Let α, β, γ be the real roots of the equation, x³ + ax² + bx + c = 0, (a, b, c ∈ R and a, b ≠ 0). If the system of equations (in u, v, w) given by αu + βv + γw = 0, βu + γv + αw = 0; γu + αv + βw = 0 has non-trivial solution, then the value of a²/b is

Solution

Maths sir · 4 steps

Non-trivial solution means the determinant is zero, and this circulant determinant factorises neatly.

1. Determinant must vanish. A homogeneous system has a non-trivial solution only when its coefficient determinant is zero. Expanding along the first row gives 3αβγ - α³ - β³ - γ³.

   |α, β, γ; β, γ, α; γ, α, β| = -(α+β+γ)(α²+β²+γ²-αβ-βγ-γα) = 0

2. Pick the right factor. By Vieta, α+β+γ = -a, and a ≠ 0, so the first factor cannot be zero. So α²+β²+γ²-αβ-βγ-γα = 0.
3. Write it using Vieta. This factor equals (α+β+γ)² - 3(αβ+βγ+γα). With αβ+βγ+γα = b, it becomes a² - 3b = 0.
4. So. a² = 3b, and since b ≠ 0, a²/b = 3. The answer is (b).

The idea to keep. A cyclic determinant in α, β, γ always gives -(α³+β³+γ³-3αβγ). Factorise it and turn each factor into Vieta sums.

Answer from the NTA final answer key. Solution by PrepWiser.