The sum of two numbers is 6 times their geometric means, show that the

1.	The sum of two numbers is 6 times their geometric means, show that the numbers are in the ratio (3 + 2√2 ) : (3 – 2√2).
Answer» Let the two numbers be a and b. GM = √ab According to the given condition, a + b = 6√ab …(1) (a + b)² = 36ab Also, (a – b)² = (a + b)² – 4ab = 36ab – 4ab = 32ab ⇒ a – b = √32ab = 4√2ab …..(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 =(3 – 2√2)√ab \(\therefore\) \(\cfrac{3 + 2\sqrt{2}}{3-2\sqrt{2}}\) Thus, the required ratio is (3+2√2) : 3–2√2.

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

Answer»

Let the two numbers be a and b.

GM = √ab

According to the given condition,

a + b = 6√ab …(1)

(a + b)² = 36ab

Also,

(a – b)² = (a + b)² – 4ab

= 36ab – 4ab

= 32ab

⇒ a – b = √32ab

= 4√2ab …..(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 =(3 – 2√2)√ab

\(\therefore\) \(\cfrac{3 + 2\sqrt{2}}{3-2\sqrt{2}}\)

Thus, the required ratio is (3+2√2) : 3–2√2.