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

Q) Solve the following quadratic equation by factorization method: $4x^2-2(a^2+b^2)x+a^2b^2=0$

Ans:

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

The Constant term $a^2b^2$ can also be presented as:

$a^2\times b^2$

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

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

Now, the original quadratic equation:

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

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

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

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

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

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

$~or~(2x-b^{2})=0$

$\implies x=\frac{a^{2}}{2}$ $~or~ x=\frac{b^{2}}{2}$

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