Question

If sum of two positive numbers $a$ and $b,$ where $a>b$ is thrice their G.M. and $\frac{a}{b}=\frac{7+\sqrt{p}}{2},$ then the value of $p$ is

Answer 1

Given :
$a+b=3\sqrt{ab}$
Squaring both sides,
$a^2+b^2+2ab=9ab\Rightarrow a^2+b^2=7ab$
Assuming $a\ne b\ne 0$
$\frac{a}{b}+\frac{b}{a}=7$
Taking $\frac{a}{b}=t$
$\Rightarrow t+\frac{1}{t}=7\Rightarrow t^2-7t+1=0\Rightarrow t=\frac{7\pm \sqrt{45}}{2}\Rightarrow t=\frac{7+\sqrt{45}}{2}(\because a>b\Rightarrow \frac{a}{b}>1)\Rightarrow \frac{a}{b}=\frac{7+\sqrt{45}}{2}\therefore p=45$