Finding Relations for Coefficients of Polynomial p(x)=0 to Get Real Roots

  • Thread starter Kadmax
  • Start date
  • Tags
    Polynomial
In summary, the conversation discusses finding the relations between coefficients of a polynomial so that the equation has only real roots. The example of a quadratic equation is used to illustrate that the discriminant must be greater than or equal to 0, and the conversation then moves on to discussing equations of higher power. It is mentioned that there may be complications due to the number of coefficients, but they can be written as functions of one parameter. However, determining the number of real roots for these equations can still be complicated. Finally, it is noted that for an nth degree equation with a leading coefficient of 1, if the second coefficient squared is less than twice the third coefficient, then all roots are complex.
  • #1
Kadmax
2
0
I'd like to know how to find relations between coefficients of a polynomial p(x) so that equation p(x) = 0 has only real roots.

For example I have quadratic equation:

x[tex]^{2}[/tex]+px+q = 0

then the Discriminant must be >= 0 so p[tex]^{2}[/tex] >= 4q

But I need to find it for equations of higher power of x (like 6), without having the formula for exact solutions - I know it's not possible to solve polynomial equations of order 6 and more.

It may be complicated if there are too many coefficients (there are 5 in 6th order pol.), but they just all depend just on one constant r, so it is like a5 = 2r, a4 = r[tex]^{2}[/tex], a3 = 1/r, and so on, and I need to know for which values of r the equation has only real roots.
 
Mathematics news on Phys.org
  • #2
It may be complicated if there are too many coefficients (there are 5 in 6th order pol.), but they just all depend just on one constant r, so it is like a5 = 2r, a4 = r, a3 = 1/r, and so on, and I need to know for which values of r the equation has only real roots.
First you had better clear up some misconceptions. In the first place, a 6[/sup]th[/sup] degree (not "order") equation would have 7 coefficients (or 6 if you always take the leading coefficient to be 1 and ignore it), not 5. In general, an nth degree equation has n+1 coefficients, not n-1. Again, you could always divide the equation by the leading coefficient and just use the other n.
Secondly, any set of number can be written as functions of one parameter. How many real roots an equation has would depend upon how each coefficient depends on r and that might be very complicated. Even in the very simple case [itex]x^3+ bx^2+ cx+ d[/itex] whether there are non-real roots would depend upon whether the local min and max are on opposite sides of the x-axis- and that alone can be complicated.
 
  • #3
i don't know about all real roots ,but i can tell you that for all complex roots
in equation of n th degree whose first coefficient is 1
second ceofficient sqaure<2*third coefficient
and if it is = then all the roots are 0
 

FAQ: Finding Relations for Coefficients of Polynomial p(x)=0 to Get Real Roots

What is the purpose of finding relations for coefficients of a polynomial?

Finding relations for coefficients of a polynomial is important because it helps us understand the behavior of the polynomial and its roots. By analyzing the coefficients, we can determine the number of real roots and their values.

How do you find the relations for coefficients of a polynomial?

To find the relations for coefficients of a polynomial, we use the properties of polynomials and their roots. These include the sum and product of roots, the relationship between the coefficients and the roots, and the fact that complex roots occur in conjugate pairs.

Can finding relations for coefficients of a polynomial help us determine if the roots are real or complex?

Yes, finding relations for coefficients of a polynomial can help us determine the nature of the roots. For example, if the polynomial has only real coefficients, then the roots must occur in conjugate pairs. This means that if one root is a+bi, the other root must be a-bi, where i is the imaginary unit.

Why is it important to get real roots for a polynomial?

Real roots of a polynomial correspond to the x-intercepts of the graph of the polynomial. These are the points where the polynomial crosses the x-axis, and they provide valuable information about the behavior of the polynomial. Additionally, real roots are easier to work with in mathematical calculations.

Are there any specific methods for finding relations for coefficients of a polynomial?

Yes, there are specific methods for finding relations for coefficients of a polynomial, including the Rational Root Theorem, Descartes' Rule of Signs, and the Fundamental Theorem of Algebra. These methods can help us determine the possible rational roots, the number of positive and negative roots, and the number of complex roots of a polynomial.

Similar threads

MHB Real Roots of Composite Polynomials: Solving P(Q(x))=0
Replies
1
Views
983
MHB Find Polynomials Fulfilling Real Coefficient Equation
Replies
1
Views
980
MHB Find Cubic Function & Polynomial of Degree 3
Replies
5
Views
1K
MHB Find Polynomials: Real Coeffs Resulting in 1
Replies
1
Views
857
MHB Roots of a Polynomial Function A²+B²+18C>0
Replies
4
Views
1K
MHB Solving a Third-Degree Polynomial with Real Coefficients
Replies
1
Views
1K
MHB Roots of Polynomial: Find $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$
Replies
1
Views
979
MHB Proving $x-1$ is a Factor of $P(x)$ with Polynomials
Replies
1
Views
862
MHB Finding the Largest Root of a Polynomial Using Synthetic Division
Replies
1
Views
1K
MHB Solving 8 Roots: Can 3 Quadratic Polynomials Fulfill $f(g(h(x)))=0$?
Replies
1
Views
935

Hot Threads

Recent Insights

Back
Top