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

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

Find the zeros of y^3+y=2

The zeros of y^3+y=2 are y=1,y=-1/2+i(sqrt(7))/2 ,y=-1/2-i(sqrt(7))/2

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y^{3} + y = 2

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

Solution steps

y^{3} + y = 2

Move

2

to the left side

y^{3} + y - 2 = 0

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

y - 1 = 0 : y = 1

Solve

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

The solutions are

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

x, (3 x y - x)^{3} = 6 y^{2}

x^{3} - x^{2} = 400

7 x^{6} + 35 x^{4} + 4 = 0

- 2 x^{4} + 3.8 x^{3} - 1 = 0

25 x^{4} - 34 x^{2} + 9 = 0

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