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

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