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

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

Find the zeros of (y(1-y))-1=0

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

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(y (1 - y)) - 1 = 0

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

Solution steps

(y (1 - y)) - 1 = 0

Expand

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

y - y^{2} - 1 = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 (- 1) (- 1) : 3 i

y_{1, 2} = \frac{- 1 \pm 3 i}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 1 + 3 i}{2 ( - 1 )}, y_{2} = \frac{- 1 - 3 i}{2 ( - 1 )}

y = \frac{- 1 + 3 i}{2 ( - 1 )} : \frac{1}{2} - i \frac{3}{2}

y = \frac{- 1 - 3 i}{2 ( - 1 )} : \frac{1}{2} + i \frac{3}{2}

The solutions to the quadratic equation are:

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

∣ x + 1 ∣ \leq 5

x^{2} + 8 x + 32 = 0

∣ 3 x - 5 ∣ < 2

(\frac{1}{27})^{\frac{4}{3}}

x^{2} - 17 x + 72 = 0

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