2A. Determine the equation of the common tangent to the parabolas $y=x^{2}+2 x+2$ and $y=x^{2}+x+2$.
Solution. Let the common tangent have the equation $y=k x+n$. Then each of the equations $x^{2}+2 x+2=k n+n$ and $x^{2}+x+2=k n+n$ has a unique solution, which means that the discriminants of these quadratic equations are equal to 0. According to this, $k^{2}-4 k+4 n-4=0$ and $k^{2}-2 k+4 n-7=0$. If we subtract the first equation from the second, we get $2 k-3=0$, from which we find $k=\frac{3}{2}$. According to this, $n=\frac{4+4 k-k^{2}}{4}=\frac{31}{16}$, which means that the equation of the common tangent to the parabolas is $y=\frac{3}{2} x+\frac{31}{16}$.