Solve for x: 4x² - 4a²x + (a⁴ - b⁴) = 0

Solution:

Given: Equation is 4x² - 4a²x + (a⁴ - b⁴) = 0

For finding the values of x use the quadratic formula

x = [-b ± √(b² - 4ac)]/2a

Here compare the given equation by ax² + bx + c = 0 to get the values of a, b, c

So we get a = 4,b = -4a² and c = a⁴ - b⁴

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

⇒ x = [-(-4a²) ± √((-4a²)² - 4(4)(a⁴ - b⁴))] /2.(4)

⇒ x = [4a² ± √(16a⁴ - 16a⁴ +16b⁴)]/8

⇒ x = [4a² + √16b⁴]/8 and [4a² - √16b⁴]/8

⇒ x = [4a² + 4b²]/8 and [4a² - 4b²]/8

⇒ x = 4[a² + b²]/8 and 4[a² - b²]/8

⇒ x = [a² + b²]/2 and [a² - b²]/2

Solve for x: 4x² - 4a²x + (a⁴ - b⁴) = 0

Summary :

The value of x for 4x² - 4a²x + (a⁴ - b⁴) = 0 are [a² + b²]/2 and [a² - b²]/2 for the equation 4x² - 4a²x + (a⁴ - b⁴) = 0