Question

Let α, β, γ be the real roots of the equation, x³ + a x² + b x + c = 0, (a, b, c ∈ R and a, b ≠ 0). If the system of equations (in u, v, w) given by α u + β v + γ w = 0; β u + γ v + α w = 0; γ u + α v + β w = 0 has non-trivial solution, then the value of a²/b is :

• A. 0
• B. 1
• C. 3
• D. 5

Hint:

The expression $\alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha = 0$ implies $\alpha=\beta=\gamma$ for real numbers.

Answer 1

The Correct Option is C

Step 1: For a non-trivial solution, the determinant of the coefficients must be zero. 
Step 2:
 
Step 3: This is a circulant determinant: $-(\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) = 0$. 
Step 4: From $x^3 + ax^2 + bx + c = 0$: $\alpha+\beta+\gamma = -a$ and $\alpha\beta+\beta\gamma+\gamma\alpha = b$. 
Step 5: Case 1: $\alpha+\beta+\gamma = 0 \implies -a = 0$ (but $a \neq 0$). 
Step 6: Case 2: $\alpha^2+\beta^2+\gamma^2 - (\alpha\beta + \beta\gamma + \gamma\alpha) = 0$. $(\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha) = 0 \implies (-a)^2 - 3b = 0 \implies a^2 = 3b$. 
Step 7: $a^2/b = 3$.