Flashback of Quadratic equation From previous Classes
Quadratic Polynomial
P(x) = ax
^{2} +bx+c where a≠0
Quadratic equation
ax
^{2} +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
aα
^{2} +bα+c=0
Some other points to remember
 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
How to Solve Quadratic equation
S.no

Method

Working

1

factorization

This method we factorize the equation by splitting the middle term b
In ax^{2}+bx+c=0
Example
6x^{2}x2=0
1) First we need to multiple the coefficient a and c.In this case =6X2=12
2) Splitting the middle term so that multiplication is 12 and difference is the coefficient b
6x^{2} +3x4x2=0
3x( 2x+1) 2(2x+1)=0
(3x2) (2x+1)=0
3) Roots of the equation can be find equating the factors to zero
3x2=0 => x=3/2
2x+1=0 => x=1/2

2

Square method

In this method we create square on LHS and RHS and then find the value.
ax^{2} +bx+c=0
1) x^{2} +(b/a) x+(c/a)=0
2) ( x+b/2a)^{2} –(b/2a)^{2} +(c/a)=0
3) ( x+b/2a)^{2}=(b^{2}4ac)/4a^{2}
4)
Example
x^{2} +4x5=0
1) (x+2)^{2} 45=0
2) (x+2)^{2}=9
3) Roots of the equation can be find using square root on both the sides
x+2 =3 => x=5
x+2=3=> x=1

3

Quadratic method

For quadratic equation
ax^{2} +bx+c=0,
roots are given by
$x=\frac{b+\sqrt{b^{2}4ac}}{2a}$
and
$x=\frac{b\sqrt{b^{2}4ac}}{2a}$
For b^{2} 4ac > 0, Quadratic equation has two real roots of different value
For b^{2}4ac =0, quadratic equation has one real root
For b^{2}4ac < 0, no real roots for quadratic equation

Nature of roots of Quadratic equation
S.no

Condition

Nature of roots

1

b^{2} 4ac > 0

Two distinct real roots

2

b^{2}4ac =0

One real root

3

b^{2}4ac < 0

No real roots

Solved examples
1) Find the roots of the quadratic equation
x
^{2} 6x=0
Solution
There is no constant term in this quadratic equation, we can x as common factor
x(x6)=0
So roots are x=0 and x=6
2) Solve the quadratic equation
x
^{2} 16=0
Solution
x
^{2} 16=0
x
^{2} =16
or x=4 or 4
3) Solve the quadratic equation by factorization method
x
^{2} x 20=0
Solution
1) First we need to multiple the coefficient a and c.In this case =1X20=20
The possible multiple are 4,5 ,2,10
2) The multiple 4,5 suite the equation
x
^{2}5x+4x20=0
x(x5)+4(x5)=0
(x+4)(x5)=0
or x=4 or 5
4) Solve the quadratic equation by
Quadratic method
x^{2} 3x18=0
Solution
For quadratic equation
ax
^{2} +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
Related Topics
Class 11 Maths
Class 11 Physics