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(2y + {6y}^(4) ) - (7y + 8y ^(2) - {4y}^(4)


1 Answer

3 votes

Answer: y = 0 or y = 1/6 ((3 i) sqrt(263343) - 43877)^(-1/3) (((6 i) sqrt(263343) - 87754)^(2/3) + 1568 2^(1/3)) - 28/3 or y = -28/3 + 784/3 (-1)^(2/3) (1/2 ((3 i) sqrt(263343) - 43877))^(-1/3) - 1/3 (-1/2)^(1/3) ((3 i) sqrt(263343) - 43877)^(1/3) or y = 1/6 ((3 i) sqrt(263343) - 43877)^(-1/3) ((-2)^(2/3) ((3 i) sqrt(263343) - 43877)^(2/3) - 1568 (-2)^(1/3)) - 28/3

Solve for y over the real numbers:

-2 y (y^3 + 28 y^2 - 1) = 0

Divide both sides by -2:

y (y^3 + 28 y^2 - 1) = 0

Split into two equations:

y = 0 or y^3 + 28 y^2 - 1 = 0

Eliminate the quadratic term by substituting x = y + 28/3:

y = 0 or -1 + 28 (x - 28/3)^2 + (x - 28/3)^3 = 0

Expand out terms of the left hand side:

y = 0 or x^3 - (784 x)/3 + 43877/27 = 0

Change coordinates by substituting x = z + λ/z, where λ is a constant value that will be determined later:

y = 0 or 43877/27 - 784/3 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

y = 0 or z^6 + z^4 (3 λ - 784/3) + (43877 z^3)/27 + z^2 (3 λ^2 - (784 λ)/3) + λ^3 = 0

Substitute λ = 784/9 and then u = z^3, yielding a quadratic equation in the variable u:

y = 0 or u^2 + (43877 u)/27 + 481890304/729 = 0

Find the positive solution to the quadratic equation:

y = 0 or u = 1/54 i (43877 i + 3 sqrt(263343))

Substitute back for u = z^3:

y = 0 or z^3 = 1/54 i (43877 i + 3 sqrt(263343))

Taking cube roots gives (i (43877 i + 3 sqrt(263343)))^(1/3)/(3 2^(1/3)) times the third roots of unity:

y = 0 or z = (i (43877 i + 3 sqrt(263343)))^(1/3)/(3 2^(1/3)) or z = -1/3 (-1/2)^(1/3) (i (43877 i + 3 sqrt(263343)))^(1/3) or z = ((-1)^(2/3) (i (43877 i + 3 sqrt(263343)))^(1/3))/(3 2^(1/3))

Substitute each value of z into x = z + 784/(9 z):

y = 0 or x = 784/(3 (1/2 i (3 sqrt(263343) + 43877 i))^(1/3)) + 1/3 (i/2 (43877 i + 3 sqrt(263343)))^(1/3) or x = (784 (-1)^(2/3))/(3 (1/2 i (3 sqrt(263343) + 43877 i))^(1/3)) - 1/3 ((-1)/2)^(1/3) (i (43877 i + 3 sqrt(263343)))^(1/3) or x = 1/3 (-1)^(2/3) (i/2 (43877 i + 3 sqrt(263343)))^(1/3) - (784 (-2)^(1/3))/(3 (i (3 sqrt(263343) + 43877 i))^(1/3))

Bring each solution to a common denominator and simplify:

y = 0 or x = ((6 i sqrt(263343) - 87754)^(2/3) + 1568 2^(1/3))/(6 (3 i sqrt(263343) - 43877)^(1/3)) or x = (784 (-1)^(2/3))/(3 (1/2 (3 i sqrt(263343) - 43877))^(1/3)) - 1/3 ((-1)/2)^(1/3) (3 i sqrt(263343) - 43877)^(1/3) or x = ((-2)^(2/3) (3 i sqrt(263343) - 43877)^(2/3) - 1568 (-2)^(1/3))/(6 (3 i sqrt(263343) - 43877)^(1/3))

Substitute back for y = x - 28/3:

Answer: y = 0 or y = 1/6 ((3 i) sqrt(263343) - 43877)^(-1/3) (((6 i) sqrt(263343) - 87754)^(2/3) + 1568 2^(1/3)) - 28/3 or y = -28/3 + 784/3 (-1)^(2/3) (1/2 ((3 i) sqrt(263343) - 43877))^(-1/3) - 1/3 (-1/2)^(1/3) ((3 i) sqrt(263343) - 43877)^(1/3) or y = 1/6 ((3 i) sqrt(263343) - 43877)^(-1/3) ((-2)^(2/3) ((3 i) sqrt(263343) - 43877)^(2/3) - 1568 (-2)^(1/3)) - 28/3


User Mike Tavish
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