To find the **solution set** of a quadratic equation, following methods are used:

- factorization
- completing square
- use of quadratic formula

## Solution by factorization

In this method, write the quadratic equation in the **standard form** as

**ax ^{2} + bx + c = 0** (i)

If two numbers r and s can be found for the equation (i) such that

**r + s = b** and **rs = ac**

then ** ax^{2} + bx + c** can be factorized into two linear factors.

### Example

Solve the quadratic equation *3x ^{2} − 6x = x + 20* by factorization.

### Solution:

*3x ^{2} − 6x = x + 20 (i)*

The standard form of (i) is

*3x ^{2} − 7x − 20 = 0 (ii)*

Here a = 3, b = −7, c = −20

and ac = 3 × (−20) = −60

As −12 + 5 = −7 and −12 × 5 = −60, so

the equation (ii) can be written as

*3x ^{2} − 12x + 5x − 20 = 0*

*or 3x (x − 4) + 5 (x − 4) = 0*

*⇒ (x − 4) (3x + 5) = 0*

Either *x − 4 = 0 or 3x + 5 = 0*

x = 4 or 3x = −5 ⇒ x = −5⁄3

∴ x = −5⁄3, 4 are the solutions of the given equation.

Thus, the solution set is {−5⁄3, 4}.

### Activity:

Factorize *x ^{2} − x − 2 = 0.*

### Solution:

*x ^{2} − x − 2 = 0 (i)*

Here a = 1, b = −1, c = −2

and ac = 1 × (−2) = −2

As −2 + 1 = −1 and −2 × 1 = −2, so

the equation (i) can be written as

*x ^{2} − 2x + x − 2 = 0*

*or x (x − 2) + 1 (x − 2) = 0*

*⇒ (x − 2) (x + 1) = 0*

∴ * (x − 2), (x + 1)* are factors of the given equation.

### Example

Solve 5*x ^{2} = 30x *by factorization.

### Solution:

*5x ^{2} = 30x
*

**Remember that cancelling of x** on both sides of *5x ^{2} = 30x* means the loss of one root, i.e., x = 0

*5x ^{2} − 30x = 0 (i)
*

The equation (i) is factorized as

*5x (x − 6) = 0*

Either 5*x = 0 or x − 6 = 0 ⇒ x = 0 or x = 6
*

∴ x = 0, 6 are the roots of the given equation.

Thus, the solution set is {0, 6}.