Problem 6
If $c$ is a real number and the negative of one of the solutions of $x^2-3x+c=0$ is a solution of $x^2+3x-c=0$, then the solutions of $x^2-3x+c=0$ are
$\textbf{(A) }1,~2\qquad \textbf{(B) }-1,~-2\qquad \textbf{(C) }0,~3\qquad \textbf{(D) }0,~-3\qquad  \textbf{(E) }\frac{3}{2},~\frac{3}{2}$

Solution
We let the roots of the first equation be $r,s$ and the roots of the second equation be $s, -t$. By Vieta's Formulas, $r+s=3$ and $s-t=-3$, $rs=c$ and $-st=c$. So, $r=t$. Thus, $t+s=3$, $s-t=3$, so $t=0$, and $s=3\Rightarrow \textbf{(C)}$.~MathJams



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