Question

If $\alpha$, $\beta$, and $\gamma$ are the roots of the equation $2x^3 + 3x^2 - 5x - 7 = 0$, then $\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} =$

• A. $\frac{17}{49}$
• B. $-\frac{23}{49}$
• C. $\frac{55}{49}$
• D. $\frac{67}{49}$

Hint:

For sums involving reciprocals of roots, use the identity $\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{(\alpha\beta\gamma)^2}$ with Vieta’s formulas.

Answer 1

The Correct Option is D

For the cubic $2x^3 + 3x^2 - 5x - 7 = 0$, use Vieta’s formulas: - Sum of roots: $\alpha + \beta + \gamma = -\frac{3}{2}$ - Sum of pairwise products: $\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{-5}{2} = \frac{5}{2}$ - Product of roots: $\alpha\beta\gamma = -\frac{-7}{2} = \frac{7}{2}$ We need $\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha^2\beta^2\gamma^2}$. The denominator is: \[ (\alpha\beta\gamma)^2 = \left(\frac{7}{2}\right)^2 = \frac{49}{4} \] The numerator is: \[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \] \[ = \left(\frac{5}{2}\right)^2 - 2 \cdot \frac{7}{2} \cdot \left(-\frac{3}{2}\right) = \frac{25}{4} + \frac{42}{4} = \frac{67}{4} \] Thus: \[ \frac{\frac{67}{4}}{\frac{49}{4}} = \frac{67}{49} \] Option (4) is correct. Options (1), (2), and (3) do not match the computed value.