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Bulgarian Mathematical Olympiad

Algebra Difficulty 5.2 AIME, harder Prove it Bulgaria

Problem:
Find all real numbers $a$ such that the roots $x_{1}$ and $x_{2}$ of the equation
$x^{2}+6x+6a-a^{2}=0$
satisfy the relation $x_{2}=x_{1}^{3}-8x_{1}$ .

Solution

Solution:
It follows from Vieta's formulae that
$-6 = x_{1} + x_{2} = x_{1} + x_{1}^{3} - 8x_{1}$
Therefore $x_{1}^{3} - 7x_{1} + 6 = 0$ and $x_{1} = -3, 1$ or $2$ .

Plugging these values of $x_{1}$ in the initial equation gives $a = 3$ for $x = -3$ , $a = -1$ and $7$ for $x = 1$ , and $a = -2$ and $8$ for $x = 2$ .

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