# Complex roots of Quadratic Equations for Class 11 ,CBSE Board, IITJEE maths and other exams

## Introduction of Complex roots

We have already studied about Complex number in previous chapter,Now it times to use in quadratic equation
We know that if
b2-4ac < 0
We dont have real  roots
Now if b2-4ac < 0 then 4ac -b2 > 0
So now we can define imaginary roots of the equation as
$k_{1}=\frac{-b+i\sqrt{4ac-b^{2}}}{2a}$
$k_{2}=\frac{-b-i\sqrt{4ac-b^{2}}}{2a}$

Complex roots occurs in conjugate pairs and it is very clear from the roots given above
The quadratic equations containing complex roots can be solved using the  Factorization method,square method and quadratic method explained above

So far we have read about  quadratic equation where the coefficent are real. Complex quadratic equation are the equations where the coefficent are complex numbers.
These quadratic equation can also be solved using Factorization method,square method and quadratic method explained above
The roots  may not conjugate pair in these quadratic equations

## Examples

x2-2x+10=0
Solution
x2-2x+1 +9=0
(x-1)2  -(3i)2=0
(x-1-3i)(x-1+3i)=0
So roots   are
(1+3i)  and (1-3i)
We can also obtain the same things using quadratic methods
$k_{1}=\frac{-b+i\sqrt{4ac-b^{2}}}{2a}$
$k_{2}=\frac{-b-i\sqrt{4ac-b^{2}}}{2a}$
$k_{1}=\frac{2+i\sqrt{40-2^{2}}}{2}$=1+3i
$k_{1}=\frac{2-i\sqrt{40-2^{2}}}{2}$=1-3i

2) For which values of k does the polynomial  x2+4x+k have two complex conjugate roots?
Solution:
For the roots to be complex conjugate ,we should have
b2 -4ac < 0
16-4k <0
or  k> 4