Common tangent(s) to two parabolas without calculus We have the parabolas: $y = 2x^{2} + 2x + 1$ and $y = 2x^{2} - 2x - 1$.
Finding the common tangents with calculus is as straight forward as just solving a system of equations with the derivatives of both parabolas and $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$; but how about in analytical geometry? 

A line $y=kx+m$ is a tangent to the parabola $y=ax^2+bx+c$ iff the quadratic equation 
$$
kx+m=ax^2+bx+c
$$
has exactly one solution, which makes the discriminant to be zero. In our case
$$
kx+m=2x^2+2x+1\iff 2x^2+(2-k)x+1-m=0,\\
kx+m=2x^2-2x-1\iff 2x^2-(2+k)x-1-m=0
$$
should each have a unique solution $x$, that is the discriminants are zeros for both equations
$$
\begin{cases}
(2-k)^2-8(1-m)&=&0,\\
(2+k)^2+8(1+m)&=&0.
\end{cases}
$$
Solve the system: $k=-2$, $m=-1$.
