Question

answer this fast please

Answer 1

hope it helps. the image solves the problem

Answer 2

Given Equation is 4x^2 - 4a^2x + (a^4 - b^4) = 0.

On comparing with ax^2 + bx + c = 0,

a = 4, b = -4a^2, c = a^4 - b^4.

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(1)

$= > x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{-(-4a^2) + \sqrt{(-4a^2)^2 - 4(4)(a^4 - b^4)}}{8}$

$= > \frac{-(-4a^2) + \sqrt{16a^4 - 16a^4 + 16b^4}}{8}$

$= > \frac{4a^2 + \sqrt{16b^4}}{8}$

$= > \frac{4a^2 + 4b^2}{8}$

$= > \frac{a^2 + b^2}{2}$

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(2)

$= > x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{-(-4a^2) - \sqrt{(4a^2)^2 - 4 * 4(a^4 - b^4)}}{8}$

$= > \frac{4a^2 - \sqrt{16a^4 - 16(a^4 - b^4)}}{8}$

$= > \frac{4a^2 - \sqrt{16a^4 - 16a^4 + 16b^4}}{8}$

$= > \frac{4a^2 - 4b^2}{8}$

$= > \frac{a^2 - b^2}{2}$

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Therefore:

$= > \boxed{x = \frac{a^2 + b^2}{2}, \frac{a^2 - b^2}{2}}$

Hope this helps!