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

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

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$\frac{3}{4}$

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

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

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