What are the solutions to the equation 8x^3+2x-2=0 ?

The solutions to the equation 8x^3+2x-2=0 are x= 1/2 ,x=-1/4+i(sqrt(7))/4 ,x=-1/4-i(sqrt(7))/4

Find the zeros of 8x^3+2x-2=0

The zeros of 8x^3+2x-2=0 are x= 1/2 ,x=-1/4+i(sqrt(7))/4 ,x=-1/4-i(sqrt(7))/4

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

x = \frac{1}{2}, x = - \frac{1}{4} + i \frac{7}{4}, x = - \frac{1}{4} - i \frac{7}{4}

Solution steps

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x - 1 = 0 : x = \frac{1}{2}

Solve

2 x^{2} + x + 1 = 0 : x = - \frac{1}{4} + i \frac{7}{4}, x = - \frac{1}{4} - i \frac{7}{4}

The solutions are

x = \frac{1}{2}, x = - \frac{1}{4} + i \frac{7}{4}, x = - \frac{1}{4} - i \frac{7}{4}

x^{2} - 4 = - x^{3} + 6

0.11 = (1 + \frac{r}{12})^{12} - 1

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

x (x^{2} - 9) = 0

x^{3} (22 - (\frac{x}{2})) = 18

What are the solutions to the equation 8x^3+2x-2=0 ?
The solutions to the equation 8x^3+2x-2=0 are x= 1/2 ,x=-1/4+i(sqrt(7))/4 ,x=-1/4-i(sqrt(7))/4
Find the zeros of 8x^3+2x-2=0
The zeros of 8x^3+2x-2=0 are x= 1/2 ,x=-1/4+i(sqrt(7))/4 ,x=-1/4-i(sqrt(7))/4