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)

$a+b=3\cdot 2\sqrt{ab}$
$\Rightarrow \frac{a}{b}-6\sqrt{\frac{a}{b}}+1=0$
$\Rightarrow \sqrt{\frac{a}{b}}=3\pm 2\sqrt{2}$
$\Rightarrow \frac{a}{b}={\left(3\pm 2\sqrt{2}\right)}^2$
$=17\pm 12\sqrt{2}$
or $\frac{a}{b}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\cdot \frac{3+2\sqrt{2}}{3+2\sqrt{2}}$
or $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}\cdot \frac{3-2\sqrt{2}}{3-2\sqrt{2}}$.
$\Rightarrow a=\left(3\pm 2\sqrt{2}\right)k$,
$b=\left(3\mp 2\sqrt{2}\right)k$
As $a^2+b^2=34$
$\Rightarrow k=1$,
thus the two numbers are $3+2\sqrt{2}$ and $3-2\sqrt{2}$.