Question

If $ A $ is the A.M. of the roots of the equation $ x^2 - 2ax + b = 0 $ and $ G $ is the G.M. of the roots of the equation $ x^2 - 2bx + a^2 = 0 $, $ a>0 $, then:

• A. \( A>G \)
• B. \( A = G \)
• C. \( A<G \)
• D. None of these

Hint:

For the quadratic equations, use Vieta's formulas to find the sum and product of the roots, which helps calculate the A.M. and G.M.

Answer 1

The Correct Option is A

Let the roots of the equation \( x^2 - 2ax + b = 0 \) be \( r_1 \) and \( r_2 \). Using Vieta's formulas, we know that: \[ r_1 + r_2 = 2a \quad \text{and} \quad r_1r_2 = b. \] The A.M. (Arithmetic Mean) of the roots is: \[ A = \frac{r_1 + r_2}{2} = \frac{2a}{2} = a. \] Next, let the roots of the equation \( x^2 - 2bx + a^2 = 0 \) be \( s_1 \) and \( s_2 \). Using Vieta's formulas again: \[ s_1 + s_2 = 2b \quad \text{and} \quad s_1s_2 = a^2. \] The G.M. (Geometric Mean) of the roots is: \[ G = \sqrt{s_1s_2} = \sqrt{a^2} = a. \] Therefore, \( A = a \) and \( G = a \). Since both the A.M. and G.M. are equal to \( a \), the correct answer is \( A = G \).