ax

A real number α is called the root or solution of the quadratic equation if

aα

we can learn in details from below links

- Quadratic Polynomial
- Graphing quadratics Polynomial
- what is a quadratic equation
- How to Solve Quadratic equations
- Factoring quadratics equations
- Solving quadratic equations by completing the square
- Solving quadratic equations by using Quadratic formula
- Nature of roots of Quadratic equation
- Problem based on discriminant of a quadratic equation
- Quadratic word problems

- 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 b
^{2}- 4ac < 0

x

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

x

Solution

x

x

or x=4 or -4

3. Solve the quadratic equation by factorization method

x

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

x

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

x

Solution

For quadratic equation

ax

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

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

We know that if

b

We dont have real roots

Now if b

So now we can define imaginary roots of the equation as

$k_{1}=\frac{-b+i\sqrt{4ac-b^{2}}}{2a}$

$k_{2}=\frac{-b-i\sqrt{4ac-b^{2}}}{2a}$

The quadratic equations containing complex roots can be solved using the Factorization method,square method and quadratic method explained above

(1) Solve the quadratic equation

x

x

(x-1)

(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

Solution:

For the roots to be complex conjugate ,we should have

b

16-4k < 0

or k> 4

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

$z= \pm r^{1/2) e^{i \theta/2}$

- 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

(1) Solve the quadratic equation

3z

3z

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

**Notes**- What is complex numbers
- Algebra Of complex Numbers
- Conjugate of Complex Numbers
- Modulus of complex numbers
- Argand Plane
- Polar Representation of the complex number
- Rotation of Complex Number
- Identities for Complex Numbers
- Eulers formula and De moivre's theorem
- Cube Root of unity
- Complex roots of Quadratic equations

**NCERT Solutions & Worksheets**