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# Solve for x: 4x^{2} - 4a^{2}x + (a^{4} - b^{4}) = 0

**Solution: **

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

For finding the values of x use the quadratic formula

x = [-b ± √(b^{2 }- 4ac)]/2a

Here compare the given equation by ax^{2 }+ bx + c = 0 to get the values of a, b, c

So we get a = 4,b = -4a^{2} and c = a^{4 }- b^{4}

Put the values of a, b , c in the formula we get,

⇒ x = [-(-4a^{2}) ± √((-4a^{2})^{2 }- 4(4)(a^{4} - b^{4}))] /2.(4)

⇒ x = [4a^{2} ± √(16a^{4 }- 16a^{4 }+16b^{4})]/8

⇒ x = [4a^{2} + √16b^{4}]/8 and [4a^{2} - √16b^{4}]/8

⇒ x = [4a^{2} + 4b^{2}]/8 and [4a^{2} - 4b^{2}]/8

⇒ x = 4[a^{2} + b^{2}]/8 and 4[a^{2} - b^{2}]/8

⇒ x = [a^{2} + b^{2}]/2 and [a^{2} - b^{2}]/2

## Solve for x: 4x^{2} - 4a^{2}x + (a^{4 }- b^{4}) = 0

**Summary : **

The value of x for 4x^{2} - 4a^{2}x + (a^{4 }- b^{4}) = 0 are [a^{2} + b^{2}]/2 and [a^{2} - b^{2}]/2 for the equation 4x^{2} - 4a^{2}x + (a^{4} - b^{4}) = 0

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