Question:
If the roots of $x^{2}-b x+c=0$ are two consecutive integers, then $b^{2}-4 c$ is
(a) 0
(b) 1
(c) 2
(d) none of these.
Solution:
(b) 1
Given equation: $x^{2}-b x+c=0$
Let $\alpha$ 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 $=\alpha(\alpha+1)=\alpha^{2}+\alpha$
So, sum of the roots $=2 \alpha+1=\frac{-\text { Coeffecient of } x}{\text { Coeffecient of } x^{2}}=\frac{b}{1}=b$
Product of the roots $=\alpha^{2}+\alpha=\frac{\text { Constant term }}{\text { Coeffecient of } x^{2}}=\frac{c}{1}=c$
Now,$b^{2}-4 c=(2 \alpha+1)^{2}-4\left(\alpha^{2}+\alpha\right)=4 \alpha^{2}+4 \alpha+1-4 \alpha^{2}-4 \alpha=1$