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Algebra Difficulty 4.3 AIME Find the answer Slovenia

Let $a$ be a real number for which the quadratic equations $x^2 + a x + 2 = 0$ and $x^2 + 2x + a = 0$ have real roots. The sum of the squares of the roots of the first equation equals the sum of the squares of the roots of the second equation. What is the value of $a$ ?

Pick one

Solution

The discriminant of the equation $x^2 + a x + 2 = 0$ is $D = a^2 - 8 \ge 0$ . Its roots $x_{1,2} = \frac{-a \pm \sqrt{a^2-8}}{2}$ satisfy $x_1^2 + x_2^2 = a^2 - 4$ .

The discriminant of the equation $x^2 + 2x + a = 0$ is $D = 4 - 4a \ge 0$ and the roots $x_{1,2} = \frac{-2 \pm \sqrt{4-4a}}{2}$ satisfy $x_1^2 + x_2^2 = 4 - 2a$ .

It follows that $a^2 - 4 = 4 - 2a$ , which can be rearranged as $(a-2)(a+4) = 0$ .

The solution $a=2$ is not the correct one because it would result in the discriminant of the second equation being negative. The solution $a=-4$ satisfies all of the conditions.

The correct answer is (A).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.