frac{sqrt{8+sqrt{28}}+sqrt{8-sqrt{28}}}{sqrt{ | vedclass

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(C) Let $x = \frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$.Rationalizing the denominator:$x = \frac{(\sqrt{8+\s

Vedclass · Accepted Answer

$\sqrt{7}$

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(C) Let $x = \frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$.Rationalizing the denominator:$x = \frac{(\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}})^2}{(\sqrt{8+\sqrt{28}})^2-(\sqrt{8-\sqrt{28}})^2}$$x = \frac{(8+\sqrt{28}) + (8-\sqrt{28}) + 2\sqrt{(8+\sqrt{28})(8-\sqrt{28})}}{(8+\sqrt{28}) - (8-\sqrt{28})}$$x = \frac{16 + 2\sqrt{64-28}}{2\sqrt{28}}$$x = \frac{16 + 2\sqrt{36}}{2\sqrt{4 	imes 7}}$$x = \frac{16 + 2(6)}{2(2\sqrt{7})}$$x = \frac{16 + 12}{4\sqrt{7}} = \frac{28}{4\sqrt{7}} = \frac{7}{\sqrt{7}} = \sqrt{7}$.

$\frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$ is equal to

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