Question

If the arithmetic mean between $a$ and $b$ is twice as great as the geometric mean, show that $a:b=2+\sqrt{3}:2-\sqrt{3}$.

Answer 1

Arithmetic mean between $a$ and $b$ is given by
$A.M=\frac{a+b}{2}$
Geometric mean between $a$ and $b$ is given by
$G.M=\sqrt{ab}$
Now, it is given that $A.M.=2G.M.$
$\frac{a+b}{2}=2\sqrt{ab}$
Squaring both sides, we get
$(a+b{)}^2=16ab$
$\Rightarrow a^2+b^2-14ab=0$
$a=\frac{14b\pm \sqrt{196b^2-4b^2}}{2}$
$\Rightarrow a=7b\pm \sqrt{49b^2-b^2}$
$\Rightarrow a=7b\pm 4b\sqrt{3}$
$\Rightarrow a:b=7\pm 4\sqrt{3}$

$\Rightarrow a:b=7+4\sqrt{3}$ or $a:b=7-4\sqrt{3}$

$\Rightarrow a:b=(2+\sqrt{3}):(2-\sqrt{3})$ or $a:b=(2-\sqrt{3}):(2+\sqrt{3})$

The hypothesis is true only when $a$ is greater of two numbers.

Then, $a:b=(2+\sqrt{3}):(2-\sqrt{3})$