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Hellenic Mathematical Olympiad ARCHIMEDES

Algebra Difficulty 5.7 AIME, harder Prove it Greece

Determine the values of the parameter α∈R\alpha \in \mathbb{R} for which the equation x2+(α−2)x−(α−1)(2α−3)=0x^2 + (\alpha - 2)x - (\alpha - 1)(2\alpha - 3) = 0 has two roots such that the one of them is equal to the square of the other.

Solution

1. We have the discriminant D=(α−2)2+4(α−1)(2α−3)=(3α−4)2D = (\alpha - 2)^2 + 4(\alpha - 1)(2\alpha - 3) = (3\alpha - 4)^2, and so the equation has the roots x1=α−1x_1 = \alpha - 1, x2=−2α+3x_2 = -2\alpha + 3.

Therefore we seek the values of α\alpha for which:
x1=x22 or x2=x12⇔α−1=(−2α+3)2 or −2α+3=(α−1)2⇔α−1=4α2−12α+9 or −2α+3=α2−2α+1 \begin{align*} x_1 = x_2^2 \text{ or } x_2 = x_1^2 &\Leftrightarrow \alpha - 1 = (-2\alpha + 3)^2 \text{ or } -2\alpha + 3 = (\alpha - 1)^2 \\ &\Leftrightarrow \alpha - 1 = 4\alpha^2 - 12\alpha + 9 \text{ or } -2\alpha + 3 = \alpha^2 - 2\alpha + 1 \end{align*}
⇔4α2−13α+10=0 or α2=2⇔α=2 or α=54 or α=2 or α=−2. \Leftrightarrow 4\alpha^2 - 13\alpha + 10 = 0 \text{ or } \alpha^2 = 2 \Leftrightarrow \alpha = 2 \text{ or } \alpha = \frac{5}{4} \text{ or } \alpha = \sqrt{2} \text{ or } \alpha = -\sqrt{2}.

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