What are the solutions to the equation solvefor y,y^2=x^3-4x ?

The solutions to the equation solvefor y,y^2=x^3-4x are y=sqrt(x^3-4x),y=-sqrt(x^3-4x)

Find the zeros of solvefor y,y^2=x^3-4x

The zeros of solvefor y,y^2=x^3-4x are y=sqrt(x^3-4x),y=-sqrt(x^3-4x)

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solve for

y, y^{2} = x^{3} - 4 x

y = x^{3} - 4 x, y = - x^{3} - 4 x

Solution steps

y^{2} = x^{3} - 4 x

For

x^{2} = f (a)

the solutions are

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

y = x^{3} - 4 x, y = - x^{3} - 4 x

f = \frac{( 62 - 2 )}{3}

12 = 32 t - 16 t^{2}

- 3 x^{2} + 18 x - 24 = 0

130 = - 16 t^{2} + 70 t + 5

\frac{1 + i}{3 + 2 i}

What are the solutions to the equation solvefor y,y^2=x^3-4x ?
The solutions to the equation solvefor y,y^2=x^3-4x are y=sqrt(x^3-4x),y=-sqrt(x^3-4x)
Find the zeros of solvefor y,y^2=x^3-4x
The zeros of solvefor y,y^2=x^3-4x are y=sqrt(x^3-4x),y=-sqrt(x^3-4x)