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Determine number of real roots of $f(x)=x^{5}+x^{3}-2x+1$.

I used Descartes's rule of sign to determine the number of positive and negative roots; there is only 1 negative (hence minimum 1 real root) and two positive roots. These 2 positive roots can be anything- a complex pair, or two real roots- but what about the other 2 roots? Is there any flaw in my understanding of rule of signs? How can I find the exact number of real roots?

Thanks in advance.

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3 Answers 3

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You have found that there is exactly one negative real root and either two or no positive real roots. Thus there are either two or four complex roots.

But how many positive real roots are there really? Let us look for points that lie between the roots (if there are roots): The roots of the derivative $f'(x)=5x^4+3x^2-2$. This is $=0$ if $x^2=\frac{-3\pm\sqrt{3^2+40}}{10}$, i.e., $x^2\in\{-1,\frac25\}$. Hence the only positive $x$ with $f'(0)$ is $x=\sqrt{\frac25}$. Now $$f\left(\sqrt{\tfrac25}\right)=\frac4{25}\sqrt{\frac25}+\frac25\sqrt{\frac25}-2\sqrt{\frac25}+1=1-\frac{36}{25}\sqrt{\frac25}$$ This is positive because $$\left(\frac{36}{25}\sqrt{\frac25} \right)^2=\frac{2592}{3125}<1.$$ We conclude that there are no positive roots.

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  • $\begingroup$ As we need to judge for positive roots, we get that the function is ever increasing for x>√(2/5) and decreasing for x<√(2/5) and we know there is exactly one negative real root hence there is no possible positive root <√(2/5) hence we have only one real root. Thanks for the answer $\endgroup$
    – Onix
    Feb 4, 2016 at 17:55
  • $\begingroup$ You could also use for the last inequality that $\frac{36}{25}\sqrt{\frac25}=\frac{24}{25}\sqrt{\frac{9}{10}}$ which is obviously smaller than $1$. $\endgroup$ Feb 4, 2016 at 19:23
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Descarte's rule doesn't give you the actual number of positive or negative roots. It just gives you an upper bound for them. As there are $2$ sign-changes there are either $2$ positive roots or the number of positive roots is less than $2$ by an even number ($0$ is the only possible number of positive roots in this situation).

Note that complex numbers that aren't purely real don't have sign as they aren't ordered. You can't say $i>2+i$ or $i$ is positive. Positiveness, negativeness are reserved for real numbers only.

Plotting the curve you can see the number of positive roots is $0$ like the other prediction of Descarte's rule. Number of negative roots is $1$. So the remaining $4$ roots are non-real.

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Are you allowed to plot the graph:

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XjUreF4zLTJ4KzEiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjEwMDB9XQ--

It is easy to see from the graph that there is only one root. The positive roots are complex.

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  • $\begingroup$ Is there a way to turn this graphical evidence into a proof ? $\endgroup$
    – user65203
    Feb 4, 2016 at 17:20
  • $\begingroup$ Not an analytic one. I would probably prove this using the fact that you can find all the local maxima with derivatives then show they are all greater than one. So there is only one crossing as you see on the graph. Another answer has this information for you. $\endgroup$
    – amcalde
    Feb 5, 2016 at 13:59

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