CBSE 10th Mathematics | Solution of a Quadratic Equation by Completing the Square

Solution of a Quadratic Equation by Completing the Square

 

    Following steps are involved in solving a quadratic equation by quadratic formula

        ·    Consider the equation \displaystyle a{{x}^{2}}+bx+c=0, where a ¹ 0

        ·    Dividing throughout by ‘a’, we get

            \displaystyle {{x}^{2}}+\frac{b}{a}x+\frac{c}{a}=0

        ·    Add and subtract \displaystyle {{\left( \frac{1}{2}\,\,\text{coefficient}\,\,\text{of}\,\,x \right)}^{2}}, we get

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

            \displaystyle {{\left( x+\frac{b}{2a} \right)}^{2}}=\frac{{{b}^{2}}}{4{{a}^{2}}}-\frac{c}{a}=\frac{{{b}^{2}}-4ac}{4{{a}^{2}}}

        ·    If \displaystyle {{b}^{2}}-4ac\ge 0 taking square root of both sides, we obtain

            \displaystyle x+\frac{b}{2a}=\frac{\pm \sqrt{{{b}^{2}}-4ac}}{2a}

            Therefore \displaystyle x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}

 

The Quadratic Formula: Quadratic equation, \displaystyle a{{x}^{2}}+bx+c=0, where a, b, c are real number and \displaystyle a\ne 0, has the roots as

            \displaystyle x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}

Problems based on Solution of a Quadratic Equation by Completing the Square with Solutions

 

Problem:     

 

Solve by completing the square : \displaystyle {{y}^{2}}+\frac{1}{2}y-1=0.

 

Solution:     

 

(i)    \displaystyle {{y}^{2}}+\frac{1}{2}y-1=0

    Þ    \displaystyle {{y}^{2}}+\frac{1}{2}y=1

    Adding \displaystyle {{\left( \frac{1}{4} \right)}^{2}} i.e. \displaystyle \frac{1}{16} on both sides, we have

    Þ    \displaystyle {{y}^{2}}+\frac{1}{2}y+\frac{1}{16}=1+\frac{1}{16}

    or    \displaystyle {{\left( y+\frac{1}{4} \right)}^{2}}=\frac{17}{16}

    Þ    \displaystyle y+\frac{1}{4}=\pm \sqrt{\frac{17}{16}}=\pm \frac{\sqrt{17}}{4}

    Þ    \displaystyle y=-\frac{1}{4}\pm \frac{\sqrt{17}}{4}=\frac{-1\pm \sqrt{17}}{4}

    So, the solutions are \displaystyle \frac{-1+\sqrt{17}}{4},\frac{-1-\sqrt{17}}{4}        

 

Problem:

 

By using the method of completing the square, show that the equation \displaystyle 2{{x}^{2}}+x+5=0 has no real roots.

 

Solution:

 

        \displaystyle 2{{x}^{2}}+x+5=0

    Þ    \displaystyle {{x}^{2}}+\frac{x}{2}+\frac{5}{2}=0            [Dividing the equation by 2]    

    Þ    \displaystyle {{x}^{2}}+\frac{x}{2}=-\frac{5}{2}

    Adding \displaystyle {{\left( \frac{1}{4} \right)}^{2}} i.e. \displaystyle \frac{1}{16} on both sides, we have

    Þ    \displaystyle {{x}^{2}}+2\,\left( \frac{1}{4} \right)\,x+{{\left( \frac{1}{4} \right)}^{2}}=-\frac{5}{2}+{{\left( \frac{1}{4} \right)}^{2}}

    Þ    \displaystyle {{\left( x+\frac{1}{4} \right)}^{2}}=-\frac{5}{2}+\frac{1}{16}

             \displaystyle =\frac{-40+1}{16}=-\frac{39}{16}

    Þ    \displaystyle {{\left( x+\frac{1}{4} \right)}^{2}}=-\frac{39}{16}

    RHS is negative but \displaystyle {{\left( x+\frac{1}{4} \right)}^{2}} cannot be negative for any real values of x.

