# Show that the equation $x^{4} + rx + s = 0$ has at most two distinct real roots.

Show that the equation $x^{4} + rx + s = 0\:$ has at most two distinct real roots. Also, find a condition on $r$ and $s$ which ensures there are two distinct roots.

This seems like it should be easy, but I have no idea where to start. This question is posed in a Calculus exam so the solution should be simple (nothing too advanced).

• Letting $p(x)=x^{4} + rx + s = 0,\: p'(x)=4x^3+r$. Therefore, $p'(x)=0$ at $x=\sqrt[3]{-r/4}$ (the only root of $p'(x)$). So $a=\sqrt[3]{-r/4}$ is a global minimum with the function decreasing to the left of a, and increasing to the right of a. So p(x) can have at most two distinct roots. So if $p(\sqrt[3]{-r/4}) < 0$, $p(x)$ will have two distinct (real) roots. The condition I found for r and s that would ensure two distinct real roots is $-3(4^{-4/3})(r^{4/3})+s < 0$ or $s < 3(4^{-4/3})(r^{4/3})$ (hopefully I did not make any mistakes simplifying). – White Rice Dec 31 '14 at 5:10
• Is there a more general condition that would guarantee two real roots? – White Rice Dec 31 '14 at 5:11

## 1 Answer

HINT: The derivative is $4x^3+r$; how many real roots does the derivative have? What does that tell you about the number of relative maxima or minima of the original function, and therefore about the number of roots?