# How many real roots has the equation?

How many real roots (depends on the parametr $a$) has the equation $x^{13}=a(x^{14}+1)?$

I ploted left and right sides and observed that for all $a \neq 0$ there is only one root. Is it true?

Edit. Correct typos.

• HINT: solve for $$a>0 \\a<0$$ – Khosrotash Jan 29 '17 at 17:50
• What happens when $a=1$? – Mark Bennet Jan 29 '17 at 17:51
• I am sorry, correct typos, the equation was $x^{13}=a(x^{14}+1).$ – Leox Jan 29 '17 at 18:14

The equation is equivalent to

$P_a(x)=ax^{14}-x^{13} +a=0$

Check out Descartes' Rule of Signs:

https://en.wikipedia.org/wiki/Descartes'_rule_of_signs

It is very useful for this kind of problems. In this specific case, it says that $P_a(x)$ has exactly one zero, because

• $P_a(x)$ has 1 change of signs for $a>0$ and 0 change of signs for $a<0$.
• $P_a(-x)$ has 0 change of signs for $a<0$ and 1 change of signs for $a>0$.
• Do you the expression "Using a sledgehammer to crack a nut" ? – Jean Marie Jan 29 '17 at 18:01
• Haha yes i know that expression but i really like this result – svelaz Jan 29 '17 at 18:06
• I am sorry, correct typos, the equation was $x^{13}=a(x^{14}+1).$ – Leox Jan 29 '17 at 18:14
• @Leox It works anyway. I added some details though – svelaz Jan 29 '17 at 18:22

Note: This is in answer to OP's original question: How many real solutions to $x^{13}=a\left(x^{13}+1\right)$.

The equation of $x^{13}=\dfrac{a}{1-a}$ will have one real solution (either positive or negative when $1\ne a\ne0$. All $13$ solutions are equally spaced around a circle in the complex plane, thus none of the remaining $12$ will be real.

This is true in general for odd exponents of $x$. If the exponent is even, then there will be exactly two real solutions of the form $\pm r$.

• I am sorry, correct typos, the equation was $x^{13}=a(x^{14}+1).$ – Leox Jan 29 '17 at 18:14

This is an answer to the original question: roots of $x^{13}=a(x^{13}+1)$.

Let us define $$f(x)=x^{13}-a^{13}-a=(1-a)x^{13}-a$$ Your function is strictly increasing if $1-a>0$, that is, $a>1$, and strictly increasing if $a<1$. In the first case, for $a>1$ we also have $$\lim_{x\to +\infty} f(x)=+\infty \qquad \lim_{x\to -\infty} f(x)=-\infty$$ so, applying Bolzano's theorem, there's a single point $c$ which satisfies $f(c)=0$. For $a<1$ is other way around, we have $\lim_{x\to -\infty} f(x)=+\infty$ and $\lim_{x\to -\infty} f(x)=+\infty$, but we arrive to the same conclusion.

For $a=1$, we get the function $f(x)=-1$, which obviously has no roots.

• I am sorry, correct typos, the equation was $x^{13}=a(x^{14}+1).$ – Leox Jan 29 '17 at 18:14

This is an answer to the original question :(

It is not necessary for such an issue to use analysis tools.

Two partiuclar cases. If $a=0$ then $x=0$. If $a=1$ then there is no solution.

Now, let us assume that $a \neq 0$ and $a\neq 1$; the initial expression is equivalent to :

$$x^{13}=b \ \text{with} \ b:=\dfrac{a}{1-a}.$$

Two cases:

• either $0<a<1 \iff \ b>0$ then $x=\sqrt{b},$

• or $a$ outside $[0,1] \ \iff \ b<0$ then $x=-\sqrt{b}.$

Conclusion: In all cases, there is one solution, but for $a=1$ where there are no solutions.

• I am sorry, correct typos, the equation was $x^{13}=a(x^{14}+1).$ – Leox Jan 29 '17 at 18:22