Question

If A and G be A.M. and G.M., respectively between two positive numbers prove that the numbers are $A\pm \sqrt{(A+G)(A-G)}.$

Answer 1

Let the two numbers be $a$ and $b$
$\therefore AM=A=\frac{a+b}{2}...\left(1\right)$ & $GM=G=\sqrt{ab}...\left(2\right)$
From (1) and (2) ,we obtain
$a+b=2A...\left(3\right)$
& $ab=G^2...\left(4\right)$
Substituting the value of $a$ and $b$ from(3) and (4) in the identity
${(a-b)}^2={(a+b)}^2-4ab,$ we obtain
${(a-b)}^2=4A^2-{4G}^2=4(A^2-G^2){(a-b)}^2=4A^2-4(A+G)(A-G)(a-b)=2\sqrt{(A+G)(A-G)}$
From (3) and (5) , we obtain $2a=2A+2\sqrt{(A+G)(A-G)}$
$\Rightarrow a=A+\sqrt{(A+G)(A-G)}$
Substituting the value of $a$ in (3), we obtain
$b=2A-A-\sqrt{(A+G)(A-G)}=A-\sqrt{(A+G)(A-G)}$
Thus , the two numbers are $A\pm \sqrt{(A+G)(A-G)}$