                    [\displaystyle \because square of any real number is non-negative]

    Hence, the given equation has no real roots.

 

Problem:

 

Find the roots of the following equation

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

    by the method of completing the square.

 

Solution:     

 

We have,    

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

    Þ    \displaystyle {{x}^{2}}+bx-\left( \frac{{{a}^{2}}-{{b}^{2}}}{4} \right)=0 [Dividing the equation by 4]

    Þ    \displaystyle {{x}^{2}}+2\,\left( \frac{b}{2} \right)x=\frac{{{a}^{2}}-{{b}^{2}}}{4}

        Adding \displaystyle {{\left( \frac{b}{2} \right)}^{2}} on both sides we have

    Þ    \displaystyle {{x}^{2}}+2\,\left( \frac{b}{2} \right)x+{{\left( \frac{b}{2} \right)}^{2}}=\frac{{{a}^{2}}-{{b}^{2}}}{4}+{{\left( \frac{b}{2} \right)}^{2}}

                 \displaystyle =\frac{{{a}^{2}}-{{b}^{2}}}{4}+\frac{{{b}^{2}}}{4}

    Þ    \displaystyle {{\left( x+\frac{b}{2} \right)}^{2}}=\frac{{{a}^{2}}}{4}

    Þ    \displaystyle x+\frac{b}{2}=\pm \frac{a}{2}        [taking square root of both sides]

    Þ    \displaystyle x=-\frac{b}{2}\pm \frac{a}{2}

    Þ    \displaystyle x=\frac{-b-a}{2},\frac{-b+a}{2}

    Hence, the roots are \displaystyle \frac{-b-a}{2} and \displaystyle \frac{-b+a}{2}

 

Problem:

 

Solve the equation \displaystyle {{x}^{2}}-(\sqrt{3}+1)\,\,x+\sqrt{3}=0 by the method of completing the square.

 

Solution:     

 

We have,    

        \displaystyle {{x}^{2}}-(\sqrt{3}+1)\,x+\sqrt{3}=0

    Þ    \displaystyle {{x}^{2}}-(\sqrt{3}+1)\,x=-\sqrt{3}

        Adding \displaystyle {{\left( \frac{\sqrt{3}+1}{2} \right)}^{2}} on both sides we have

    Þ    \displaystyle {{x}^{2}}-2\,\left( \frac{\sqrt{3}+1}{2} \right)\,x+{{\left( \frac{\sqrt{3}+1}{2} \right)}^{2}}=-\sqrt{3}+{{\left( \frac{\sqrt{3}+1}{2} \right)}^{2}}

    Þ     \displaystyle {{\left( x-\frac{\sqrt{3}+1}{2} \right)}^{2}}=-\sqrt{3}+\frac{3+1+2\sqrt{3}}{4}

                \displaystyle =\frac{-4\sqrt{3}+3+1+2\sqrt{3}}{4}

                \displaystyle =\frac{3+1-2\sqrt{3}}{4}

    Þ    \displaystyle {{\left( x-\frac{\sqrt{3}+1}{2} \right)}^{2}}={{\left( \frac{\sqrt{3}-1}{2} \right)}^{2}}

    Þ    \displaystyle x-\frac{\sqrt{3}+1}{2}=\pm \frac{\sqrt{3}-1}{2}

    Þ    \displaystyle x=\frac{\sqrt{3}-1}{2}+\frac{\sqrt{3}+1}{2}=\frac{2\sqrt{3}}{2}=\sqrt{3}

    or     \displaystyle x-\frac{\sqrt{3}+1}{2}=-\frac{\sqrt{3}-1}{2}

        \displaystyle x=-\left( \frac{\sqrt{3}-1}{2} \right)+\frac{\sqrt{3}+1}{2}=\frac{2}{2}=1

    Þ    \displaystyle x=\sqrt{3},\,\,1

    Hence, the roots are \displaystyle \sqrt{3} and 1.

