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, 2013

Algebra Difficulty 6.6 National olympiad Find the answer Saudi Arabia

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.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

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,
$r_{1}^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}-2 r_{1} r_{2}=1+2 \times 2013=4027.$

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