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  1. $x^9+x^7+x^5+x^3+x+1=0$
  2. $x^3+x-1=0$

I know that the complex roots exist in pairs and also that the number of distinct roots has to do something with change in sign of the equation, but how? How should i proceed in such type of questions?

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  • $\begingroup$ See Lemma of Descartes, Theorem of Budan-Fourier, etc. $\endgroup$ Oct 27, 2016 at 17:40
  • $\begingroup$ @LutzL can u provide me with any link so i can learn the concept? $\endgroup$
    – Parul
    Oct 27, 2016 at 18:09

2 Answers 2

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for the equation $x^9+x^7+x^5+x^3+x+1=0$ we take the function f(x)=$x^9+x^7+x^5+x^3+x+1$...now see f ' (x)=$9x^8+7x^6+5x^4+3x^2+1$ >0 for all x$\in$R. So f(x) is stricktly increasing for all x$\in$R...so it wil obviously cut the X axis at exactly one point.so the equation will have only one real root as f(x) is continuous for all x$\in$R ( as f(0)=1>0 and f(-1)=-4<0 )

  1. similarly for $x^3+x-1=0$ f(x)=$x^3+x+1$ and f ' (x)=$3x^2+1$ >0 for all x$\in$R so as f(0)=-1 and f(1)=1 so f(x) will have a root between x=0 and x=1 and exactly one real root.
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  • $\begingroup$ Nice answer, I suspected that there already was something special with the odd powers (xcept for the constant term) $\endgroup$ Oct 28, 2016 at 19:14
  • $\begingroup$ what if the function is in both odd and even powers, say $x^5+3x^4-6x^3+2x^2$ without the constant term, how are we suppose to find the real roots then? $\endgroup$
    – Parul
    Nov 3, 2016 at 19:31
  • $\begingroup$ for the functions with both even and odd powers actually it depends on the function; you have to give your intuition. moreover you can take several intervals by your own choice and see the nature of the function in the intervals ... $\endgroup$ Nov 4, 2016 at 2:02
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Descartes' rule of signs.

To count the possible number of positive roots, count the number of sign changes in $f(x).$ That is, count the number of times the signs go from plus to minus or minus to plus between terms.

For the possible number of negative roots, count the number of sign changes in $f(-x).$ Evaluate the function at $-x$ by replacement, be mindful of powers, and do the same process.

For each of these two numbers, subtract $2$ repeatedly until you arrive at $1$ or $0$. This gives possible root combinations. So, if I ended up with a maximal number of $5$ positive roots and $4$ negative roots, I may have $5, 3,$ or $1$ positive roots and $4, 2, 0$ negative.

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  • $\begingroup$ by positive and negative roots both, u mean real roots? or is it like we will have 5 positive roots(real roots) and 4 negative roots (imaginary roots)? $\endgroup$
    – Parul
    Oct 27, 2016 at 18:03

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