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).

 

[CellCoree]

Hi Ann,

Thank you for your question. To show that the equation x^4 + 4x + c has at most 2 roots, we can use the fundamental theorem of algebra. This theorem states that a polynomial of degree n has at most n distinct roots.

In this case, the degree of the polynomial is 4. Therefore, it can have at most 4 distinct roots. However, we can see that the coefficient of the x^3 term is 0, which means that there is no x^3 term in the polynomial. This means that one of the roots must be 0.

Now, we can use the rational root theorem to determine the other possible roots. This theorem states that if a polynomial has integer coefficients and a rational root p/q, where p and q are relatively prime integers, then p must be a factor of the constant term and q must be a factor of the leading coefficient.

In this case, the constant term is c and the leading coefficient is 1. Therefore, the only possible rational roots are factors of c. Since we are only concerned with at most 2 roots, we can assume that c is not equal to 0, as this would result in more than 2 possible roots.

Thus, we can conclude that the equation x^4 + 4x + c has at most 2 roots, one of which is 0 and the other is a factor of c. I hope this helps to answer your question. Please let me know if you need any further clarification.

Best,