India — Team Selection Test, 2007

Suppose $a, b \in [-2, 2]$ and $\alpha$ is a real root of $x^2 + ax + b = 0$. Then
$$\alpha = \frac{-a \pm \sqrt{a^2 - 4b}}{2},$$
which shows that
$$\alpha \leq \frac{2 + \sqrt{4 + 8}}{2} = 1 + \sqrt{3}.$$
(Take $a = -2$, $b = -2$.) Similarly, we see that $\alpha \ge -1 - \sqrt{3}$, by taking $a = 2$ and $b = -2$. Take any real number $\lambda$ such that $|\lambda| \le 1$. We have
$$(\lambda\alpha)^2 + (\lambda a)(\lambda\alpha) + \lambda^2 b = \lambda^2 (\alpha^2 + a\alpha + b) = 0.$$
Thus $\lambda\alpha$ is also a root of the equation $x^2 + cx + d$, where
$c = \lambda a, \ d = \lambda^2 b.$
Note that $|c| \le |a|$ and $|d| \le |b|$. Thus for each real root $\alpha$ of $x^2 + ax + b = 0$ and each real $\lambda$ with $|\lambda| \le 1$, we see that $\lambda\alpha$ is also a root of $x^2 + cx + d$, where $c, d \in [-2, 2]$. Since $1 + \sqrt{3}$ is a root of $x^2 - 2x - 2 = 0$, it follows that $\lambda(1 + \sqrt{3})$ is in the range for any $\lambda$ with $|\lambda| \le 1$. Thus the range is $[-1 - \sqrt{3}, 1 + \sqrt{3}]$.