sqrt {(3 + sqrt 5 )} = . . . | vedclass

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Question

(c) Let $\sqrt {3 + \sqrt 5 } = \sqrt x + \sqrt y $

$3 + \sqrt 5 = \,x + y + 2\sqrt {xy} $. Obviously $x + y = 3$

and $4xy = 5$. So ${(x - y)^2}

Vedclass · Accepted Answer

$(\sqrt 5 + 1)/\sqrt 2 $

Vedclass · Answer

(c) Let $\sqrt {3 + \sqrt 5 } = \sqrt x + \sqrt y $

$3 + \sqrt 5 = \,x + y + 2\sqrt {xy} $. Obviously $x + y = 3$

and $4xy = 5$. So ${(x - y)^2} = 9 - 5 = 4$ or $(x - y) = 2$

After solving $x = {5 \over 2},y = {1 \over 2}$.

Hence, $\sqrt {3 + \sqrt 5 } = \sqrt {{5 \over 2}} + \sqrt {{1 \over 2}} = {{\sqrt 5 + 1} \over {\sqrt 2 }}$.

$\sqrt {(3 + \sqrt 5 )} = . .$ .

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