Question

If A be A.M. and G be the G.M. between two numbers, show that the number are given by $A\pm \sqrt{A^2-G^2}$.

Answer 1

Let the numbers be a, b

$A=\frac{a+b}{2}G=\sqrt{ab}$

$a+b=2A,ab=G^2$

$\Rightarrow (a-b{)}^2=(a+b{)}^2-4ab$

$\Rightarrow (a-b{)}^2=4A^2-4G^2$

$\Rightarrow a-b=2\sqrt{A^2-G^2}$

$\Rightarrow (a+b)+(a-b)=2A+2\sqrt{A^2-G^2}\Rightarrow 2a=2(A+\sqrt{A^2-G^2})\Rightarrow a=A+\sqrt{A^2-G^2}$

$\Rightarrow (a+b)-(a-b)\Rightarrow b=A-\sqrt{A^2-G^2}$

Hence $a,b=A\pm \sqrt{A^2-G^2}$