A puzzle via @CmonMattTHINK (Matt Enlow):

(I think we may have used this as a @WrongButUseful puzzle).

## Double roots

I set this up as a double roots puzzle: we need to find where $y = mx + c$ and $y = x^4 - x^3$ have two double solutions.

Now, I don’t know about you, but I sort of dislike finding tangent points on sloping lines. It’s so much easier when things are equal to zero.

So, I’m going to rewrite what I’m looking for by eliminating $y$:

$x^4 - x^3 - mx - c = 0$

This needs to have two double roots - which is equivalent to saying it can be written as $(x-a)^2(x-b)^2$ for some values of $a$ and $b$. (In principle, it could be multiplied by a constant, but the $x^4$ term needs to be 1, so it isn’t.)

Expanding the brackets one level gives $(x^2 - 2ax + a^2)(x^2 - 2bx + b^2)$, and going one step further leads to $x^4 - 2(a+b)x^3 + (a^2 +4ab + b^2)x^2 - 2ab(a+b)x + a^2b^2$.

Now we can match coefficients!

We have $(a+b) = \frac{1}{2}$, from the $x^3$ term, $a^2 + 4ab + b^2 = 0$ from the $x^2$ term, $2ab(a+b) = -m$ from the $x$s and $a^2b^2 = c$ from the units.

Now. I don’t especially care what $a$ and $b$ are. I can rewrite $a^2 + 4ab + b^2=0$ as $(a+b)^2 + 2ab=0$ and state that, since $a+b = \frac{1}{2}$, $ab = - \frac{1}{8}$.

Then $2ab(a+b) = -\frac{1}{8} = m$ and $a^2b^2 = \frac{1}{64} = -c$.

So, the line $y = -\frac{1}{8}x - \frac{1}{64}$ is tangent to the curve in two places.

I thought that was a neat puzzle! I’d be interested to see how you approached it - with or without calculus!