CBSE Tutorials

Introduction of Quadratic Equation

A polynomial of degree 2 (i.e.) is called a quadratic polynomial where a ¹ 0 and a, b, c are real numbers.

Any equation of the form, p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. Therefore, is called the standard form of a quadratic equation.

e.g.

Zeros of Quadratic Equation

For a quadratic polynomial , those values of x for which is satisfied, are called zeros of quadratic polynomial , i.e. if then a is called the zero of quadratic polynomial.

Roots of Quadratic Equation

If a, b are zeros of polynomial then a, b are called roots (or solutions) of corresponding equation which implies that

i.e., and

Problems based on Quadratic Equation:

Example:

Check whether the following equations are quadratic or not.

(i)     (ii)

(iii)             (iv)

Solution:

(i)     can be rewritten as

i.e.,            

It is of the form

Therefore, the given equation is a quadratic equation.

(ii)    As and

Therefore, we have        

Þ                

It is not of the form

Therefore, the given equation is not a quadratic equation.

(iii)    

As is a quadratic polynomial

\ is a quadratic equation.

(iv)    

Þ    

Since is not a quadratic polynomial.

\ is not a quadratic equation

Example:

In each of the following, determine the value of k for which the given value is a solution of the equation.

(i)         (ii)

Solution:

(i)                 …(i)

Since is a solution of the equation

Now substituting x = 2     in equation (i), we get

Þ    

Þ    

Þ    

(ii)    Since x = –2 is a solution of the equation

            …(i)

Now substituting x = – 2 in (i), we get

Þ    

or    

Þ    

Example:

In each of the following, determine whether the given values of x are the solutions of the given equation or not:

(i)         (ii)

Solution:

(i)    Consider =        …(i)

Substituting x = –1 in (i), we get

= 1 – 6 + 5

= .

Hence x = –1 is the solution of given equation.

Again substituting x = –5 in (i), we get

=

= 25 – 30 + 5

= 30 – 30 = 0

Hence x = –5 is the solution of given equation.

\    x = –1 and x = –5 are the solutions of given equation.

(ii)    Consider p(x)            …(i)

Substituting in (i), we get

=

= =

= 2 – 2 = 0

Þ     is the solution of given equation.

Again substituting in (i), we get

= =

Þ         is the solution of given equation.

\     and are the solutions of quadratic equation.

Example:

Determine whether the given values of x are solutions of the given equation or not:

Solution:

Consider         …(i)

Substituting in (i)

=

\     is a solution of the given quadratic equation.

Substituting in (i) we get

=

= 27 – 18 – 9 = 0 = RHS

\     is a solution of the given quadratic equation.

Example:

If x = 2 and x = 3 are roots of find the values of p and q.

Solution:

Since x = 2 is a root of the given equation,

\ we have

Þ             Þ            …(i)

Since x = 3 is a root of the given equation,

\ we have    

Þ             Þ            …(ii)

Subtracting equation (i) from equation (ii), we get

or                 Þ    

Substituting in equation (i), we get

Þ         Þ     Þ    q = 9

Hence and q = 9.