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

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(D) We need to find the square root of $\sqrt{50} + \sqrt{48}$.First,simplify the expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.Let $\sq

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

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(D) We need to find the square root of $\sqrt{50} + \sqrt{48}$.First,simplify the expression: $\sqrt{50} + \sqrt{48} = 5\sqrt{2} + 4\sqrt{3}$.Let $\sqrt{5\sqrt{2} + 4\sqrt{3}} = \sqrt{x} + \sqrt{y}$.Squaring both sides,we get $5\sqrt{2} + 4\sqrt{3} = x + y + 2\sqrt{xy}$.This does not simplify easily into the form $\sqrt{a} + \sqrt{b}$.Let us factor out $2^{1/2}$ from the expression: $\sqrt{50} + \sqrt{48} = \sqrt{2}(5 + \sqrt{24}) = \sqrt{2}(5 + 2\sqrt{6})$.Now,we need the square root of $\sqrt{2}(5 + 2\sqrt{6})$.This is equal to $2^{1/4} 	imes \sqrt{5 + 2\sqrt{6}}$.We know that $5 + 2\sqrt{6} = 3 + 2 + 2\sqrt{3 	imes 2} = (\sqrt{3} + \sqrt{2})^2$.Therefore,$\sqrt{5 + 2\sqrt{6}} = \sqrt{3} + \sqrt{2}$.Thus,the expression becomes $2^{1/4}(\sqrt{3} + \sqrt{2})$.

What is the square root of $\sqrt{50} + \sqrt{48}$?

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