# 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]

âˆ´ 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

âˆ´ Corresponding quadratic factor is

Let another quadratic factor be

Equating

*x*

^{2}term in both sides, we get -6

*p*+ 10 + 2 = 24

â‡’ -6

*p*= 24 - 12 = 12

âˆ´

*p*= -2

âˆ´ The other factor is

*x*

^{2}- 2

*x*+ 2

âˆ´ The roots are 3 Â±

*i*, 1 Â±

*i*

**Some Theorems stated (without proof)**

- A polynomial equation with real coefficients of degree
*n*has*n*roots. - 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

*x*

^{3}= 1 (Ï‰ is the symbol used to represent the cube root of unity)

*x*

^{3}= 1

*x*

^{3}âˆ’ 1 = 0

4. Fourth roots of unity are obtained by solving

*x*

^{4}= 1