Question

If \( \alpha, \beta \) are the roots of the equation \( ax^2 + bx + c = 0 \), then \[ \frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b} = \]

• A. \( \frac{2}{a} \)
• B. \( \frac{2}{b} \)
• C. \( \frac{2}{c} \)
• D. \( -\frac{2}{a} \)

Hint:

Use Vieta's relations for the sum and product of the roots of quadratic equations to simplify such problems effectively.

Answer 1

The Correct Option is D

Step 1: Using Vieta's formulas, the sum and product of the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are given by: \[ \alpha + \beta = -\frac{b}{a}, \quad \alpha \beta = \frac{c}{a}. \] Step 2: We need to find the value of the expression \[ \frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b}. \] We combine the terms into a single fraction: \[ \frac{\alpha(a\alpha + b) + \beta(a\beta + b)}{(a\beta + b)(a\alpha + b)}. \] Simplifying the numerator: \[ \alpha(a\alpha + b) + \beta(a\beta + b) = a\alpha^2 + b\alpha + a\beta^2 + b\beta = a(\alpha^2 + \beta^2) + b(\alpha + \beta). \] Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \), we substitute: \[ a((\alpha + \beta)^2 - 2\alpha \beta) + b(\alpha + \beta) = a\left(\left(-\frac{b}{a}\right)^2 - 2 \times \frac{c}{a}\right) + b\left(-\frac{b}{a}\right). \] Simplifying further: \[ = a\left(\frac{b^2}{a^2} - \frac{2c}{a}\right) - \frac{b^2}{a} = \frac{b^2}{a} - \frac{2ac}{a} - \frac{b^2}{a} = -\frac{2ac}{a}. \] Step 3: Now, for the denominator: \[ (a\beta + b)(a\alpha + b) = a^2 \alpha \beta + ab(\alpha + \beta) + b^2. \] Using the same values for \( \alpha \beta \) and \( \alpha + \beta \), this becomes: \[ = a^2 \times \frac{c}{a} + ab \times \left(-\frac{b}{a}\right) + b^2 = ac - \frac{b^2}{a} + b^2 = ac. \] Step 4: Combining the numerator and denominator, we get: \[ \frac{-\frac{2ac}{a}}{ac} = -\frac{2}{a}. \] Therefore, the value of the expression is \( -\frac{2}{a} \).