Complex roots of Quadratic Equations for Class 11 ,CBSE Board, IITJEE maths and other exams
Flashback of Quadratic equation From previous Classes
Quadratic equation
ax2 +bx+c =0 where a≠0 Solution or root of the Quadratic equation
A real number α is called the root or solution of the quadratic equation if
aα2 +bα+c=0
we can learn in details from below links
The root of the quadratic equation is the zeroes of the polynomial p(x).
We know from chapter two that a polynomial of degree can have max two zeroes. So a quadratic equation can have maximum two roots
A quadratic equation has no real roots if b2- 4ac < 0
Solved examples
1.Find the roots of the quadratic equation
x2 -6x=0
Solution
There is no constant term in this quadratic equation, we can x as common factor
x(x-6)=0
So roots are x=0 and x=6
2. Solve the quadratic equation
x2 -16=0
Solution
x2 -16=0
x2 =16
or x=4 or -4
3. Solve the quadratic equation by factorization method
x2 -x -20=0
Solution
1) First we need to multiple the coefficient a and c.In this case =1X-20=-20
The possible multiple are 4,5 ,2,10
2) The multiple 4,5 suite the equation
x2-5x+4x-20=0
x(x-5)+4(x-5)=0
(x+4)(x-5)=0
or x=-4 or 5
4. Solve the quadratic equation by Quadratic method x2 -3x-18=0 Solution
For quadratic equation
ax2 +bx+c=0,
roots are given by
$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$
and
$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$
Here a=1
b=-3
c=-18
Substituting these values,we get
x=6 and -3
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
Examples
(1) Solve the quadratic equation
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
Square root of Complex Number
if $z = \sqrt {a+ ib}$
Then z can be easy derived by putting z=x+iy, squaring both the side and comparing real and imaginary parts, we get
$ \pm \left ( \sqrt {\frac {\sqrt {a^2 + b^2} +a}{2}} + i \frac {b}{|b|}\sqrt {\frac {\sqrt {a^2 + b^2} -a}{2}} \right )$
where $b -ne 0$
We can write this in polar form as
$z = \sqrt {re^{i \theta}}$
then this can be derived using De Movire theorem
$z= \pm r^{1/2) e^{i \theta/2}$
Complex Quadratic quation
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
The above derivation of square will come handy while evaluating the roots
Examples
(1) Solve the quadratic equation
3z2+6iz -3=0 Solution
3z2+6iz -3=0
Using Quadratic formula
$k_{1}=\frac{-b+\sqrt{b^{2} -4ac}}{2a}$
$k_{2}=\frac{-b-\sqrt{b^{2} -4ac}}{2a}$
or
$k_{1}=\frac{-6i+\sqrt{-36+36}}{6}$=-i
$k_{1}=\frac{-6i-\sqrt{-36+36}}{6}$=-i
So repeatable roots is -i
