What are the solutions to the equation solvefor y,y^2=x^3-3x+1 ?

The solutions to the equation solvefor y,y^2=x^3-3x+1 are y=sqrt(x^3-3x+1),y=-sqrt(x^3-3x+1)

Find the zeros of solvefor y,y^2=x^3-3x+1

The zeros of solvefor y,y^2=x^3-3x+1 are y=sqrt(x^3-3x+1),y=-sqrt(x^3-3x+1)

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solve for

y, y^{2} = x^{3} - 3 x + 1

y = x^{3} - 3 x + 1, y = - x^{3} - 3 x + 1

Solution steps

y^{2} = x^{3} - 3 x + 1

For

x^{2} = f (a)

the solutions are

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

y = x^{3} - 3 x + 1, y = - x^{3} - 3 x + 1

- 6 x^{2} + 96 x = 0

- 6 n^{2} + 15 n = 0

s^{2} + 2 s + 1

8 x^{2} - 4 x + 19 = 0

10 x^{2} - 7 x - 3 = 0

What are the solutions to the equation solvefor y,y^2=x^3-3x+1 ?
The solutions to the equation solvefor y,y^2=x^3-3x+1 are y=sqrt(x^3-3x+1),y=-sqrt(x^3-3x+1)
Find the zeros of solvefor y,y^2=x^3-3x+1
The zeros of solvefor y,y^2=x^3-3x+1 are y=sqrt(x^3-3x+1),y=-sqrt(x^3-3x+1)