$x^3+ax+b=0$ has only one solution Let $a,b \in \mathbb{R}$ with $a>0$. Show that there exists only one $x \in \mathbb{R}$ so that $x^3+ax+b=0$. 
Is it possible to prove this using the mean value theorem?
 A: Since $f'(x)>0$ the given function is increasing, hence injective. 
A cubic polynomial always has a real root, hence the claim is straightforward.
A: The derivative of the function is $3x^2+a$, Since the derivative is always positive the function is increasing and thus injective.
So it cannot repeat values, in particular it cannot have two roots.
A: Alternatively, the discriminant of a reduced cubic $x^3 + ax + b$ is $-4a^3 - 27b^2$, hence negative in this case. So from the three roots, two are complex conjugates and one is real (there is always at least one real root for a cubic with real coefficients).
A: let $f(x) = x^3 + ax+b$. 


*

*Existence
Since $\lim\limits_{x\to\pm\infty} f(x) = \pm\infty$ and $f(x)$ is continuous. By IVT, $f(x) = 0$ has at least one real root.

*Uniqueness
For any $x > y$, we have
$$\begin{align}f(x)-f(y) &= (x-y)(x^2+xy+y^2 + a)\\
&= (x-y)\left(\left(x+\frac{y}{2}\right)^2 + \frac34 y^2 + a\right)\\
&\ge a(x-y) > 0
\end{align}
$$
So $f(x)$ is strictly increasing and $f(x) = 0$ has at most one real root. 
A: Another alternative approach is using Descartes' rule of signs.
Case $1: b =0 $. Then clearly the equation has exactly one real root, namely $0$.
Case $2: b \neq 0$. As the degree of equation is odd therefore there exists at least one real root, say $x$. Suppose it has another real root $y \neq x$. 
Then $$x^3 + ax + b=0 \\y^3+ay + b=0$$ This gives $$x^3 - y^3 +a(x-y) =0$$ This implies $x^2+y^2 +xy+a=0.$ This gives $$xy= -(x^2+y^2+a) <0$$ Thus $x$ and $y$ have different signs. Without loss of generality assume $x>0, y<0$. 
Thus the equation has at least one positive and one negative root. Applying descartes rule of signs this gives a contradiction.
A: A solution using Rolle's Theorem is as follows:  Suppose $x^3+ax+b = 0$ has more than one real solution.  Call the solutions $a$ and $b$.  Then by Rolle's Theorem we know that $3x^2+a = 0$ for some $c$ in $(a,b)$, but with $a>0$ this is impossible.  Hence $x^3+ax+b$ cannot have more than one real solution.
