TITLE: Given that $3a^2 < 8b$, prove that the equation $x^4 + ax^3 + bx^2 + 2018x = 207$ has exactly two real roots.
QUESTION [5 upvotes]: $${x^4 + ax^3 + bx^2 + 2018x = 207} $$
$$3a^2 < 8b$$
While trying to prove that this equation has exactly two solutions, I defined the function $$f(x) = {x^4 + ax^3 + bx^2 + 2018x - 207}$$ and then evaluated it at $f(0)$ to show that it is less than zero. I was going to use the Intermediate Value Theorem to prove that at some large positive and negative value result in values greater than $0.$ But then I don't know how to prove that the function doesn't cross more than twice.

REPLY [2 votes]: An elementary solution, only Vieta formulas
$$0 = {x^4 + ax^3 + bx^2 + 2018x - 207}$$ Since $$x_1x_2x_3x_4 = -207$$ it can not have 4 nonreal solution.
Suppose all solution are real. Since $x_1+x_2+x_3+x_4 =-a$ and $$x_1x_2+x_1x_3+...x_3x_4 = b$$
we have $$3a^2 = 3x_1^2+3x_2^2+...+6(x_1x_2+...) = 3x_1^2+3x_2^2+...+6b$$
Now since $$x_1^2+x_2^2\geq 2x_1x_2$$
$$x_1^2+x_3^2\geq 2x_1x_3$$
$$\vdots $$
we get $$3x_1^2+3x_2^2+...\geq 2b\implies 3a^2\geq 8b$$
a contradiction.