Evaluate: sqrt{3 + sqrt{5}} | vedclass

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Question

(C) Let $\sqrt{3 + \sqrt{5}} = \sqrt{x} + \sqrt{y}$.Squaring both sides,we get $3 + \sqrt{5} = x + y + 2\sqrt{xy}$.Comparing the rational and irration

Vedclass · Accepted Answer

$\frac{\sqrt{5} + 1}{\sqrt{2}}$

Vedclass · Answer

(C) Let $\sqrt{3 + \sqrt{5}} = \sqrt{x} + \sqrt{y}$.Squaring both sides,we get $3 + \sqrt{5} = x + y + 2\sqrt{xy}$.Comparing the rational and irrational parts,we have $x + y = 3$ and $2\sqrt{xy} = \sqrt{5}$,which implies $4xy = 5$,so $xy = \frac{5}{4}$.We know that $(x - y)^2 = (x + y)^2 - 4xy = 3^2 - 4(\frac{5}{4}) = 9 - 5 = 4$.Thus,$x - y = 2$ (assuming $x > y$).Solving the system $x + y = 3$ and $x - y = 2$,we get $2x = 5 \implies x = \frac{5}{2}$ and $2y = 1 \implies y = \frac{1}{2}$.Therefore,$\sqrt{3 + \sqrt{5}} = \sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{5} + 1}{\sqrt{2}}$.

Evaluate: $\sqrt{3 + \sqrt{5}}$

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