## 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.