Question

If the sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is equal to the sum of the square of their reciprocals, then \( \frac{a}{c}, \frac{b}{a}, \frac{c}{b} \) are in:

• A. AP
• B. GP
• C. HP
• D. None of the above

Hint:

Use root-sum and product formulas with care when transforming conditions involving reciprocals or squares of roots.

Answer 1

The Correct Option is C

Step 1: Use root properties of a quadratic equation.
Let \( \alpha, \beta \) be roots. Then: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a} \] Step 2: Use given condition.
\[ \alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2} \Rightarrow -\frac{b}{a} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \] Now, \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{b^2}{a^2} - \frac{2c}{a} \right), \quad \alpha^2 \beta^2 = \left(\frac{c}{a} \right)^2 \] Substitute: \[ -\frac{b}{a} = \frac{b^2 - 2ac}{a^2} \cdot \frac{a^2}{c^2} = \frac{b^2 - 2ac}{c^2} \Rightarrow \frac{b}{a} = -\frac{b^2 - 2ac}{c^2} \] This relation leads to: \[ \frac{1}{\frac{a}{c}}, \frac{1}{\frac{b}{a}}, \frac{1}{\frac{c}{b}} \text{ are in A.P.} \Rightarrow \frac{a}{c}, \frac{b}{a}, \frac{c}{b} \text{ are in H.P.} \]