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Question

# If the roots of ${x}^{2}-bx+c=0$ are two consecutive integers, then b2 − 4 c is (a) 0 (b) 1 (c) 2 (d) none of these.

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Solution

## (b) 1 Given equation: ${x}^{2}-bx+c=0$ Let $\alpha \mathrm{and}\alpha +1$ be the two consecutive roots of the equation. Sum of the roots = $\alpha +\alpha +1=2\alpha +1$ Product of the roots = $\alpha \left(\alpha +1\right)={\alpha }^{2}+\alpha$ $\mathrm{So},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{roots}=2\alpha +1=\frac{-\mathrm{Coeffecient}\mathrm{of}x}{\mathrm{Coeffecient}\mathrm{of}{x}^{2}}=\frac{b}{1}=b\phantom{\rule{0ex}{0ex}}\mathrm{Product}\mathrm{of}\mathrm{the}\mathrm{roots}={\alpha }^{2}+\alpha =\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coeffecient}\mathrm{of}\mathit{}{x}^{2}}=\frac{c}{1}=c\phantom{\rule{0ex}{0ex}}\mathrm{Now},{b}^{2}-4c={\left(2\alpha +1\right)}^{2}-4\left({\alpha }^{2}+\alpha \right)=4{\alpha }^{2}+4\alpha +1-4{\alpha }^{2}-4\alpha =1$

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