Answer
The equation $~~x^4+4x+c = 0~~$ has at most two real roots.
Work Step by Step
$x^4+4x+c = 0$
Let $f(x) = x^4+4x+c$
$f'(x) = 4x^3+4$
We can find the points $x$ where $f'(x) = 0$:
$f'(x) = 4x^3+4 = 0$
$x^3+1 = 0$
$x^3 = -1$
$x = -1$
The function $f(x)$ has exactly one point where $f'(x) = 0$
The function $f(x)$ is continuous and differentiable for all $x$.
Let's assume that the equation has at least three real roots $a_1$, $a_2$, and $a_3$ where $a_1 \lt a_2 \lt a_3$. Then $f(a_1) = f(a_2) = f(a_3) = 0$. According to Rolle's Theorem, there is a number $c_1$ in the interval $(a_1,a_2)$ such that $f'(c_1) = 0$, and there is a number $c_2$ in the interval $(a_2,a_3)$ such that $f'(c_2) = 0$.
However, this contradicts the fact that $f'(x) = 0$ at exactly one point.
Therefore, the equation has at most two real roots.
