What are the solutions to the equation y=(y^2)^2 ?

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

Find the zeros of y=(y^2)^2

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

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y = (y^{2})^{2}

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

Solution steps

y = (y^{2})^{2}

Expand

(y^{2})^{2} : y^{4}

y = y^{4}

Switch sides

y^{4} = y

Move

y

to the left side

y^{4} - y = 0

Factor

y^{4} - y : y (y^{3} - 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

The solutions are

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

x (x + 2) (x + 4) = 4

n^{4} - 5 n^{2} = - 4

x^{3} + 7 x^{2} + 25 x + 175 = 0

x (x - 8) (x - 4)^{2} = 225

z^{5} = - 2

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