# 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:

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

### 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:

(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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