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

Let the two positive numbers be a and b.

Let A and G be the arithmetic mean (A.M.) and geometric mean (G.M.) of a and b respectively.

Then, \[ A = \frac{a + b}{2} \quad \text{and} \quad G = \sqrt{ab} \]

From the definition of A.M., we have \[ a + b = 2A \quad \text{… (1)} \]

From the definition of G.M., squaring both sides gives \[ ab = G^2 \quad \text{… (2)} \]

The two numbers a and b are the roots of the quadratic equation \[ x^2 - (a + b)x + ab = 0 \]

Substituting the values from (1) and (2), we get \[ x^2 - 2Ax + G^2 = 0 \]

Solving this quadratic equation, we have \[ x = \frac{2A \pm \sqrt{(2A)^2 - 4G^2}}{2} \]

\[ x = A \pm \sqrt{A^2 - G^2} \]

\[ x = A \pm \sqrt{(A + G)(A - G)} \]

Hence, the two positive numbers are \[ A \pm \sqrt{(A + G)(A - G)}. \]

Thus, the given result is proved.