HMMT February

After dividing the equation by $x^{2}$, we can rearrange it as $(x+\frac{1}{x})^{2}+a(x+\frac{1}{x})-b-2=0$. Let $y=x+\frac{1}{x}$. We can check that the range of $x+\frac{1}{x}$ as $x$ varies over the nonzero reals is $(-\infty,-2] \cup[2, \infty)$. Thus, the following equation needs to have a real root: $y^{2}+a y-b-2=0$. Its discriminant, $a^{2}+4(b+2)$, is always positive since $a, b \geq 0$. Then, the maximum absolute value of the two roots is $\frac{a+\sqrt{a^{2}+4(b+2)}}{2}$. We need this value to be at least 2. This is equivalent to $\sqrt{a^{2}+4(b+2)} \geq 4-a$. We can square both sides and simplify to obtain $2 a \geq 2-b$. This equation defines the region inside $[0,1] \times[0,1]$ that is occupied by $S$, from which we deduce that the desired area is $1 / 4$.