
1. Find all real $c$ such that the equation $x^{2}+\frac{5}{2} x+c=0$ has two real solutions which can be together with $c$ arranged into a three-member arithmetic sequence.

(Pavel Calábek, Jaroslav Švrček)


Solution. Let $c$ meets the criteria of the problem. Let us denote by $d$ the difference of the corresponding arithmetic sequence, and by $x_{1}, x_{2}$ the solutions of the equation.

a) If $c$ is a middle term of the arithmetic sequence, then $x_{1}=c-d$ and $x_{2}=c+d$. Further Monsieur Viète says $-\frac{5}{2}=x_{1}+x_{2}=2 c$, that is $c=-\frac{5}{4}$. Moreover for any negative $c$ the equation has two real solutions (especially for $c=-\frac{5}{4}$ we have $x_{1,2}=-\frac{5}{4} \pm \frac{3}{4} \sqrt{5}$ ).

b) If $c$ is the first or the last member of the sequence we have (in an appropriate notation of the solutions of the equation) $x_{1}=c+d, x_{2}=c+2 d$. Thus we get $-\frac{5}{2}=x_{1}+x_{2}=2 c+3 d$, which gives $d=-\frac{5}{6}-\frac{2}{3} c$ and substituting into $x_{1}=c+d$ and $x_{2}=c+2 d$ we get $x_{1}=\frac{1}{6}(2 c-5)$ and $x_{2}=-\frac{1}{3}(c+5)$. Monsieur helps again, since $x_{1} x_{2}=c$ and we get $2 c^{2}+23 c-25=0$, with solutions 1 and $-\frac{25}{2}$. (If $c=1$ the solutions are $x_{1}=-\frac{1}{2}, x_{2}=-2$; if $c=-\frac{25}{2}$ the solutions are $x_{1}=-5, x_{2}=\frac{5}{2}$.)

Conclusion. The conditions of the problem are met for $c \in\left\{-\frac{25}{2} ;-\frac{5}{4} ; 1\right\}$.
