Question

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A±√(A+G)(A-G).

Answer 1

It is given that A and G are A.M. and G.M. between two positive numbers. Let these two positive numbers be a and b.
$∴ AM = A = \frac{a + b }{2}$ ...(1)
$GM = G = √ab$ ...(2)
From (1) and (2), we obtain
$a + b = 2A … (3)$
$ab = G^ 2 … (4)$
Substituting the value of a and b from (3) and (4) in the identity (a-b) 2 =(a+b) 2- 4ab, we obtain
$(a- b)^ 2 = 4A ^2- 4G ^2 = 4 (A ^2-G ^2 )$
$(a- b) ^2 = 4 (A+G) (A-G)$
$(a - b) = 2 \sqrt{(A + G)(A - G)}$
From (3) and (5), we obtain
$2a = 2a + 2 \sqrt{(A + G)(A - G)}$
$⇒ a = A + \sqrt{(A + G)(A - G)}$
Substituting the value of a in (3), we obtain
$b = 2A - A - \sqrt{(A + G)(A - G)} = A - \sqrt{√(A + G)(A - G)}$
Thus , the two numbers are $A ± \sqrt{(A + G)(A - G).}$