What is the answer to solvefor x,x^2+(y-4/(x^2))2=1 ?

The answer to solvefor x,x^2+(y-4/(x^2))2=1 is

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solvefor x,x^2+(y-4/(x^2))2=1

solve for

x, x^{2} + (y - \frac{4}{x ^{2}}) 2 = 1

x = \frac{1 - 2 y + 4 y ^{2} - 4 y + 33}{2}, x = - \frac{1 - 2 y + 4 y ^{2} - 4 y + 33}{2}, x = \frac{1 - 2 y - 4 y ^{2} - 4 y + 33}{2}, x = - \frac{1 - 2 y - 4 y ^{2} - 4 y + 33}{2}

Solution steps

x^{2} + (y - \frac{4}{x ^{2}}) \cdot 2 = 1

Simplify

(y - \frac{4}{x ^{2}}) \cdot 2 : 2 (y - \frac{4}{x ^{2}})

x^{2} + 2 (y - \frac{4}{x ^{2}}) = 1

Expand

x^{2} + 2 (y - \frac{4}{x ^{2}}) : x^{2} + 2 y - \frac{8}{x ^{2}}

x^{2} + 2 y - \frac{8}{x ^{2}} = 1

Move

2 y

to the right side

x^{2} - \frac{8}{x ^{2}} = 1 - 2 y

Multiply both sides by

x^{2}

x^{4} - 8 = x^{2} - 2 y x^{2}

Solve

x^{4} - 8 = x^{2} - 2 y x^{2} : x = \frac{1 - 2 y + 4 y ^{2} - 4 y + 33}{2}, x = - \frac{1 - 2 y + 4 y ^{2} - 4 y + 33}{2}, x = \frac{1 - 2 y - 4 y ^{2} - 4 y + 33}{2}, x = - \frac{1 - 2 y - 4 y ^{2} - 4 y + 33}{2}

x = \frac{1 - 2 y + 4 y ^{2} - 4 y + 33}{2}, x = - \frac{1 - 2 y + 4 y ^{2} - 4 y + 33}{2}, x = \frac{1 - 2 y - 4 y ^{2} - 4 y + 33}{2}, x = - \frac{1 - 2 y - 4 y ^{2} - 4 y + 33}{2}

\frac{1}{( 1 + x ^{a - b} )} + \frac{1}{( 1 + x ^{b - a} )} = 1

\frac{x ^{2} + 1}{( x ^{2} ) - 5 ( x + \frac{1}{x} ) + 8} = 0

1 = \frac{1.5 h}{800 + h}

\frac{x ^{2} + 1}{x ^{2} - 5 ( x + \frac{1}{x} ) + 8} = 0

\frac{x + 1}{x} = \frac{21}{20}

What is the answer to solvefor x,x^2+(y-4/(x^2))2=1 ?
The answer to solvefor x,x^2+(y-4/(x^2))2=1 is