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

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

Vedclass · Accepted Answer

$2$

Vedclass · Answer

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

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

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