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

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

Find the zeros of 2=y+y^3

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

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2 = y + y^{3}

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

Solution steps

2 = y + y^{3}

Switch sides

y + y^{3} = 2

Move

2

to the left side

y + y^{3} - 2 = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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} - 8 x^{2} + x + 42 = 0

(x - 4)^{4} = 81

- \frac{1}{4} x^{5} = 8

(2 x - 4)^{2} + 6 x - 3 = 4 x^{3} - (3 x - 1)

x^{3} - 12 x^{2} - 60 x + 400 = 0

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