If the arithmetic mean of two positive numbers x and y , where (x > y), is twice their geometric mean then a : b = $_____$

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Solution

$A r i t h m e t i c M e a n o f x a n d y = \frac{x + y}{2} G e o m e t r i c M e a n o f x a n d y = \sqrt{x y} \frac{x + y}{2} = 2 \sqrt{x y} => {(x + y)}^{2} = 16 x y => {(x - y)}^{2} = 12 x y => x^{2} - 14 x y + y^{2} = 0 x = \frac{14 y \pm \sqrt{192 y^{2}}}{2} => x = 7 y \pm 4 \sqrt{3} y x = (7 + 4 \sqrt{3}) y (∵ x > y) ∴ x : y = (7 + 4 \sqrt{3}) : 1$

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