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Ifthe roots of the equation $x^{2}-b x+c=0$ are two consecutiv integers, then what is the value of $b^{2}-4 c$ ?
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Let $\alpha$ and $(\alpha+1)$ be the roots of equation $x^{2}-b x+c=0$
$\therefore \alpha+(\alpha+1)=b$
$\Rightarrow 2 \alpha+1=b$
and $\alpha(\alpha+1)=c$
Now, $b^{2}-4 c=(2 \alpha+1)^{2}-4[\alpha(\alpha+1)]$
$=4 \alpha^{2}+1+4 \alpha-4 \alpha^{2}-4 \alpha=1$
$\therefore \alpha+(\alpha+1)=b$
$\Rightarrow 2 \alpha+1=b$
and $\alpha(\alpha+1)=c$
Now, $b^{2}-4 c=(2 \alpha+1)^{2}-4[\alpha(\alpha+1)]$
$=4 \alpha^{2}+1+4 \alpha-4 \alpha^{2}-4 \alpha=1$
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