Question

The arithmetic mean of two positive numbers is $2$ more than its geometric mean. If their difference is $8,$ then the numbers are

• A. $10,2$
• B. $9,1$
• C. $8,3$
• D. $12,4$

Answer 1

The correct option is B $9,1$

Let the numbers be $a$ and $b.$
According to the given condition,
$|a-b|=8\Rightarrow (a+b{)}^2-4ab=64\Rightarrow (a+b{)}^2=4ab+64\cdots \left(1\right)$

Given, $AM-GM=2$
$\Rightarrow \frac{a+b}{2}-\sqrt{ab}=2\Rightarrow a+b=2\sqrt{ab}+4\Rightarrow (a+b{)}^2=4ab+16+16\sqrt{ab}$
Using equation $\left(1\right),$ we get
$4ab+64=4ab+16+16\sqrt{ab}\Rightarrow \sqrt{ab}=3$

Now, $a+b=10$
and $|a-b|=8$
$\Rightarrow a=9,b=1$ or $a=1,b=9$

Therefore, the required numbers are $9,1$