Question

The sum of two positive numbers is $6$ times their geometric mean, show that the numbers are in the ratio $(3+2\sqrt{2}):(3-2\sqrt{2})$.

Answer 1

$a+b=6\sqrt{ab}$ …….(1)

$\Rightarrow a^2+b^2+2ab=36ab$

$\Rightarrow (a-b{)}^2=32ab$

$\Rightarrow a-b=4\sqrt{2}\sqrt{ab}$ ………….(2)

Adding $\left(1\right)$ & $\left(2\right)$

$2a=\sqrt{ab}(6+4\sqrt{2})$

$\Rightarrow a=\sqrt{ab}(3+2\sqrt{2})$

& $b=\sqrt{ab}(3-2\sqrt{2})$

Dividing, $\frac{a}{b}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$.