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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 1 = (1 - y) (2 - y)

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

Solution steps

- 1 = (1 - y) (2 - y)

Expand

(1 - y) (2 - y) : 2 - 3 y + y^{2}

- 1 = 2 - 3 y + y^{2}

Switch sides

2 - 3 y + y^{2} = - 1

Move

1

to the left side

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

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 3}{2 \cdot 1}

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 2 x^{2} + 40 x - 72 = 0

x, z^{2} = x^{2} + y^{2} z

- 5 x^{2} + 15 x + 50 = 0

3 (x - 1) (x + 2) = 81

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

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