16x^4-y^12=0

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Solution for 16x^4-y^12=0 equation: 


Simplifying
16x4 + -1y12 = 0

Solving
16x4 + -1y12 = 0

Solving for variable 'x'.

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

Add 'y12' to each side of the equation.
16x4 + -1y12 + y12 = 0 + y12

Combine like terms: -1y12 + y12 = 0
16x4 + 0 = 0 + y12
16x4 = 0 + y12
Remove the zero:
16x4 = y12

Divide each side by '16'.
x4 = 0.0625y12

Simplifying
x4 = 0.0625y12

Combine like terms: 0.0625y12 + -0.0625y12 = 0.0000
x4 + -0.0625y12 = 0.0000

Factor a difference between two squares.
(x2 + 0.25y6)(x2 + -0.25y6) = 0.0000

Factor a difference between two squares.
(x2 + 0.25y6)((x + 0.5y3)(x + -0.5y3)) = 0.0000


Subproblem 1
Set the factor '(x2 + 0.25y6)' equal to zero and attempt to solve:

Simplifying
x2 + 0.25y6 = 0

Solving
x2 + 0.25y6 = 0

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

Add '-0.25y6' to each side of the equation.
x2 + 0.25y6 + -0.25y6 = 0 + -0.25y6

Combine like terms: 0.25y6 + -0.25y6 = 0.00
x2 + 0.00 = 0 + -0.25y6
x2 = 0 + -0.25y6
Remove the zero:
x2 = -0.25y6

Simplifying
x2 = -0.25y6

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 + 0.5y3)' equal to zero and attempt to solve:

Simplifying
x + 0.5y3 = 0

Solving
x + 0.5y3 = 0

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

Add '-0.5y3' to each side of the equation.
x + 0.5y3 + -0.5y3 = 0 + -0.5y3

Combine like terms: 0.5y3 + -0.5y3 = 0.0
x + 0.0 = 0 + -0.5y3
x = 0 + -0.5y3
Remove the zero:
x = -0.5y3

Simplifying
x = -0.5y3

Subproblem 3
Set the factor '(x + -0.5y3)' equal to zero and attempt to solve:

Simplifying
x + -0.5y3 = 0

Solving
x + -0.5y3 = 0

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

Add '0.5y3' to each side of the equation.
x + -0.5y3 + 0.5y3 = 0 + 0.5y3

Combine like terms: -0.5y3 + 0.5y3 = 0.0
x + 0.0 = 0 + 0.5y3
x = 0 + 0.5y3
Remove the zero:
x = 0.5y3

Simplifying
x = 0.5y3
Solution
x = {-0.5y3, 0.5y3}

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