If arithmetic mean of two distinct positive real numbers a and b (a b) be twice their geometric mean, then a:b =

Question

If arithmetic mean of two distinct positive real numbers $a$ and $b$ $(a > b)$ be twice their geometric mean, then $a : b =$

Smart Achievers · Accepted Answer

(2)

$By the given condition,$

$(a > b)$

$Now, {(a - b)}^{2} = {(a + b)}^{2} - 4 a b = 16 a b - 4 a b = 12 a b$

$∴$ $a - b = \sqrt{12 a b} = 2 \sqrt{3} \sqrt{a b}$ $(Taking +ve sign only as a > b)$

$∴$ $\frac{a + b}{a - b} = \frac{4 \sqrt{a b}}{2 \sqrt{3} \sqrt{a b}} = \frac{2}{\sqrt{3}}$

$By componendo and dividendo, we get$

$\frac{2 a}{2 b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \Rightarrow \frac{a}{b} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}}$

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