sqrt{2+sqrt{5}-sqrt{6-3 sqrt{5}+sqrt{14-6 sqr | vedclass

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Question

(B) We simplify the expression from the innermost radical outward: $\sqrt{14-6 \sqrt{5}} = \sqrt{9+5-2(3)(\sqrt{5})} = \sqrt{(3-\sqrt{5})^2} = 3-\sqrt

Vedclass · Accepted Answer

$2$

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(B) We simplify the expression from the innermost radical outward: $\sqrt{14-6 \sqrt{5}} = \sqrt{9+5-2(3)(\sqrt{5})} = \sqrt{(3-\sqrt{5})^2} = 3-\sqrt{5}$. Substituting this back into the expression: $\sqrt{6-3 \sqrt{5} + (3-\sqrt{5})} = \sqrt{9-4 \sqrt{5}}$. We simplify $\sqrt{9-4 \sqrt{5}}$ as $\sqrt{9-2(2)(\sqrt{5})} = \sqrt{(\sqrt{5}-2)^2} = \sqrt{5}-2$. Now, substitute this into the main expression: $\sqrt{2+\sqrt{5}-(\sqrt{5}-2)} = \sqrt{2+\sqrt{5}-\sqrt{5}+2} = \sqrt{4} = 2$.

$\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{14-6 \sqrt{5}}}}$ is equal to

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