CBSE 10th Mathematics | Solution of a Quadratic Equation by Factorisation

Solution of a Quadratic Equation by Factorisation

 

    Consider the following products.

        \displaystyle 6\times 0=0;\,\,-b\times 0=0,\,\,0\times a=0

    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 \displaystyle a{{x}^{2}}+bx+c=0. 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,    \displaystyle a{{x}^{2}}+bx+c=0

    Þ    \displaystyle (px+q)\,(rx+s)=0

    \    Either \displaystyle (px+q)=0    or    \displaystyle (rx+s)=0

        \displaystyle px=-q         or    rx = –s

        \displaystyle x=-\frac{q}{p}        or     \displaystyle x=-\frac{s}{r}

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

        ·    Transform the equation into standard form, if necessary.

        ·    Factorise \displaystyle a{{x}^{2}}+bx+c.

        ·    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) \displaystyle {{(x-2)}^{2}}-36=0        (ii) \displaystyle {{x}^{2}}+2\sqrt{2}x+3=0

 

Solution:     

 

(i)    \displaystyle {{(x-2)}^{2}}-36=0

    Þ    \displaystyle {{(x-2)}^{2}}-{{6}^{2}}=0

    Þ    \displaystyle (x-2-6)(x-2+6)=0

    Þ    \displaystyle (x-8)\,(x+4)=0

    either    \displaystyle x-8=0 or \displaystyle x+4=0

    either    \displaystyle x=8 or \displaystyle x=-4

    Solution is x = 8, –4

    (ii)     \displaystyle {{x}^{2}}+2\sqrt{3}x+3=0

    or     \displaystyle {{x}^{2}}+\sqrt{3}x+\sqrt{3}x+3=0

         \displaystyle x(x+\sqrt{3})+\sqrt{3}(x+\sqrt{3})=0

         \displaystyle (x+\sqrt{3})\,\,(x+\sqrt{3})=0

    either    \displaystyle x+\sqrt{3}=0    or \displaystyle x+\sqrt{3}=0

    either     \displaystyle x=-\sqrt{3}    or \displaystyle x=-\sqrt{3}

    Solution is \displaystyle x=-\sqrt{3}

 

Problem:     

 

Solve : \displaystyle {{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0

 

Solution:

 

        \displaystyle {{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0

    Þ    \displaystyle {{b}^{2}}x({{a}^{2}}x+1)-1({{a}^{2}}x+1)=0

    Þ    \displaystyle ({{a}^{2}}x+1)\left( {{b}^{2}}x-1 \right)=0

    Either \displaystyle ({{a}^{2}}x+1)=0    or    \displaystyle ({{b}^{2}}x-1)=0

    Þ    \displaystyle {{a}^{2}}x=-1    or    \displaystyle {{b}^{2}}x=1

    Þ    \displaystyle x=-\frac{1}{{{a}^{2}}}    or    \displaystyle x=\frac{1}{{{b}^{2}}}

    Hence,     \displaystyle x=-\frac{1}{{{a}^{2}}},\frac{1}{{{b}^{2}}} are the required solutions.

 

Problem:    

 

Solve: \displaystyle 4{{x}^{2}}-4ax+({{a}^{2}}-{{b}^{2}})=0, a, b are real numbers

 

Solution:         

 

\displaystyle 4{{x}^{2}}-4ax+({{a}^{2}}-{{b}^{2}})=0

        Þ    \displaystyle 4{{x}^{2}}-[2\,(a+b)\,x+2\,(a-b)x]+{{a}^{2}}-{{b}^{2}}=0

        [\displaystyle \because \,\,\,\,2\left( a+b \right)+2\left( a-b \right)=2a+2b+2a-2b=4a]

         \displaystyle \left[ also\,\,\,2\left( a+b \right)\times 2\left( a-b \right)=4\left( {{a}^{2}}-{{b}^{2}} \right) \right]

        Þ    \displaystyle [4{{x}^{2}}-2(a+b)x]-[2(a-b)x-({{a}^{2}}-{{b}^{2}})]=0

        Þ    \displaystyle 2x[2x-(a+b)]-(a-b)[2x-(a+b)]=0

        Þ    \displaystyle [2x-(a+b)]\,\,[2x-(a-b)]=0

        Þ     either \displaystyle [2x-(a+b)]=0

        or    \displaystyle [2x-(a-b)]=0

            \displaystyle 2x=a+b or \displaystyle 2x=a-b

            either        \displaystyle x=\frac{a+b}{2} or \displaystyle x=\frac{a-b}{2}

        Hence, \displaystyle x=\frac{a+b}{2},\frac{a-b}{2} are the required solutions.

 

Problem:    

 

Solve the following quadratic equation by factorization method:

        \displaystyle {{x}^{2}}+\left( \frac{a}{a+b}+\frac{a+b}{a} \right)\,x+1=0

 

Solution:     

 

We have,    

            \displaystyle {{x}^{2}}+\left( \frac{a}{a+b}+\frac{a+b}{a} \right)\,x+1=0

        Þ    \displaystyle {{x}^{2}}+\frac{a}{a+b}x+\frac{a+b}{a}x+\frac{a}{a+b}\times \frac{a+b}{a}=0 \displaystyle \left[ \because \,\,\,\frac{a}{a+b}\times \frac{a+b}{a}=1 \right]

        Þ    \displaystyle x\,\left\{ x+\frac{a}{a+b} \right\}+\frac{a+b}{a}\,\left\{ x+\frac{a}{a+b} \right\}=0

        Þ    \displaystyle \left\{ x+\frac{a}{a+b} \right\}\,\,\left\{ x+\frac{a+b}{a} \right\}=0

            either    \displaystyle x+\frac{a}{a+b}=0

        or,         \displaystyle x+\frac{a+b}{a}=0

        Þ    \displaystyle x=-\frac{a}{a+b}

        or,     \displaystyle x=-\frac{a+b}{a}

 

Problem:    

 

Solve the following quadratic equations by factorization method:

        \displaystyle \frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x},a+b\ne 0

 

Solution:     

 

We have,

            \displaystyle \frac{1}{a+b+x}-\frac{1}{x}=\frac{1}{a}+\frac{1}{b}

        Þ    \displaystyle \frac{x-(a+b+x)}{x(a+b+x)}=\frac{a+b}{ab}

        Þ    \displaystyle \frac{x-a-b-x}{x\left( a+b+x \right)}=\frac{a+b}{ab}

        Þ    \displaystyle \frac{-(a+b)}{x(a+b+x)}=\frac{a+b}{ab}

        Þ    \displaystyle \frac{-1}{x(a+b+x)}=\frac{1}{ab}        \displaystyle [\because \,a+b\ne 0]

        Þ    ab =\displaystyle \,x(a+b+x)    

        Þ    \displaystyle x\,(a+b+x)+ab=0                

        Þ    \displaystyle {{x}^{2}}+ax+bx+ab=0

        Þ    \displaystyle x\,(x+a)\,+b\,(x+a)=0

        Þ    \displaystyle (x+a)\,(x+b)=0

        Þ    \displaystyle x+a=0 or, \displaystyle x+b=0

        Þ    \displaystyle x=-a or, \displaystyle x=-b

        \    Solution is x = -a, -b

 

        

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