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

$\therefore$ A.M. of a and b $=\frac{a+b}{2}$

G.M. of a and b $=\sqrt{ab}$

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

By componendo and dividendo, we get

$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{m+n}{m-n}$

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

$\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}$

Again by componendo and dividendo, we have:

$\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

$\Rightarrow \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

Squaring both sides,

$\frac{a}{b}=\frac{(\sqrt{m+n}+\sqrt{m-n}{)}^2}{(\sqrt{m+n}-\sqrt{m-n}{)}^2}$

$\Rightarrow \frac{a}{b}=\frac{m+n+m-n+2\sqrt{(m+n)(m-n)}}{m+n+m-n-2\sqrt{(m+n)(m-n)}}$

$\Rightarrow \frac{a}{b}=\frac{2m+2\sqrt{m^2-n^2}}{2m-2\sqrt{m^2-n^2}}$

$\Rightarrow \frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}$

Thus,

$a:b=(m+\sqrt{m^2-n^2}):(m-\sqrt{m^2-n^2})$