The square root of sqrt{50} + sqrt{48} is | vedclass

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(D) Given expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.We want to find $\sqrt{\sqrt{50} + \sqrt{48}}$.Note that $\sqrt{50} + \sqrt{48}

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$2^{1/4}(\sqrt{3} + \sqrt{2})$

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(D) Given expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.We want to find $\sqrt{\sqrt{50} + \sqrt{48}}$.Note that $\sqrt{50} + \sqrt{48} = \sqrt{2}(25 + \sqrt{24}) = \sqrt{2}(5^2 + (\sqrt{24})^2 + 2 	imes 5 	imes \sqrt{24})$ is not quite right. Let us rewrite $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.Consider the square of $(\sqrt{3} + \sqrt{2})$: $(\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6}$.Then $\sqrt{2}(\sqrt{3} + \sqrt{2})^2 = \sqrt{2}(5 + 2\sqrt{6}) = 5\sqrt{2} + 2\sqrt{12} = 5\sqrt{2} + 4\sqrt{3} = \sqrt{50} + \sqrt{48}$.Therefore,$\sqrt{\sqrt{50} + \sqrt{48}} = \sqrt{\sqrt{2}(\sqrt{3} + \sqrt{2})^2} = 2^{1/4}(\sqrt{3} + \sqrt{2})$.

The square root of $\sqrt{50} + \sqrt{48}$ is

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