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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 solution of $x^2-3x+c=0$ are....

i cant do this problem due to language barrier

Iqbal Jurist
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  • .. then the solution of ... is what? – mvw Jun 02 '16 at 13:36
  • If $\alpha$ is a root of $x^2-3x+c=0$ and $-\alpha$ is a root of $x^2+3x-c=0$, then we have $\alpha^2-3\alpha+c=0,\alpha^2-3\alpha-c=0$, so $c=0$ and hence the roots of $x^2-3x+c=0$ are $0,3$. – almagest Jun 02 '16 at 14:15

1 Answers1

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Suppose $x_1$ and $x_2$ are roots of $x^2-3x+c$, and $-x_1$ and $x_3$ are roots of $x^2+3x-c$, then we know that $$x_1^2-3x_1+c=0=(-x_1)^2+3(-x_1)-c$$

so $c=?$. Then the roots of both equations become clear (hopefully).

snulty
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