factorization method: 4x^{2}-4a^{2}x+(a^{4}-b^{4})=0

Q) Solve the following quadratic equation by factorization method:

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

Ans:

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

The Constant term $a^{4}-b^{4}$ can also be presented as:

$(a^{2}-b^{2})(a^{2}+b^{2})$

Similarly, Coefficient of the middle term can be presented as:

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

Now, the original quadratic equation:

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

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

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

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

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

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

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

$\Rightarrow x=\frac{a^{2}+b^{2}}{2}$ and $x = \frac {a^2 - b^2}{2}$

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