Find the number of real roots of given equation.

Question

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?

Answer 1

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.

Answer 2

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.