Question

If $ \alpha, \beta, \gamma $ are the roots of the equation $ x^3 - 12x^2 + kx - 18 = 0 $ and one of them is thrice the sum of the other two, then $$ \alpha^2 + \beta^2 + \gamma^2 - k = ? $$

• A. 115
• B. 41
• C. 56
• D. 57

Hint:

Use symmetry and Vieta’s formulas to express roots in terms of each other, especially when a root is a multiple of others.

Answer 1

The Correct Option is D

Let \( \alpha + \beta + \gamma = 12 \), by Vieta's formula. Suppose \( \alpha = 3(\beta + \gamma) \) Then: \[ \alpha + \beta + \gamma = 12 \Rightarrow 3(\beta + \gamma) + \beta + \gamma = 12 \Rightarrow 4(\beta + \gamma) = 12 \Rightarrow \beta + \gamma = 3 \Rightarrow \alpha = 9 \] So roots are \( \alpha = 9,\ \beta + \gamma = 3,\ \beta \gamma = \frac{18}{\alpha} = 2 \) Now: \[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \Rightarrow (12)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \] We know: \[ \alpha\beta + \beta\gamma + \gamma\alpha = k \Rightarrow \alpha^2 + \beta^2 + \gamma^2 - k = 144 - 2k - k = 144 - 3k \] From polynomial, use identity \( \alpha\beta + \beta\gamma + \gamma\alpha = k \) Try \( k = 29 \Rightarrow 144 - 3(29) = 57 \)