What are the solutions to the equation 0=2x^3-3x+1 ?

The solutions to the equation 0=2x^3-3x+1 are x=1,x=(-1+sqrt(3))/2 ,x=-(1+sqrt(3))/2

Find the zeros of 0=2x^3-3x+1

The zeros of 0=2x^3-3x+1 are x=1,x=(-1+sqrt(3))/2 ,x=-(1+sqrt(3))/2

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0 = 2 x^{3} - 3 x + 1

x = 1, x = \frac{- 1 + 3}{2}, x = - \frac{1 + 3}{2}

Solution steps

0 = 2 x^{3} - 3 x + 1

Switch sides

2 x^{3} - 3 x + 1 = 0

Factor

2 x^{3} - 3 x + 1 : (x - 1) (2 x^{2} + 2 x - 1)

(x - 1) (2 x^{2} + 2 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 2 x^{2} + 2 x - 1 = 0

Solve

x - 1 = 0 : x = 1

Solve

2 x^{2} + 2 x - 1 = 0 : x = \frac{- 1 + 3}{2}, x = - \frac{1 + 3}{2}

The solutions are

x = 1, x = \frac{- 1 + 3}{2}, x = - \frac{1 + 3}{2}

0 = (x - 1)^{3} - 2

x^{3} - 15 x = 0

- 3 x^{2} x = - 7

x (1 - x)^{2} = \frac{1}{10}

t^{4} - 5 t^{3} + 5 t^{2} + 17 t - 42 = 0

What are the solutions to the equation 0=2x^3-3x+1 ?
The solutions to the equation 0=2x^3-3x+1 are x=1,x=(-1+sqrt(3))/2 ,x=-(1+sqrt(3))/2
Find the zeros of 0=2x^3-3x+1
The zeros of 0=2x^3-3x+1 are x=1,x=(-1+sqrt(3))/2 ,x=-(1+sqrt(3))/2