Question

The arithmetic mean of two numbers is $3$ times their geometric mean and the sum of the squares of the two numbers is $34$. The two numbers are

• A. $2\sqrt{3} +\sqrt{5}, 2\sqrt{3} -\sqrt{5}$
• B. $3+2\sqrt{2}, 3- 2\sqrt{2}$
• C. $\sqrt{10} +\sqrt{7}, \sqrt{10} - \sqrt{7} $
• D. $\sqrt{5} -2\sqrt{2}, \sqrt{5} +2\sqrt{2}$

Answer 1

Answer: (B)

$a+b = 3\cdot2 \sqrt{ab} $
$ \Rightarrow \frac{a}{b} -6 \sqrt{\frac{a}{b}}+1 = 0 $
$ \Rightarrow \sqrt{\frac{a}{b}} = 3 \pm2\sqrt{2} $
$ \Rightarrow \frac{a}{b} = \left(3\pm2\sqrt{2}\right)^{2}$
$ = 17 \pm12\sqrt{2}$
or $\frac{a}{b} =\frac{ 3+2\sqrt{2}}{3-2\sqrt{2}}\cdot\frac{ 3+2\sqrt{2}}{3+2\sqrt{2}} $
or $\frac{ 3-2\sqrt{2}}{3+2\sqrt{2}}\cdot\frac{ 3-2\sqrt{2}}{3-2\sqrt{2}} $.
$ \Rightarrow a= \left(3\pm 2\sqrt{2}\right)k$,
$ b = \left(3\mp2\sqrt{2}\right)k $
As $a^{2}+b^{2} = 34$
$ \Rightarrow k = 1$,
thus the two numbers are $3+2\sqrt{2}$ and $3 - 2\sqrt{2}$.