Question

The ratio of the A.M. and G.M. of two positive numbers $a$ and $b$, is $m:n$. Show that $a:b=(m+\sqrt{m^2-n^2}):(m-\sqrt{m^2-n^2})$

Answer 1

Let the two numbers be $a$ and $b$.
$\therefore A.M=\frac{a+b}{2}\text{and}G.M=\sqrt{ab}$
According to the given condition,
$\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\Rightarrow \frac{{(a+b)}^2}{4\left(ab\right)}=\frac{m^2}{n^2}\Rightarrow {(a+b)}^2=\frac{4ab{m}^2}{n^2}\Rightarrow (a+b)=\frac{2\sqrt{ab}m}{n}....\left(1\right)$
Using this in the identity ${(a-b)}^2={(a+b)}^2-4ab,$ we obtain
${(a-b)}^2=\frac{{4abm}^2}{n^2}-4ab=\frac{4ab(m^2-n^2)}{n^2}\Rightarrow (a-b)=\frac{2\sqrt{ab}\sqrt{m^2-n^2}}{n}...\left(2\right)$
Adding (1) and (2) we obtain
$2a=\frac{2\sqrt{ab}}{n}(m+\sqrt{m^2+n^2})\Rightarrow a=\frac{\sqrt{ab}}{n}(m+\sqrt{m^2+n^2})$
Substituting the value of $a$ in (1), we obtain
$b=\frac{2\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}(m+\sqrt{m^2+n^2})=\frac{\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}\sqrt{m^2+n^2}=\frac{\sqrt{ab}}{n}(m-\sqrt{m^2+n^2})\therefore a:b=\frac{a}{b}=\frac{\frac{\sqrt{ab}}{n}(m+\sqrt{m^2+n^2})}{\frac{\sqrt{ab}}{n}(m-\sqrt{m^2+n^2})}=\frac{(m+\sqrt{m^2+n^2})}{(m-\sqrt{m^2+n^2})}\text{Thus}a:b=(m+\sqrt{m^2+n^2}):(m-\sqrt{m^2+n^2})$