# 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