 # Flashback of Quadratic equation From previous Classes

P(x) = ax2 +bx+c   where a≠0
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
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 b2- 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 ax2+bx+c=0 Example 6x2-x-2=0 1) First we need to multiple the coefficient a and c.In this case =6X-2=-12 2) Splitting the middle term so that multiplication is 12 and difference is the coefficient b 6x2 +3x-4x-2=0 3x( 2x+1) -2(2x+1)=0 (3x-2) (2x+1)=0 3) Roots of the equation can be find equating the factors to zero 3x-2=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. ax2 +bx+c=0 1) x2 +(b/a) x+(c/a)=0 2) ( x+b/2a)2 –(b/2a)2 +(c/a)=0 3) ( x+b/2a)2=(b2-4ac)/4a2 4) Example x2 +4x-5=0 1)  (x+2)2 -4-5=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 ax2 +bx+c=0, roots are given by $x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$ For b2 -4ac  > 0, Quadratic equation has two real roots of different value For b2-4ac =0, quadratic equation has one real root For b2-4ac < 0, no real roots for quadratic equation

## Nature of roots of Quadratic equation

 S.no Condition Nature of roots 1 b2 -4ac  > 0 Two distinct real roots 2 b2-4ac =0 One real root 3 b2-4ac < 0 No real roots

## 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
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
x2 -3x-18=0
Solution
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