Characteristic Equations with Complex Roots

Author: John J Weber III, PhD Corresponding Textbook Sections:

Expected Educational Results

Algebra

Theorem: Fundamental Theorem of Algebra

A polynomial of degree n, P(x)=anxn+an1xn1++a2x2+a1x+a0, an0, has exactly n factors, counting multiplicities.

Corollary

By Fundamental Theorem of Algebra, P(x)=anxn+an1xn1++a2x2+a1x+a0=0, where an0, has exactly n solutions.

Solving Polynomial Equations

Distributive Property and Zero Product Property of Complex Numbers

*Example 01

x3+x=0x(x2+1)=0x=0 or x2+1=0x=0 or x2=1x=0 or x=i or x=i

Quadratic Equations

Quadratic Formula

ax2+bx+c=0x=b±b24ac2a

Discriminant

The discriminant of a quadratic equation ax2+bx+c=0 is b24ac.

The discriminant is used to determine the types of roots to a quadratic equation:

Example 02

x2x+1=0x=(1)±(1)24(1)(1)2(1)x=1±142x=1±32x=1i32 or x=1+i32

Quadratic-Like Equations

A quadratic-like equation has the form: ax2n+bxn+c=0,nR.

Method to Solve Quadratic-Like Equations
  1. Let u=xn and substitute into the quadratic-like equation;

  2. Factor au2+bu+c=0(u+s)(u+t)=0, where (s)(t)=c and s+t=b;

  3. Solve in terms of u: (u+s)(u+t)=0(u+s)(u+t)=0u=s or u=t;

  4. Solve in terms of x: u=sxn=sx=(s)1n\newline u=txn=tx=(t)1n

Example 03

x4+3x2+4=0 Let u=x2u2+3u+4=0u=3±324(1)(4)2(1)u=3±9162(1)u=3i52 or u=3+i52x2=3i52 or x2=3+i52x=3i52 or x=3+i52

In order to write the solutions in standard cartesian form, we will need to use technology. For example, in Mathematica:

Difference of Two Cubes

a3x3b3=0(axb)(a2x2+abx+b2)=0

Example 06

8x31=023x313=0(2x1)(22x2+(2)(1)x+12)=0(2x1)(4x2+2x+1)=02x1=0 or 4x2+2x+1=0x=12 or x=1i34 or x=1+i34

Sum of Two Cubes

a3x3+b3=0(ax+b)(a2x2abx+b2)=0

Finding Zeros Using Rational Root Theorem and Polynomial or Synthetic Division

Given the polynomial anxn+an1xn1++a2x2+a1x+a0=0,an0

Let p1,p2,p3, be the factors of a0. Let q1,q2,q3, be the factors of an. Then the set of potential rational roots of the polynomial are:

{p1q1,p1q1,p1q2,p1q2,p2q1,p2q1,}

Example 04

Find all three real roots of: x42x3+3x22x+2=0.

The factors of a0=2 are p=1,1,2,2.

The factors of a4=1 are q=1,1.

Then the set of distinct potential rational roots of the polynomial are:

{11,11,21,21}

Now use polynomial or synthetic division to find the three factors.

Using Polynomial Division

a. Divide the polynomial by x( rational root), i.e., divide by x(1):

x42x3+3x22x+2x+1

Since the above polynomial division has a non-zero remainder, then x+1 is not a factor of x42x3+3x22x+2=0.

The same is true for all the other potential rational roots. Thus, there are no real solutions to the polynomial equation.

Since all roots are non-real numbers, we will need a method for solving a quartic (4th-degree polynomial) equation with [all] four complex roots which you probably do not have experience.

NOTE: Potential methods for solving polynomial equations that have only complex roots:

  1. Substitute x=y24, where the 2 is the coefficient of the cubic term to reduce the quartic equation to a cubic equation;

  2. Muller's method;

  3. Numerical methods to approximate the complex roots;

  4. If one of the complex roots is known, then polynomial or synthetic division can be used to reduce the degree of the polynomial.

Thus, we will will use technology to compute all roots of polynomial equations.

Practice 01

Solve, i.e., find all solutions, to the following equations:

  1. x2+3x=0

  2. x28=0

  3. x2+4=0

  4. x25x+8=0

  5. x2+2x=4

  6. x3+3x2+2=0

  7. x45x36x2=0

  8. x38=0

  9. x3=8

  10. x4+12x=0

  11. x49=0

  12. 2x3x25x=3

  13. x37x6=0

  14. x4+4x3+3x24x4=0

  15. x4+5x3+9x2+7x+2=0

Determinants

See: PRE_10_Homogeneous_Equations.html.

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Last Modified: Monday, 6 September 2020 13:33 EDT