What are the solutions to the equation solvefor y,y^2=x^2-x^4 ?

The solutions to the equation solvefor y,y^2=x^2-x^4 are y=xsqrt(1-x^2),y=-xsqrt(1-x^2)

Find the zeros of solvefor y,y^2=x^2-x^4

The zeros of solvefor y,y^2=x^2-x^4 are y=xsqrt(1-x^2),y=-xsqrt(1-x^2)

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solve for

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

y = x 1 - x^{2}, y = - x 1 - x^{2}

Solution steps

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

Simplify

x^{2} - x^{4} : x 1 - x^{2}

Simplify

- x^{2} - x^{4} : - x 1 - x^{2}

y = x 1 - x^{2}, y = - x 1 - x^{2}

2 x^{5} - 2 x^{3} = 0

x^{3} + 5 x^{2} + 9 x + 5

x^{4} + 8 = - 9 x^{2}

x^{3} + 5 x^{2} + 7 x + 3 = 0

12 x^{4} - 32 x^{3} + 13 x^{2} + 8 x + 1 = 0

What are the solutions to the equation solvefor y,y^2=x^2-x^4 ?
The solutions to the equation solvefor y,y^2=x^2-x^4 are y=xsqrt(1-x^2),y=-xsqrt(1-x^2)
Find the zeros of solvefor y,y^2=x^2-x^4
The zeros of solvefor y,y^2=x^2-x^4 are y=xsqrt(1-x^2),y=-xsqrt(1-x^2)