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Algebra Difficulty 5.3 AIME, harder Prove it

Let aa be a real positive number such that a3=6(a+1)a^{3}=6(a+1). Prove that the equation x2+ax+a2−6=0x^{2}+a x+a^{2}-6=0 has no solution in the set of the real number.

Solution

The discriminant of the equation is Δ=3(8−a2)\Delta=3\left(8-a^{2}\right). If we accept that Δ≥0\Delta \geq 0, then a≤22a \leq 2 \sqrt{2} and 1a≥24\frac{1}{a} \geq \frac{\sqrt{2}}{4}, from where a2≥6+6⋅24=6+6a≥6+322>8a^{2} \geq 6+6 \cdot \frac{\sqrt{2}}{4}=6+\frac{6}{a} \geq 6+\frac{3 \sqrt{2}}{2}>8 (contradiction).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.