What are the solutions to the equation 0=-v^2-1+v ?

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

Find the zeros of 0=-v^2-1+v

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Switch sides

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 2 x^{2} + 64 = 0

2 x^{2} + x - 4 = 0

- 12 x^{2} - 12 x - 3 = 0

x, 3 x^{2} y = λ x

3 y^{2} = - 16

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