Question

If the AM of two positive numbers be three times their geometric mean then the ratio of the numbers is

• A. $3\pm 2\sqrt{2}$
• B. $\sqrt{2}\pm 1$
• C. $17+12\sqrt{2}$
• D. ${(3-2\sqrt{2})}^{-2}$

Answer 1

Answer: (C), (D)

The correct options are
C $17+12\sqrt{2}$
D ${(3-2\sqrt{2})}^{-2}$

Let the numbers be $a$ and $b$.
Given
$\frac{a+b}{2}=3\sqrt{ab}$
$\Rightarrow \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=6$
$\Rightarrow t+\frac{1}{t}=6$ (where $t=\sqrt{\frac{a}{b}}$)
$\Rightarrow t^2-6t+1=0$
$\Rightarrow t=\frac{6\pm \sqrt{6^2-4}}{2}$
$\Rightarrow t=3\pm 2\sqrt{2}$
$\Rightarrow \frac{a}{b}=17\pm 12\sqrt{2}$
$\Rightarrow \left(i\right)\frac{a}{b}=17+12\sqrt{2}\therefore$ option C is correct.
or $\left(ii\right)\frac{a}{b}=17-12\sqrt{2}$
$\Rightarrow \frac{b}{a}=\frac{1}{17-12\sqrt{2}}=\frac{1}{(3-2\sqrt{2}{)}^2}\therefore$ option D is correct.