 

Problem:

 

Using quadratic formula, solve the following quadratic equation for x:

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

 

Solution:     

 

Comparing the given equation

        \displaystyle {{x}^{2}}-2ax+({{a}^{2}}-{{b}^{2}})=0 with the standard quadratic equation

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

    We have        a = 1

                b = –2a

    And            \displaystyle c={{a}^{2}}-{{b}^{2}}

    Now Discriminant    \displaystyle D={{b}^{2}}-4ac

        \displaystyle ={{(-2a)}^{2}}-4(1)({{a}^{2}}-{{b}^{2}})=4{{a}^{2}}-4{{a}^{2}}+4{{b}^{2}}

    Þ    \displaystyle D={{(2b)}^{2}}    Þ    \displaystyle \sqrt{D}=2b

    Using the quadratic formula, we get

        \displaystyle x=\frac{-b\pm \sqrt{D}}{2a}=\frac{2a\pm 2b}{2\times 1}=a\pm b

    Hence    \displaystyle x=a+b or \displaystyle x=a-b.

 

Problem:

 

Using quadratic formula solve the following quadratic equations:

    (i) \displaystyle {{p}^{2}}{{x}^{2}}+({{p}^{2}}-{{q}^{2}})\,x-{{q}^{2}}=0    (ii) \displaystyle 9{{x}^{2}}-9\,(a+b)\,x\,+(2{{a}^{2}}+5ab+2{{b}^{2}})=0

 

Solution:     

 

(i)    We have,    

            \displaystyle {{p}^{2}}{{x}^{2}}+({{p}^{2}}-{{q}^{2}})\,x-{{q}^{2}}=0

        Comparing the equation with \displaystyle a{{x}^{2}}+bx+c=0, we have

            \displaystyle a={{p}^{2}},\,\,b={{p}^{2}}-{{q}^{2}} and \displaystyle c=-{{q}^{2}}

    \ \displaystyle D={{b}^{2}}-4ac={{({{p}^{2}}-{{q}^{2}})}^{2}}-4\times {{p}^{2}}\times -{{q}^{2}}={{({{p}^{2}}-{{q}^{2}})}^{2}}+4{{p}^{2}}{{q}^{2}}

     \displaystyle ={{p}^{4}}+{{q}^{4}}-2{{p}^{2}}{{q}^{2}}+4{{p}^{2}}{{q}^{2}}={{({{p}^{2}}+{{q}^{2}})}^{2}}>0

                    [\displaystyle \because square of any real number is non-negative]

    So, the given equation has real roots given by

        \displaystyle \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{-({{p}^{2}}-{{q}^{2}})+({{p}^{2}}+{{q}^{2}})}{2{{p}^{2}}}=\frac{2{{q}^{2}}}{2{{p}^{2}}}=\frac{{{q}^{2}}}{{{p}^{2}}}

    and,    \displaystyle \beta =\frac{-b-\sqrt{D}}{2a}=\frac{-({{p}^{2}}-{{q}^{2}})-({{p}^{2}}+{{q}^{2}})}{2{{p}^{2}}}=-1

    Hence the roots are \displaystyle \frac{{{q}^{2}}}{{{p}^{2}}} and -1

    (ii)    We have,

            \displaystyle 9{{x}^{2}}-9\,(a+b)\,x+(2{{a}^{2}}+5ab+2{{b}^{2}})=0

        Comparing this equation with \displaystyle a{{x}^{2}}+bx+c=0, we have

        \displaystyle a=9,\,\,b=-9\,\,(a+b) and \displaystyle c=2{{a}^{2}}+5ab+2{{b}^{2}}

    \    \displaystyle D={{b}^{2}}-4ac

    Þ    \displaystyle D=81\,{{(a+b)}^{2}}-36(2{{a}^{2}}+5ab+2{{b}^{2}})

    Þ    \displaystyle D=81\left( {{a}^{2}}+{{b}^{2}}+2ab \right)-72{{a}^{2}}-180ab-72{{b}^{2}}

