Question

If the roots of $x^2-bx+c=0$ are two consecutive integers, then $b^2-4c$ is

• A. $2$
• B. $1$
• C. none of these
• D. $0$

Answer 1

The correct option is B $1$

Let $\alpha$ and $\alpha +1$ are roots of $x^2-bx+c=0$

Where $\alpha \in Z$

Now,

$\text{sum of roots}=-\frac{\text{coefficient of}x}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha +(\alpha +1)=b\dots \left(1\right)$

$\text{Product of roots}=\frac{\text{constant term}}{\text{coefficient of}{x}^2}$

$\Rightarrow \alpha (\alpha +1)=c\dots \left(2\right)$

So, $b^2-4c$

$=(2\alpha +1{)}^2-4\alpha (\alpha +1)$

$\left[\text{using equation}\right(1),(2\left)\right]$

$=\left[\right(2\alpha {)}^2+2\left(2\alpha \right)\left(1\right)+1^2]-4{\alpha }^2-4\alpha$

$=4{\alpha }^2+4\alpha +1-4{\alpha }^2-4\alpha$

$=1$

$\therefore b^2-4c=1$

Hence, Option (B) is correct.