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 'a' and 'b' be two numbers

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

Now, $A\pm \sqrt{\sqrt{(A+G)(A-G)}}=A\pm \sqrt{A^2-G^2}$

= $\frac{a+b}{2}\pm \sqrt{{\left(\frac{a+b}{2}\right)}^2+\left({\sqrt{ab}}^2\right)}$

$\frac{a+b}{2}\pm \sqrt{\frac{a^2+b^2+2ab}{4}-ab}$

=$\frac{a+b}{2}\pm \sqrt{\frac{a^2+b^2+2ab-4ab}{4}}$

= $\frac{a+b}{2}\pm \sqrt{\frac{(a-b{)}^2}{4}}=\frac{(a+b)}{2}\pm \frac{(a-b)}{2}$

Now, $A+\sqrt{(A+G)(A-G)}$

= $\frac{a+b}{2}+\frac{a-b}{2}=\frac{a+b+a-b}{2}=\frac{2a}{2}=b$

and $A-\sqrt{(A+G)(A-G)}$

= $\frac{a+b}{2}-\frac{a-b}{2}$

= $\frac{a+b-a+b}{2}=\frac{2b}{2}=b$