If the A.M. of two positive numbers a and b (a > b) is twice their geometric mean. Prove that:

$\mathrm{AM}=2 \mathrm{GM}$

$\therefore \frac{a+b}{2}=2 \sqrt{a b}$

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

Squaring both the sides:

$\Rightarrow(a+b)^{2}=(4 \sqrt{a b})^{2}$

$\Rightarrow a^{2}+2 a b+b^{2}=16 a b$

$\Rightarrow a^{2}-14 a b+b^{2}=0$

Using the quadratic formula:

$\Rightarrow a=\frac{-(-14 b)+\sqrt{(-14 b)^{2}-4 \times 1 \times b^{2}}}{2 \times 1} \quad[\because a$ is positive number $]$

$\Rightarrow a=\frac{14 b+2 b \sqrt{49-1}}{2}$

$\Rightarrow a=b(7+4 \sqrt{3})$

$\Rightarrow \frac{a}{b}=7+4 \sqrt{3}$

$\Rightarrow \frac{a}{b}=4+3+2 \times 2 \times \sqrt{3}$

$\Rightarrow \frac{a}{b}=(2+\sqrt{3})^{2}$

$\Rightarrow \frac{a}{b}=\frac{(2+\sqrt{3})^{2}(2-\sqrt{3})}{(2-\sqrt{3})}$

$\Rightarrow \frac{a}{b}=\frac{(2+\sqrt{3})(4-3)}{(2-\sqrt{3})}$

$\therefore \frac{a}{b}=\frac{(2+\sqrt{3})}{(2-\sqrt{3})}$