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.

Answer 1

(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 = $\alpha (\alpha +1)={\alpha }^2+\alpha$


$$\begin{gathered} \mathrm{So},\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{roots}=2\alpha +1=\frac{-\mathrm{Coeffecient}\mathrm{of}x}{\mathrm{Coeffecient}\mathrm{of}{x}^2}=\frac{b}{1}=b \\ \mathrm{Product}\mathrm{of}\mathrm{the}\mathrm{roots}={\alpha }^2+\alpha =\frac{\mathrm{Constant}\mathrm{term}}{\mathrm{Coeffecient}\mathrm{of}\mathit{}{x}^2}=\frac{c}{1}=c \\ \mathrm{Now},b^2-4c={\left(2\alpha +1\right)}^2-4\left({\alpha }^2+\alpha \right)=4{\alpha }^2+4\alpha +1-4{\alpha }^2-4\alpha =1 \end{gathered}$$