If the real polynomial $x^4+ax^3+bx^2+cx+d$ satisfies $d < 0$ and $3 a^2 < 8 b$, it has precisely $2$ real roots

Question

Let $a,b,c,d$ be real numbers. If $d < 0$ and $3a^2 < 8b$, show that $$x^4+ax^3+bx^2+cx+d = 0$$ has exactly two roots.

I know that you have to use IVT to prove that there are at least $2$ distinct roots, but I don't really know how to go about it.

Answer 1

Hint Let $p(x)$ denote the polynomial on the l.h.s. of the equation. Since $p(0) = d < 0$, we can use information about the leading term of $p(x)$ together with the IVT to conclude that $p(x)$ has at least two real roots, and in fact that one is positive and that another is negative.

On the other hand, consider the discriminant of the quadratic polynomial $p''(x)$. What does this tell you about $p'(x)$? About $p(x)$?