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
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 
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 
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