physicscatalyst.com logo







Flashback of Quadratic Equations for Class 11 ,CBSE Board, IITJEE maths and other exams




Flashback of Quadratic equation From previous Classes

Quadratic Polynomial

P(x) = ax2 +bx+c   where a≠0

 

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

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

 

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



Go Back to Class 11 Maths Home page Go Back to Class 11 Physics Home page


link to us