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

Let the two consecutive integer roots of the equation $x^2-bx+c=0$ be $\alpha$ and $\alpha +1$.From the properties of a quadratic equation:Sum of roots:
$$\alpha +(\alpha +1)=b⟹2\alpha +1=b$$
Product of roots:
$$\alpha (\alpha +1)=c⟹{\alpha }^2+\alpha =c$$
We need to find the value of the discriminant expression $b^2-4c$:Substitute the values of $b$ and $c$ in terms of $\alpha$:
$$b^2-4c=(2\alpha +1{)}^2-4({\alpha }^2+\alpha )$$
Expanding the terms:
$$b^2-4c=(4{\alpha }^2+4\alpha +1)-(4{\alpha }^2+4\alpha )$$
$$b^2-4c=4{\alpha }^2+4\alpha +1-4{\alpha }^2-4\alpha$$
$$b^2-4c=1$$
Alternatively, since the roots are consecutive integers, their difference is always $|\alpha -\beta |=1$.The formula for the difference of roots is:
$$|\alpha -\beta |=\frac{\sqrt{b^2-4ac}}{|a|}$$
Given $a=1$:
$$1=\frac{\sqrt{b^2-4c}}{1}⟹1=b^2-4c$$