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

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

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$\frac{3}{4}$

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(A) Given the equation: $\sqrt{2x+1} - \sqrt{2x-1} = 1$ $(1)$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$Solving for $x$:$-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 $\sqrt{2x+1} - \sqrt{2x-1} = 1$ for $x \geq \frac{1}{2}$,then $\sqrt{4x^2-1}$ is equal to:

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