A line L is tangent to two curves $x^2 + 1$ and $-2x^2 - 3$, find L's equation Some things I'd like to clarify beforehand:
From my understanding, for this situation to occur these conditions need to be fulfilled:
1) $f(a) = g(a)$
2) $f'(a) = g'(a)$
However, both of these conditions are not fulfilled.
What I first did:
$a^2 + 1 = -2a^2 - 3$
I get $3a^2 + 4 = 0$.
Am I on the right track here? The thing is my class doesn't allow calculators, which is why I believe I'm wrong as the equation would get too complicated following this logic. 
Thanks.

$f(x) = x^2 +1\\
g(x) = -2x^2 - 3$
both are tangent to the line $L: y = mx+b$
If the curve is tangent, when we subtract one from the other, we have a root of multiplicity.  The discriminant in the quadratic formula is zero.
$f(x) = x^2 -mx +1-b\\
m^2 - 4(1-b) = 0$
$g(x) = -2x^2 -mx -3-b\\
m^2 - (-2)(-3-b)(4) = 0$
$4-4b = 24 + 8b\\
12b = -20\\
b = -\frac {5}{3}\\
m = \pm \sqrt {\frac {32}3}$
There are two lines:
$y_1 = 4\sqrt {\frac 23} x - \frac {5}{3}\\
y_2 = -4\sqrt {\frac 23} x - \frac {5}{3}$
Now it is probably wroth checking that each line is tangent to each curve
$y_1$ is tangent to $f(x)$ at $2\sqrt{\frac 23}$ and $g(x)$ at $2\sqrt{\frac 23}$ 
At the points of tangency $f(x) = y(x)$ and $f'(x) = y'(x).$  
