What are the solutions to the equation (y^2-y+1)^2=0 ?

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

Find the zeros of (y^2-y+1)^2=0

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

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

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

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

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

The solutions are

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

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

9 x^{4} + 82 x^{2} + 9 = 0

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

25 x^{4} - 26 x^{2} + 1 = 0

x^{3} - 3 x^{2} - 33 x + 35 = 0

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