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Problem 1782

National Olympiad, first round
Algebra Difficulty 6.6 Find the answer Selection tests for the Balkan Mathematical Olympiad · Saudi Arabia · 2013

Let kk be a real number such that the product of real roots of the equation
X4+2X3+(2+2k)X2+(1+2k)X+2k=0 X^{4}+2 X^{3}+(2+2 k) X^{2}+(1+2 k) X+2 k=0
is −2013-2013. Find the sum of the squares of these real roots.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Notice first that
X4+2X3+(2+2k)X2+(1+2k)X+2k=(X2+X+1)(X2+X+2k). X^{4}+2 X^{3}+(2+2 k) X^{2}+(1+2 k) X+2 k = (X^{2}+X+1)(X^{2}+X+2k).
Because the factor X2+X+1X^{2}+X+1 has no real roots, we deduce from Vieta relations that r1+r2=−1r_{1}+r_{2}=-1 and r1r2=2k=−2013r_{1} r_{2}=2k=-2013, where r1,r2r_{1}, r_{2} are the real roots of the equation X4+2X3+(2+2k)X2+(1+2k)X+2k=0X^{4}+2 X^{3}+(2+2 k) X^{2}+(1+2 k) X+2 k=0. Therefore,
r12+r22=(r1+r2)2−2r1r2=1+2×2013=4027. r_{1}^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}-2 r_{1} r_{2}=1+2 \times 2013=4027.

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