    Þ    \displaystyle D=81{{a}^{2}}+81{{b}^{2}}+162ab-72{{a}^{2}}-72{{b}^{2}}-180ab

    Þ    \displaystyle D=9{{a}^{2}}+9{{b}^{2}}-18ab    

    Þ    \displaystyle D=9\,{{(a-b)}^{2}}>0    [\displaystyle \because square of any real number is non-negative]

    So, the roots of the given equation are real and are given by

        \displaystyle \alpha =\frac{-b+\sqrt{D}}{2a}=\frac{9\,(a+b)+3(a-b)}{18}=\frac{12a+6b}{18}=\frac{2a+b}{3}

    and,     \displaystyle \beta =\frac{-b-\sqrt{D}}{2a}=\frac{9(a+b)-3(a-b)}{18}=\frac{6a+12b}{18}=\frac{a+2b}{3}

        \     Solution is \displaystyle x=\frac{2a+b}{3} and \displaystyle \frac{a+2b}{3}

 

Problem:

 

Using the quadratic formula, solve the equation

    \displaystyle {{a}^{2}}{{b}^{2}}{{x}^{2}}-(4{{b}^{4}}-3{{a}^{4}})x-12{{a}^{2}}{{b}^{2}}=0.

 

Solution:     

 

\displaystyle {{a}^{2}}{{b}^{2}}{{x}^{2}}-(4{{b}^{4}}-3{{a}^{4}})x-12{{a}^{2}}{{b}^{2}}=0

    Comparing the given equation with \displaystyle a{{x}^{2}}+bx+c=0,

    We have \displaystyle a={{a}^{2}}{{b}^{2}},\,b=-(4{{b}^{4}}-3{{a}^{4}}),\,\,c=-12{{a}^{2}}{{b}^{2}}

    Discriminant    \displaystyle D={{b}^{2}}-4ac

            \displaystyle D={{(3{{a}^{4}}-4{{b}^{4}})}^{2}}-4{{a}^{2}}{{b}^{2}}(-12{{a}^{2}}{{b}^{2}})

        \displaystyle =9{{a}^{8}}+16{{b}^{8}}-24{{a}^{4}}{{b}^{4}}+48{{a}^{4}}{{b}^{4}}

        \displaystyle =9{{a}^{8}}+16{{b}^{8}}+24{{a}^{4}}{{b}^{4}}

        \displaystyle ={{(3{{a}^{4}})}^{2}}+{{(4{{b}^{4}})}^{2}}+2\times 3{{a}^{4}}\times 4{{b}^{4}}

        \displaystyle ={{(3{{a}^{4}}+4{{b}^{4}})}^{2}}\ge 0    [\displaystyle \because square of any real number is non-negative]

    Hence, the given equation has real roots given by

        \displaystyle x=\frac{4{{b}^{4}}-3{{a}^{4}}\pm \sqrt{{{(3{{a}^{4}}+4{{b}^{4}})}^{2}}}}{2{{a}^{2}}{{b}^{2}}}        \displaystyle \left[ \because \,\,x=\frac{-b\pm \sqrt{D}}{2a} \right]

    The roots are

        \displaystyle \frac{4{{b}^{4}}-3{{a}^{4}}+3{{a}^{4}}+4{{b}^{4}}}{2{{a}^{2}}{{b}^{2}}} and     \displaystyle \frac{4{{b}^{4}}-3{{a}^{4}}-3{{a}^{4}}-4{{b}^{4}}}{2{{a}^{2}}{{b}^{2}}}

    or    \displaystyle \frac{8{{b}^{4}}}{2{{a}^{2}}{{b}^{2}}}         and     \displaystyle \frac{-6{{a}^{4}}}{2{{a}^{2}}{{b}^{2}}}

    or    \displaystyle \frac{4{{b}^{2}}}{{{a}^{2}}}             and \displaystyle -\frac{3{{a}^{2}}}{{{b}^{2}}}

        Hence, \displaystyle x=\frac{4{{b}^{2}}}{{{a}^{2}}}, \displaystyle -\frac{3{{a}^{2}}}{{{b}^{2}}} are the required solutions.

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