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Bulgarian Mathematical Olympiad

Algebra Difficulty 4.9 AIME Prove it Bulgaria

Problem:
Find all integers aa such that the equation
x4+2x3+(a2−a−9)x2−4x+4=0 x^{4} + 2x^{3} + (a^{2} - a - 9)x^{2} - 4x + 4 = 0
has at least one real root.

Solution

Solution:
Set u=x−2xu = x - \frac{2}{x}. Then the equation becomes
u2+2u+a2−a−5=0 u^{2} + 2u + a^{2} - a - 5 = 0
Since the equation x2−ux−2=0x^{2} - u x - 2 = 0 has real solutions for any real uu, it suffices to find the integer values of aa for which the equation (1) has a real root. The last holds when D=−a2+a+6≥0D = -a^{2} + a + 6 \geq 0, i.e. (a−3)(a+2)≤0(a - 3)(a + 2) \leq 0 giving a∈[−2,3]a \in [-2, 3]. Therefore a=−2,−1,0,1,2,3a = -2, -1, 0, 1, 2, 3.

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