Question

The arithmetic mean between $a$ and $b$ is twice the geometric mean between $a$ and $b$.
Prove that $:\frac{a}{b}=7+4\sqrt{3}$ or $7-4\sqrt{3}$.

Answer 1

Given that

A.M of $,b=2\times$ G.M of $a,b$

$\Rightarrow$ $\frac{a+b}{2}=2\sqrt{ab}$

$\Rightarrow$ $a+b=4\sqrt{ab}$

Squaring both sides, we get

$\Rightarrow$ $a^2+b^2+2ab=16ab$

$\Rightarrow$ $a^2+b^2=14ab$

$\Rightarrow$ $\frac{a}{b}+\frac{b}{a}=14$

Let $\frac{a}{b}=x$

$\Rightarrow$ $x+\frac{1}{x}=14$

$\Rightarrow$ $x^2-14x+1=0$

$\Rightarrow$ $x=\frac{-(-14)\pm \sqrt{{14}^2-4}}{2}$

$\Rightarrow$ $\frac{14\pm \sqrt{192}}{2}$

$\therefore$ $x=\frac{a}{b}=7+4\sqrt{3}$ or $7-4\sqrt{3}$