Solution of Quadratic Equation by factorization

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

ax² + 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 ax² + bx + c can be factorized into two linear factors.

Example

Solve the quadratic equation 3x² − 6x = x + 20 by factorization.

Solution:

3x² − 6x = x + 20          (i)

The standard form of (i) is

3x² − 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

3x² − 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 x² − x − 2 = 0.

Solution:

x² − 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

x² − 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 5x² = 30x by factorization.

Solution:

5x² = 30x

Remember that cancelling of x on both sides of 5x² = 30x means the loss of one root, i.e., x = 0

5x² − 30x = 0          (i)

The equation (i) is factorized as

5x (x − 6) = 0

Either 5x = 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}.