Real numbers such that $|x^2 + ax + b| \le {1\over2}$?

Question

What are real numbers $a$, $b$ such that, for every $x$ satisfying $0 \le x \le 2$, it holds that $|x^2 + ax + b| \le {1\over2}$?

Answer 1

From $x^2+ax+b\leq\frac{1}{2}$, we have $x_1\leq x\leq x_2$, where $$x_1=\frac{-a-\sqrt[]{a^2-4b+2}}{2}$$ and $$x_2=\frac{-a+\sqrt[]{a^2-4b+2}}{2}$$ From $x^2+ax+b\geq\frac{-1}{2}$, we have $x\leq x_3$ or $x\geq x_4$, where $$x_3=\frac{-a-\sqrt[]{a^2-4b-2}}{2}$$ and $$x_4=\frac{-a+\sqrt[]{a^2-4b-2}}{2}$$ It is easy to see that $x_1\leq x_3 \leq x_4 \leq x_2$.
Hence, the solution of the inequation is $x_1\leq x \leq x_3$ or $x_4\leq x \leq x_2$.
So, it must be: $x_1=0$ and $x_2=2$, from where $b=0.5$ and $a=-2$, and, in that case, $x_3=1$ and $x_4=1$ .

Answer 2

Hint:

Denote $f(x)=x^2+ax+b$. It achieves its minimum at $x=-\frac a2$ and $f(-\frac a2)=b-\frac{a^2}{4}$.

If $-\frac a2 < 0$ or $-\frac a2 > 2$, $f(x)$ is monotone increasing or decreasing in $[0,2]$, then $|f(x)|\le \frac 12 \iff |f(0)| \le \frac 12 \text{ and } |f(2)| \le \frac 12$.

If $0 \le -\frac a2 \le 2$, then $|f(x)|\le \frac 12 \iff |f(0)| \le \frac 12 \text{ and } |f(2)| \le \frac 12 \text{ and } |f(-\frac a2)|\le \frac 12$.

Can you start from here? It would be helpful if you plot all these inequalities on an $a,b$-plane.