On solving $4 x^{2} - 4 a^{2} x + (a^{4} - b^{4})$ = 0 we get value of $x$ is equal to

\frac{a^{2} \pm b^{2}}{2}

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\frac{a^{2}}{2}

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\frac{b^{2}}{2}

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\frac{a^{2} \div b^{2}}{2}

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Solution

The correct option is A $\frac{a^{2} \pm b^{2}}{2}$
On comparing the equation with $A x^{2} + B x + C = 0$ ,
We get, $A = 4, B = - 4 a^{2}, C = a^{4} - b^{4}$
D = $B^{2} - 4 A C$ = $(- 4 a^{2})^{2}$ $-$ 44 $(a^{4} - b^{4})$
= $16 a^{4} - 16 a^{4} + 16 b^{4}$ = $16 b^{4}$
$x = \frac{- B \pm \sqrt{D}}{2 A}$ = $\frac{- 4 (- 4 a^{2}) \pm \sqrt{16 b^{4}}}{2 \times 4}$
= $\frac{4 a^{2} \pm 4 b^{2}}{8}$ = $\frac{a^{2} + b^{2}}{2}$ or $\frac{a^{2} - b^{2}}{2}$

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