Question

Solve the following quadratic equations by factorization method: [4 MARKS]

$4x^2-4ax+(a^2-b^2)=0$

Answer 1

Concept : 1 Mark
Application : 1 Mark
Calculation : 2 Marks

We have,

$4x^2-4ax+(a^2-b^2)=0$

Here, Constant term $=(a^2-b^2)=(a-b)(a+b)$

and, coefficient of middle term $=-4a$

Also, Coefficient of the middle term $-4a=-\left\{2\right(a+b)+2(a-b\left)\right\}$

$\therefore 4x^2-4ax+(a^2-b^2)=0$

$\Rightarrow 4x^2-\left\{2\right(a+b)+2(a-b\left)\right\}x+(a+b)(a-b)=0$

$\Rightarrow 4x^2-2(a+b)x-2(a-b)x+(a+b)(a-b)=0$

$\Rightarrow \{4x^2-2(a+b\left)x\right\}-\left\{2\right(a-b)x-(a+b\left)\right(a-b\left)\right\}=0$

$\Rightarrow 2x\{2x-(a+b\left)\right\}-(a-b)\{2x-(a+b\left)\right\}=0$

$\Rightarrow \{2x-(a+b\left)\right\}\{2x-(a-b\left)\right\}=0$

$\Rightarrow \{2x-(a+b\left)\right\}=0or,\{2x-(a-b\left)\right\}=0$

$\Rightarrow 2x=a+bor,2x=a-b\Rightarrow x=\frac{a+b}{2}or,x=\frac{a-b}{2}$