Question

The roots $\alpha, \beta$ of the equation \[ x^2 - 6(k-1)x + 4(k-2) = 0 \] are equal in magnitude but opposite in sign. If $\alpha>\beta$, then the product of the roots of the equation \[ 2x^2 - \alpha x + 6\beta (\alpha + 1) = 0 \] is

• A. 12
• B. -12
• C. 16
• D. -18

Hint:

Use sum and product relations of roots and given conditions carefully to find parameters before calculating desired quantities.

Answer 1

The Correct Option is D

Given roots $\alpha$ and $\beta$ satisfy: \[ x^2 - 6(k-1)x + 4(k-2) = 0, \] and $\alpha = -\beta$ (equal magnitude, opposite sign). Sum of roots: \[ \alpha + \beta = 6(k - 1), \] Product of roots: \[ \alpha \beta = 4(k - 2). \] Since $\alpha = -\beta$, \[ \alpha + \beta = 0 \implies 6(k - 1) = 0 \implies k = 1. \] Product: \[ \alpha \beta = 4(k - 2) = 4(1 - 2) = -4. \] Since $\alpha = -\beta$, \[ \alpha \beta = -\alpha^2 = -4 \implies \alpha^2 = 4 \implies \alpha = 2, \beta = -2. \] The equation: \[ 2x^2 - \alpha x + 6\beta (\alpha + 1) = 0. \] Calculate the product of its roots: \[ \frac{c}{a} = \frac{6\beta(\alpha + 1)}{2} = 3\beta(\alpha + 1). \] Substitute values: \[ = 3 \times (-2) \times (2 + 1) = 3 \times (-2) \times 3 = -18. \] Hence, the product of roots is $-18$.