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

The solutions to the equation 2=y^3+2y-1 are y=1,y=-1/2+i(sqrt(11))/2 ,y=-1/2-i(sqrt(11))/2

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

The zeros of 2=y^3+2y-1 are y=1,y=-1/2+i(sqrt(11))/2 ,y=-1/2-i(sqrt(11))/2

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

y = 1, y = - \frac{1}{2} + i \frac{11}{2}, y = - \frac{1}{2} - i \frac{11}{2}

Solution steps

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

Switch sides

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

Move

2

to the left side

y^{3} + 2 y - 3 = 0

Factor

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

(y - 1) (y^{2} + y + 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

y - 1 = 0 or y^{2} + y + 3 = 0

Solve

y - 1 = 0 : y = 1

Solve

y^{2} + y + 3 = 0 : y = - \frac{1}{2} + i \frac{11}{2}, y = - \frac{1}{2} - i \frac{11}{2}

The solutions are

y = 1, y = - \frac{1}{2} + i \frac{11}{2}, y = - \frac{1}{2} - i \frac{11}{2}

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

x^{3} + 2 x^{2} - 200 x - 20 = 0

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

\frac{7 x ^{6} - 14 + 11 + 8}{2} = 21

20 x^{3} - 6 x = 0

What are the solutions to the equation 2=y^3+2y-1 ?
The solutions to the equation 2=y^3+2y-1 are y=1,y=-1/2+i(sqrt(11))/2 ,y=-1/2-i(sqrt(11))/2
Find the zeros of 2=y^3+2y-1
The zeros of 2=y^3+2y-1 are y=1,y=-1/2+i(sqrt(11))/2 ,y=-1/2-i(sqrt(11))/2