2007 JBMO Problems/Problem 1

Problem

Let $a$ be positive real number such that $a^{3}=6(a+1)$ . Prove that the equation $x^{2}+ax+a^{2}-6=0$ has no real solution.

Solution

The discriminant of the equation $x^{2}+ax+a^{2}-6=0$ is $a^2 - 4(a^2 - 6) = -3a^2 + 24.$ In order for the quadratic equation to have no real solution, the discriminant must be less than zero, so we need to show that $-3a^2 + 24 < 0.$ That means we need to show that $a > 2\sqrt{2}.$

Assume that $a \le 2\sqrt{2}.$ Rearranging the equation $a^{3}=6(a+1)$ results in $a(a^2 - 6) = 6.$ If $0 < a < \sqrt{6},$ then $a(a^2 - 6)$ would be negative, making the equality fail. If $\sqrt{6} \le a \le 2\sqrt{2},$ then $0 \le a^2 - 6 \le 2$ , making $0 \le a(a^2 - 6) \le 4\sqrt{2}.$ However, that means $a(a^2 - 6) \le 4\sqrt{2} < 6,$ so the equality also fails.

Thus, by proof by contradiction, $a$ must be greater than $2\sqrt{2}$ , so the discriminant of the equation $x^{2}+ax+a^{2}-6=0$ is negative. That means the equation $x^{2}+ax+a^{2}-6=0$ has no real solution.

Solution 2

Alternatively it is possible to show that $3a^2 > 24$ or $a^2 > 8$ . However, the original expression also rearranges to $a^2 = \frac{6(a+1)}{a}$ . Therefore it is enough to prove that $6(a+1)>8a$ or $a < 3$ . Evaluating $a^3$ and $6(a+1)$ at $a=2$ and $a=3$ shows the intersection point is between these values, and there are clearly no more because the cube is convex on positive numbers. That means the quadratic in the question has no real solution.

2007 JBMO (Problems • Resources)
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2007 JBMO Problems/Problem 1

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Problem

Solution

Solution 2

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