We have already studied about Complex number in previous chapter,Now it times to use in quadratic equation

We know that if

b^{2}-4ac < 0

We dont have real roots

Now if b^{2}-4ac < 0

then 4ac -b^{2} > 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

1) Solve the quadratic equation

x^{2}-2x+10=0

Solution

x^{2}-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 x^{2}+4x+k have two complex conjugate roots?

Solution:

For the roots to be complex conjugate ,we should have

b^{2} -4ac < 0

16-4k <0

or k> 4

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