Question

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+2√2) : (3-2√2).

Answer 1

Let the two numbers be a and b.
G.M. = √ab
According to the given condition,
a + b = 6√ab ...(1)
⇒ (a + b)2 = 36(ab)
Also,
(a - b)2 = (a + b)2 - 4ab = 36ab - 4ab = 32ab
⇒ a - b = √32 √ab
= 4√2 √ab ....(2)
Adding (1) and (2), we obtain
2a = (6 + 4√2) √ab
⇒ a = (3 + 2√2) √ab
Substituting the value of a in (1), we obtain
b = 6 √ab -(3 + 2 √2) √ab
⇒ b = (3 - 2√2) √ab
$\frac{a }{ b} = \frac{(3 + 2√2) √ab }{ (3 - 2√2) √ab} = \frac{3 + 2 √2 }{3 - 2 √2}$
Thus, the required ratio is (3+2 √2) : (3-2√2).