Selection tests for the Balkan Mathematical Olympiad 2013

Question

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

Accepted Answer

Notice first that 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 X^2+X+1 has no real roots, we deduce from Vieta relations that r_1+r_2=-1 and r_1 r_2=2k=-2013, where r_1, r_2 are the real roots of the equation X^4+2 X^3+(2+2 k) X^2+(1+2 k) X+2 k=0. Therefore, r_1^2+r_2^2=(r_1+r_2)^2-2 r_1 r_2=1+2 2013=4027.