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

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Ans: Let the two numbers be a and b.

Thus their 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= 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

»​ a/b = (3+2√2) √ab / (3-2√2) √ab

          = (3+2√2) / (3-2√2)

Thus , the required ratio of the no. are (3+2√2) / (3-2√2). 

Alternatively, 

Let the two no. be a and b.

» a+b=6 √ab

» a+b/2√ab = 3

Using dividendo & componendo, we get:

» a+b-2√ab /  a+b+2√ab = 3-1 / 3+1

» (√a-√b)² / (√a+√b)² = 1/2

» (√a-√b) / (√a+√b) = 1/√2

Using dividendo & componendo, we get:

» √a+√b-√a-√b / √a+√b+√a+√b = 1-√2 / 1+√2

» -2√b / 2√a = 1-√2 / 1+√2

Reversing both sides, we get:

» √a/√b = 1+√2 / 1-√2

Squaring both sides, we get:

» a/b = (1+√2)² / (1-√2)² = 1+2+2√2 /  1+2-2√2  

          = (3+2√2) / (3-2√2)

Thus , the required ratio of the no. are (3+2√2) / (3-2√2). 

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