Ann's question at Yahoo Answers (At most two roots)

[Fernando Revilla]

Here is a link to the question:

Show that the equation x^4 +4x+c has at most 2 roots? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Ann,

Suppose there are three real roots $a<b<d$, then all the hypothesis of the Rolle's theorem are satisfied for the function $f(x)=x^4 +4x+c$ on the intervals $[a,b]$ and $[b,d]$ so, there exists $\xi_1\in (a,b)$ and $\xi_2\in (b,d)$ such that $f'(\xi_1)=f'(\xi_2)=0$. But $$f'(x)=4x^3+4=0\Leftrightarrow x^3=-1\Leftrightarrow x=-1\quad\mbox{ (in }\mathbb{R})$$ We get a contradiction: $f'$ has at the 'same tine' only one real root and more than one real roots ($\xi_1\neq \xi_2)$.

 

[zzephod]

Here is a link to the question:

Show that the equation x^4 +4x+c has at most 2 roots? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

Assume \(c\) is positive Descartes rule of signs tell you this has no positive roots, and changing the signs of the coeficient of the odd power of x shows that it has at most two negative roots.

Now assume \(c\) is negative then Descartes rule of signs shows it has one positive and one negative root.
.

 

[Fernando Revilla]

Anoher way:

Denoting $g(x)=x^4+4x$, we have $$\lim_{x\to -\infty}g(x)=+\infty,\;\lim_{x\to +\infty}g(x)=+\infty,\;g'(x)=4(x-1)(x^2-x+1)$$ which implies $g$ has an absolute minimum at $(-1,-3)$. The intersection of the grpah of $g(x)$ with the graph of $y=-c$, easily provides the following information: $$\begin{aligned}&c\in (-\infty,0):\mbox{Two real roots (one positive and one negative)}\\&c=0: \mbox{ Two real roots }(x=-\sqrt[3]{4},x=-1)\\&c\in (0,3):\mbox{Two real roots (both negative)}\\&c=3:\mbox{One real roots }(x=-1)\\&c\in (3,+\infty):\mbox{No real roots}\end{aligned}$$ At any rate, and according to the question, it seems that the 'spirit' is to apply the Rolle Theorem (they only ask for the number of roots and not for more information).