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Number of real roots of a quartic

CaptainBlack

Well-known member
Jan 26, 2012
890
rayman's question from another place:

Could someone help me with this problem, I have no idea how to start with it

How many real roots does this polynomial have p(x)=x^4-x^3-1?

Clearly state the argument that explains the number of real roots.


Thank you for any help



Descartes rule of signs tells you this has exactly 1 positive root, and exactly 1 negative root, so it has two real roots.

CB
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
The Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\frac{3}{4}$ and here is $q(x)=- \frac{27}{256}$. Now if we consider the quartic equation $\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\frac{27}{256}$, one real root of order 2 for $a=\frac{27}{256}$ and no real roots for $a>\frac{27}{256}$...

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
The Descartes rule establishes the maximum number of positive/negative real roots that a polynomial can have but it gives no information about the effective number of the real roots of a polynomial. In our case is $\displaystyle p(x)= x^{4}-x^{3} -1$ and, in my opinion, the number of its real root can be found considering the polynomial $\displaystyle q(x)= x^{4}-x^{3}$. It is easy enough to see that $q(x)$ has a root of order 3 in x=0 and a root of order 1 in x=1. Furthermore q(x) has a minimum in $x=\frac{3}{4}$ and here is $q(x)=- \frac{27}{256}$. Now if we consider the quartic equation $\displaystyle q(x)+a=0$, on the basis of consideration we have just done, it is easy to find that the quartic equation has two real roots for $a<\frac{27}{256}$, one real root of order 2 for $a=\frac{27}{256}$ and no real roots for $a>\frac{27}{256}$...

Kind regards

$\chi$ $\sigma$
In this case Descartes rule of signs does tell us exactly how many real roots we have.

The number of positive roots is equal to the number of changes of signs of the coefficients less a multiple of 2. In this case the number of sign changes is 1, and as there is no multiple of 2 other than 0 which leaves the number of roots non-negative there is exactly one positive real root. The same argument applies to the negative roots.

CB