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If $p$ and $q$ are the roots of the equation $x^{2}+p x+q=0$, then

1) $p=1, q=-2$
2) $p=0, q=1$
3) $p=-2, q=0$
4) $p=-2, q=1$
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(1)

Explanation

$p+q=-p$ and $p q=q=>q(p-1)=0 \Rightarrow q=0$ or $p=1$.

If $q=0$, then $p=0$. i.e., $p-q \backslash p=1$ and $q=-2$

by Diamond (75,934 points)

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