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IRL ABooklet

Algebra Difficulty 6.1 National Olympiad Prove it Ireland

There are real numbers α\alpha, β\beta, such that the cubic functions
f(x)=x3−3x2+αx+βand f(x) = x^3 - 3x^2 + \alpha x + \beta \quad \text{and}
g(x)=x3+(α−3)x−6 g(x) = x^3 + (\alpha - 3)x - 6
have exactly two distinct non-zero roots in common. Find α\alpha and β\beta.

Solution

The two distinct non-zero common roots of ff and gg are also roots of g−fg - f. These are the two roots of the quadratic
g(x)−f(x)=3(x2−x−β3−2), g(x) - f(x) = 3 \left( x^2 - x - \frac{\beta}{3} - 2 \right),
hence we must have
g(x)=(x−λ)(x2−x−β3−2)=x3−(1+λ)x2+(λ−2−β3)x+λ(2+β3) \begin{aligned} g(x) &= (x - \lambda) \left( x^2 - x - \frac{\beta}{3} - 2 \right) \\ &= x^3 - (1 + \lambda)x^2 + \left( \lambda - 2 - \frac{\beta}{3} \right)x + \lambda \left( 2 + \frac{\beta}{3} \right) \end{aligned}
for some real number λ\lambda. Using that g(x)=x3+(α−3)x−6g(x) = x^3 + (\alpha - 3)x - 6 and comparing coefficients gives λ=−1\lambda = -1, α=−β3\alpha = -\frac{\beta}{3} and β=12\beta = 12, and hence α=−4\alpha = -4.

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