Explore BrainMass

# Find Real Roots of an Equation Given Two of the Complex Conjugate Roots

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

The equation x^4 - 18x^3 + 121x^2 - 368x + 420=0

has complex conjugate roots (4+j2) & (4-j2). By a process of division and solving a quadratic equation, find all the roots and hence write down all the factors of :

x^4 - 18x^3 + 121x^2 - 368x + 420

Please can you show all working out including the division part of this problem as I would like to understand the process behind this problem for future use.

Please see the attached file for the fully formatted problem.

© BrainMass Inc. brainmass.com September 26, 2022, 9:56 pm ad1c9bdddf
https://brainmass.com/math/basic-algebra/equation-given-two-complex-conjugate-roots-42648

#### Solution Preview

Please see the attached file for the complete solution.
Thanks for using BrainMass.

The equation x - 18x + 121x - 368x + 420=0

has complex conjugate roots (4+j2) & (4-j2). By a process of division and solving a quadratic equation, find all the roots and hence write down all the factors of :

x - 18x + 121x - 368x + 420

Please can you show all working out including the division part of this problem as I would like to understand the process behind this problem for future use.

Solution. Since the equation x^4- 18x^3 + 121x^2- 368x + ...

#### Solution Summary

The real roots for an equation are found given two of the complex conjugate roots. The solution is detailed and well presented. The response received a rating of "5" from the student who originally posted the question.

\$2.49