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Solution of Quadratic equations with Complex Roots


Given
 

Example 15:
Solve the following equation

Solution:


Example 16.
Solve

Solution:-


Example17.
Solve
[Recollect complex roots always occur as a conjugate pair]
if 3 + i is a root, then 3 - i is also a root.
Sum of roots = 6
Product of roots = (3 + i) (3 - i) = 10
Corresponding quadratic factor is
Let another quadratic factor be

Equating x2 term in both sides, we get -6p + 10 + 2 = 24
-6p = 24 - 12 = 12
p = -2
The other factor is x2 - 2x + 2

The roots are 3 ± i, 1 ± i

Some Theorems stated (without proof)
  1. A polynomial equation with real coefficients of degree n has n roots.
  2. For any polynomial equation P(x) = 0 with real coefficients, imaginary roots  occur in pairs, irrational roots occur in pairs.
Note:- If there are odd number of roots, one or odd number of roots can be    real, but the number of imaginary roots cannot be odd.

Cube roots of unity '1' are obtained by solving
x3 = 1 (ω is the symbol used to represent the cube root of unity)
x3 = 1
x3 1 = 0
 
4. Fourth roots of unity are obtained by solving x4 = 1
 




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