CBSE Tutorials

Solution of a Quadratic Equation by Factorisation

    Consider the following products.

    Above example illustrate that whenever the product is 0, at least one of the factors is 0.

Imp: If a and b are numbers, then ab = 0, iff a = 0 or b = 0.

    Above principle is used in solving quadratic equation by factorisation. Let given quadratic equation be Let the quadratic polynomial be expressed as product of two linear factors i.e. (px + q) and (rx + s), where p, q, r, s are real numbers and p ¹ 0, r ¹ 0,

    Then,    

    Þ    

    \    Either     or    

                 or    rx = –s

                or     

    Following steps are involved in solving a quadratic equation by factorisation.

        ·    Transform the equation into standard form, if necessary.

        ·    Factorise .

        ·    Put each factor containing variable = 0.

        ·    Solve each of the resulting equation.

Solved Problems Based on Solution of a Quadratic Equation by Factorisation

Problem:

Using factorisation, solve following equations.

    (i)         (ii)

Solution:     

(i)    

    Þ    

    Þ    

    Þ    

    either     or

    either     or

    Solution is x = 8, –4

    (ii)    

    or    

    either        or

    either         or

    Solution is

Problem:     

Solve :

Solution:

    Þ    

    Þ    

    Either     or    

    Þ        or    

    Þ        or    

    Hence,     are the required solutions.

Problem:    

Solve: , a, b are real numbers

Solution:         

        Þ    

        []

        Þ    

        Þ    

        Þ    

        Þ     either

        or    

             or

            either         or

        Hence, are the required solutions.

Problem:    

Solve the following quadratic equation by factorization method:

Solution:     

We have,    

        Þ    

        Þ    

        Þ    

            either    

        or,         

        Þ    

        or,     

Problem:    

Solve the following quadratic equations by factorization method:

Solution:     

We have,

        Þ    

        Þ    

        Þ    

        Þ            

        Þ    –ab =    

        Þ                    

        Þ    

        Þ    

        Þ    

        Þ     or,

        Þ     or,

        \    Solution is x = -a, -b