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$${x^4 + ax^3 + bx^2 + 2018x = 207} $$ $$3a^2 < 8b$$
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.
As you already observed, $f(0)=-207<0$, whereas $\lim_{x\to \pm \infty}f(x)=+\infty$. Therefore, by the Intermediate Value Theorem, there are at least two real roots: one is positive and another one is negative.
Now note that $f$ is strictly convex: for all real $x$, $$f''(x)=12x^2+6ax+2b>0$$ because $\Delta=36a^2-96b=12(3a^2-8b)<0$. Hence if $f$ has at least three real roots then, by the Mean Value Theorem, $f'$ has at least two roots and $f''$ has at least one root. Contradiction.
It follows that the real roots of $f$ are exactly two.
$f'(x) = 4x^3 + 3ax^2 + 2bx+2018$, $f''(x) = 12x^2 + 6ax + 2b = 2(6x^2+3ax+b)$.
Consider $f''(x)$ as a quadratic polynomial in $x$: $\Delta = (3a)^2-4(6)(b) = 3(3a^2-8b) < 0$, so $f''(x) > 0$ for all $x \in \Bbb{R}$.
If you allow me to borrow some concepts in convexity, we can eliminate the possibility of having four real roots: $f''$ is positive implies that $f'$ is strictly increasing and $f$ opening up. If we allowed four real roots, rolles-theorem would imply the existence of three stationary points (at which $f'$ is zero), which would contradict the fact that $f'$ is strictly increasing.
As OP has observed, $f(0) < 0$. Since $f(x)$ has a leading term with positive coefficient, as $|x|$ becomes sufficiently large, $f(x) > 0$, so by the Intermediate Value Theorem, $f(x)$ have two real roots.
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.
Since $3a^2-8b<0$, $f''(x)>0 \forall x\in \mathbb{R}.$ Therefore, the equation $f(x)=0$ has at most 2 solutions. Moreover, since $f(0)>0$ and $\displaystyle \lim_{x\to \infty}f(x)=\infty$, the equation has at least one solution.
So, the equation has exactly 2 solutions.
