Question

Find two positive numbers whose difference is 12 and whose A.M. exceeds the G.M. by 2

Answer 1

We have,

$a-b=12$

$Am=Gm+2$

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

${\left(a+b\right)}^2={\left(2\sqrt{ab}+4\right)}^2$

${\left(a-b\right)}^2+4ab=4ab+16+16\sqrt{ab}$

$144=16+16\sqrt{ab}$

$9=1+\sqrt{ab}$

$ab=64$

$b^2+12b-64=0$

$\left(b+16\right)\left(6-4\right)=0$

$b=4$

Hence,

$a=16,b=4$