Difference between revisions of "2007 JBMO Problems/Problem 1"

Revision as of 17:31, 17 August 2018

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.

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Difference between revisions of "2007 JBMO Problems/Problem 1"

Revision as of 17:31, 17 August 2018

Problem

Solution

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