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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?

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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.

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Since $f'(x)>0$ the given function is increasing, hence injective.
A cubic polynomial always has a real root, hence the claim is straightforward.

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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.

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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).

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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.

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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.

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  • $\begingroup$ (+1) Rolle's theorem is the standard approach for this kind of questions. So it is good there is atleast one answer using it. $\endgroup$ – Yanko Jun 9 '17 at 21:10

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