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

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

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

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

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

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

Solution steps

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

Expand

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

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

Move

y^{2}

to the left side

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

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

Simplify

2^{2} - 4 (- 2) (- 1) : 2 i

y_{1, 2} = \frac{- 2 \pm 2 i}{2 ( - 2 )}

Separate the solutions

y_{1} = \frac{- 2 + 2 i}{2 ( - 2 )}, y_{2} = \frac{- 2 - 2 i}{2 ( - 2 )}

y = \frac{- 2 + 2 i}{2 ( - 2 )} : \frac{1}{2} - i \frac{1}{2}

y = \frac{- 2 - 2 i}{2 ( - 2 )} : \frac{1}{2} + i \frac{1}{2}

The solutions to the quadratic equation are:

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

x^{2} - 5 = \frac{4 x ^{2}}{5}

3 x^{2} - 24 x - 36 = 0

- 3 y^{2} + 120 y - 900 = 0

2 x^{2} - 8 x - 2 = 3

2 x^{2} - 5 x + 2

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