Question

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

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

Answer 1

Concept: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

We have,

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

Here, Constant term $=a^4-b^4=(a^2-b^2)(a^2+b^2)$

and, Coefficient of middle term $=-4a^2$

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

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

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

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

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

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

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

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