32x^4-2y^4=0

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Solution for 32x^4-2y^4=0 equation: 


Simplifying
32x4 + -2y4 = 0

Solving
32x4 + -2y4 = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '2y4' to each side of the equation.
32x4 + -2y4 + 2y4 = 0 + 2y4

Combine like terms: -2y4 + 2y4 = 0
32x4 + 0 = 0 + 2y4
32x4 = 0 + 2y4
Remove the zero:
32x4 = 2y4

Divide each side by '32'.
x4 = 0.0625y4

Simplifying
x4 = 0.0625y4

Combine like terms: 0.0625y4 + -0.0625y4 = 0.0000
x4 + -0.0625y4 = 0.0000

Factor a difference between two squares.
(x2 + 0.25y2)(x2 + -0.25y2) = 0.0000

Factor a difference between two squares.
(x2 + 0.25y2)((x + 0.5y)(x + -0.5y)) = 0.0000


Subproblem 1
Set the factor '(x2 + 0.25y2)' equal to zero and attempt to solve:

Simplifying
x2 + 0.25y2 = 0

Solving
x2 + 0.25y2 = 0

Move all terms containing x to the left, all other terms to the right.

Add '-0.25y2' to each side of the equation.
x2 + 0.25y2 + -0.25y2 = 0 + -0.25y2

Combine like terms: 0.25y2 + -0.25y2 = 0.00
x2 + 0.00 = 0 + -0.25y2
x2 = 0 + -0.25y2
Remove the zero:
x2 = -0.25y2

Simplifying
x2 = -0.25y2

The solution to this equation could not be determined.

This subproblem is being ignored because a solution could not be determined.

Subproblem 2
Set the factor '(x + 0.5y)' equal to zero and attempt to solve:

Simplifying
x + 0.5y = 0

Solving
x + 0.5y = 0

Move all terms containing x to the left, all other terms to the right.

Add '-0.5y' to each side of the equation.
x + 0.5y + -0.5y = 0 + -0.5y

Combine like terms: 0.5y + -0.5y = 0.0
x + 0.0 = 0 + -0.5y
x = 0 + -0.5y
Remove the zero:
x = -0.5y

Simplifying
x = -0.5y

Subproblem 3
Set the factor '(x + -0.5y)' equal to zero and attempt to solve:

Simplifying
x + -0.5y = 0

Solving
x + -0.5y = 0

Move all terms containing x to the left, all other terms to the right.

Add '0.5y' to each side of the equation.
x + -0.5y + 0.5y = 0 + 0.5y

Combine like terms: -0.5y + 0.5y = 0.0
x + 0.0 = 0 + 0.5y
x = 0 + 0.5y
Remove the zero:
x = 0.5y

Simplifying
x = 0.5y
Solution
x = {-0.5y, 0.5y}

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