Solving Quadratic Equations by Factorisation

The process of writing an expression as a product of two or more common factors is called method of factorization. e.g.

  1. {x^2} + 5x + 6 = \left( {x + 2} \right)\left( {x + 3} \right)
  2. 5{x^2} + 8x = x\left( {5x + 8} \right)
  3. 30 = 2 \times 3 \times 5

In the above examples, \left( {x + 2}  \right)\left( {x + 3} \right)are the factors of expression {x^2} + 5x + 6, x\left( {5x + 8} \right)are the factors of 5{x^2} + 8x and 2 \times 3 \times 5are the factors of 30.

While solving the quadratic equation by the method of factorization, we have the following steps:

  • Convert the quadratic equation in standard form, if necessary i.e. a{x^2} + bx + c = 0,    where a \ne 0
  • Multiply coefficient of {x^2} with constant terms, we get a \times c = ac.
  • Now try to find two numbers whose products is ac and sum or difference is equal to b (coefficient of x).
  • Factorise the given expression on L.H.S.
  • Equate each factor equal to zero.
  • We get the required roots, say{x_1}, {x_2}.

 Example:

Solve the equation by factorization method.

5{x^2} - 11x + 6 = 0


Solution:

The given equation in standard form is 5{x^2} - 11x + 6 = 0

Multiply coefficient of{x^2}and the constant term, we get 5 \times 6 = 30

Divide 30 into two parts such that their difference or sum is 11

Possible factors of 30

Sum or Difference of factors

30 \times 1 = 30

30 - 1 = 29,{\text{ }}30 + 1 = 31 (not possible)

15 \times 2 = 30

15 - 2 = 13,{\text{ 15}} + 2 = 17 (not possible)

10 \times 3 = 30

10 - 3 = 7,{\text{ 1}}0 + 3 = 13    (not possible)

6 \times 5 = 30

6 - 5 = 1,{\text{ 6}} + 5 = 11          (possible)

Therefore,        5{x^2} - 11x + 6 = 0
                        5{x^2} - \left( {5 + 6} \right)x + 6 = 0
                        5{x^2} - 5x - 6x + 6 = 0
                        5x\left( {x - 1} \right) - 6\left( {x - 1} \right) = 0
                        \left( {x - 1} \right)\left( {5x - 6} \right) = 0
Either              x - 1 = 0   or   5x - 6 = 0
                        x = 1         or    x = \frac{6}{5}
Example:

Solve the equation by factorization method.

8x + 4{x^2} = 0


Solution:

The given equation in standard form is 4{x^2} + 8x = 0

Here, the constant term is absent; its factorization is very simple.

Taking common 4x, we get
                        4x\left( {x + 2} \right) = 0
Either              4x = 0    or   x + 2 = 0
                        x = 0      or     x = - 2

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