Quadratic Equations - How to Solve Quickly

Today I am sharing quick technique to solve Quadratic Equations quickly. I will also share practice questions on same to;

A polynomial of degree 2 is called Quadratic Equation.

For example:
  • 2x2-5x+1=0
  • x2-5=0

A general form of quadratic equation is ax2+bx+c=0

Where a, b, c all belong to real numbers

Now if you compare the equations I and II with the general form.
  • a=2, b=-5 and c=1
  • a=1, b=0 and c=-5

Zeros or the solutions of Quadratic Equation

The real value of x for which the value of the P(x) = ax2+bx+c becomes zero is known as the root of 
the quadratic equation  ax2+bx+c =0.

  • Determine whether 3and 4 are the zeros of the polynomial P(x) =x2-7x+12
                         P (3)   = 32-7.3+12
                                    = 9-21+12
                                    = -12+12
                                    = 0

                       P (4) = 42-7.4+12
                                = 16-28+12
                                = -12+12
                                = 0
Therefore 3 and 4 are the zeros or the solutions or the roots of the polynomial P(x)=x2-7x+12

Kinds of a Quadratic Equation

There are two types of quadratic equations 
  • Pure quadratic equation
  • Adfected quadratic equation

Pure quadratic equation:

An equation of the form ax2+c=0 , a ≠0 is known as the pure quadratic equation .It means that the 

quadratic equation ax2+bx+c =0, having no term containing single power of x, known as pure 

quadratic equation . Clearly in a quadratic equation a ≠0 and b=0.

For example
  • x2-4=0
  • 3/8x2=5

Adfected quadratic equation:

A quadratic equation of the form ax2+bx+c =0, a ≠0 is known adfected quadratic equation or general 
quadratic equation. An adfected quadratic equation has also a term containing single power of x .In 
adfected quadratic equation

ax2+bx+c =0, a ≠0, b ≠0

For example:
  • x2-5x+11=0
  • 0.3x2+17x-2.3=0

Solving a pure quadratic equation

There are two methods for solving a pure quadratic equation of the form ax2+c=0:
  • By square root 
  • By factorization

                To solve a quadratic equation by square root

ax2+c=0 is a pure quadratic equation. To solve it , bring the constant term the RHS (right hand side) 

and divide both side by a, coefficient of x2 and take the square root.

For example :




x = +7/2

To solve a pure quadratic equation by factorization

Bring the equation ax2+c=0 in the form p2-q2 =0. Use the p2-q2=(p+q)(p-q). Equate each factor to 

zero  and find the values of x in each case, the two values of x so obtained are roots of the equation 


For example :


(4x)2-(5)2 = 0

(4x+5) (4x-5) =0

4x-5 =0

x= 5/4

4x+5 =0

x =-5/4

Solving an Adfected Quadratic Equation

There are two methods for solving an adfected quadratic equation ax2 + bx+c=0, a ≠0, b ≠0.

(i)                  By factorization                                         (ii)           By completing square

To solve the quadratic equation ax2 + bx+c=0 by the method factorization

In this method the middle term (i.e term containing single power of x) is broken into two suitable 

parts so that the factors are formed.

Example:             solve.    8x2 + 7x – 15 = 0
Solution:                                 8x2 +7x- 15 =0

                Or                           8x2-8x + 15x-15=0

                Or                           8x (x-1) +15(x-1) =0

                Or                           (8x +15)(x-1)=0

Therefore                                      8x+15=0


                                x= -15/8
And                        x-1 =0

Hence   -15/8 and 1 are the required roots.

To solve ax2 + bx+ c=0 by completion of  square

The famous Indian mathematician ShreedharAcharya had invented a formula for solving the 

quadratic equation ax2 +bx+c=o.

If the equation ax2 +bx +c =0 has roots α and β, then


Quadratic Equations

Where      a = coefficient of x2

                b = coefficient of x,

                c = constant term
Join 40,000+ readers and get free notes in your email


Post a Comment

Thanks for commenting. It's very difficult to answer every query here, it's better to post your query on IBPSToday.com