Question

If the A.M of two positive number a and b (a >b) is twice their G.M, then a : b is:

• A. $2:\sqrt{3}$
• B. $2:7+4\sqrt{3}$
• C. $2+\sqrt{3}:2-\sqrt{3}$
• D. $7+4\sqrt{3}:7-4\sqrt{3}$

Answer 1

The correct option is C $2+\sqrt{3}:2-\sqrt{3}$

Given that A.M = 2 G.M
$\Rightarrow \frac{a+b}{2}=2\sqrt{ab}$
$\Rightarrow a+b=4\sqrt{ab}$
Now, let us consider choice (c)
$\therefore (2+\sqrt{3})+(2-\sqrt{3})=4\sqrt{(2+\sqrt{3})(2-\sqrt{3})}$
$4=4.1$
4 = 4
Hence choice (c) is correct.
Alternatively: $a+b=4\sqrt{ab}$
$\Rightarrow (a+b{)}^2=16ab$
$\Rightarrow a^2+b^2+2ab=16ab$
$\Rightarrow a^2+b^2=14ab$
$\Rightarrow \frac{a}{b}+\frac{b}{a}=14$
$\Rightarrow x+\frac{1}{x}=14$ $(\because \frac{a}{b}=x)$
$\Rightarrow x^2-14x+1=0$
$\Rightarrow x=\frac{14\pm \sqrt{192}}{2}$
$\Rightarrow x=7\pm 4\sqrt{3}$
$\Rightarrow \frac{a}{b}=7+4\sqrt{3}=(2{)}^2+(\sqrt{3}{)}^2+4\sqrt{3}$
$\frac{a}{b}=(2+\sqrt{3}{)}^2=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Now $x=7-4\sqrt{3}$
$\Rightarrow x=\frac{2-\sqrt{3}}{2+\sqrt{3}}\ne \frac{a}{b}$, since a should be greater than b.
Hence, $x\ne 7-4\sqrt{3}$