The sum of two numbers is $6$ times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$
The sum of two numbers is $6$ times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$
Le the two numbers be $a$ and $b$
$G.M.$ $=\sqrt{a b}$
According to the given condition,
$a+b=6 \sqrt{a b}$ ..........$(1)$
$\Rightarrow(a+b)^{2}=36(a b)$
Also,
$(a-b)^{2}=(a+b)^{2}-4 a b=36 a b-4 a b=32 a b$
$\Rightarrow a-b=\sqrt{32} \sqrt{a b}$
$=4 \sqrt{2} \sqrt{a b}$ .........$(2)$
Adding $(1)$ and $(2),$ we obtain
$2 a=(6+4 \sqrt{2}) \sqrt{a b}$
$a=(3+2 \sqrt{2}) \sqrt{a b}$
Substituting the value of $a$ in $(1),$ we obtain
$b=6 \sqrt{a b}-(3+2 \sqrt{2}) \sqrt{a b}$
$\Rightarrow b=(3-2 \sqrt{2}) \sqrt{a b}$
$\frac{a}{b}=\frac{(3+2 \sqrt{2}) \sqrt{a b}}{(3-2 \sqrt{2}) \sqrt{a b}}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}$
Thus, the required ratio is $(3+2 \sqrt{2}):(3-2 \sqrt{2})$
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