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Solve the quadratic equation:

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

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

x = \frac{a^{2} \pm b^{2}}{3}

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

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

answer is A.

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Detailed Solution

We are given to solve

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

.
The standard form of the quadratic equation is represented as

{ax}^{2} + bx + c = 0,

where a \neq 0

Compare the quadratic equation

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

with the general equation

{ax}^{2} + bx + c = 0 .

So, a = 4, b = {- 4 a}^{2}, and c = (a^{4} - b^{4})

So,

x = − b ± b 2 − 4 a c 2 a ⇒ x = − − 4 a 2 ± − 4 a 2 2 − 44 a 4 − b 4 24 ⇒ x = 4 a 2 ± 16 a 4 − 16 a 4 + 16 b 4 8 ⇒ x = 4 a 2 ± 16 b 4 8

⇒ x = 4 a 2 ± 4 b 2 8

Solve further,

⇒ x = 4 a 2 ± b 2 4 × 2

x = a 2 ± b 2 2

So, the roots are

x = a 2 ± b 2 2

.
Therefore, option 1 is correct.

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