Question

The sum of two positive numbers is $6$ times their geometric mean. Then the ratio of these two numbers is

• A. $(3+2\sqrt{2}):(3-2\sqrt{2})$
• B. $(6+\sqrt{2}):(6-\sqrt{2})$
• C. $(3+4\sqrt{2}):(3-4\sqrt{2})$
• D. $(3+\sqrt{2}):(3-\sqrt{2})$

Answer 1

The correct option is A $(3+2\sqrt{2}):(3-2\sqrt{2})$

Let the two numbers be $a$ and $b.$
Thus, their $G.M.=\sqrt{ab}$
According to the given condition, $a+b=6\sqrt{ab}$
$\Rightarrow (a+b{)}^2=36ab...(1)$

Also, $(a-b{)}^2=(a+b{)}^2-4ab=36ab-4ab$
$\Rightarrow (a-b{)}^2=32ab...(2)$

$\left(1\right)÷\left(2\right),$
$\frac{(a+b{)}^2}{(a-b{)}^2}=\frac{36}{32}$
$\Rightarrow \frac{a+b}{a-b}=\frac{6}{4\sqrt{2}}$
Using componendo and dividendo, we get
$\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{6+4\sqrt{2}}{6-4\sqrt{2}}$
$\Rightarrow \frac{a}{b}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$
Thus, the required ratio is $(3+2\sqrt{2}):(3-2\sqrt{2})$