Let $f(x)=x^2+ax+b$. If the roots of $f(x)=0$ are real and differ by at most 1, then the minimum value of $a^2-4b$ is:

Solution

1. Let the roots be $\alpha$ and $\beta$.
2. The discriminant $$\begin{array}{rl}D& =a^2-4b\\ & =(\alpha -\beta {)}^2\end{array}$$.
3. We are given that the roots are real, so $D\ge 0$.
4. We are given $|\alpha -\beta |\le 1$, which implies $(\alpha -\beta {)}^2\le 1$.
5. Thus $0\le D\le 1$.
6. The minimum value of $D$ is 0.
7. Hence the answer is (A).