Question

solve for x 4x2 -4a2 +(a4 -b4 ) = 0

Answer 1

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Answer 2

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

/*Splitting middle term , we get */

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

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

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

$\implies [2x - (a^{2} - b^{2})] = 0 \:Or \: [2x-(a^{2} + b^{2})] = 0$

$\implies 2x = (a^{2} -b^{2}) \:Or \: 2x = (a^{2} + b^{2})$

$\implies x = \frac{(a^{2} -b^{2})}{2} \:Or \: x = \frac{(a^{2} + b^{2})}{2}$

Therefore.,

$\green { x = \frac{(a^{2} -b^{2})}{2} \:Or \: x = \frac{(a^{2} + b^{2})}{2} }\\\green{ are \: roots \: of \: given \: Quadratic \: Equation }$

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