What are the solutions to the equation z^2=4+4r^2 ?

The solutions to the equation z^2=4+4r^2 are z=2sqrt(1+r^2),z=-2sqrt(1+r^2)

Find the zeros of z^2=4+4r^2

The zeros of z^2=4+4r^2 are z=2sqrt(1+r^2),z=-2sqrt(1+r^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

z^{2} = 4 + 4 r^{2}

z = 2 1 + r^{2}, z = - 2 1 + r^{2}

Solution steps

z^{2} = 4 + 4 r^{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

z = 4 + 4 r^{2}, z = - 4 + 4 r^{2}

Simplify

4 + 4 r^{2} : 2 1 + r^{2}

Simplify

- 4 + 4 r^{2} : - 2 1 + r^{2}

z = 2 1 + r^{2}, z = - 2 1 + r^{2}

(1 - 2 x) (1 - 3 x) = (5 x - 1) x - 2 (2 x - 5)

2 x^{2} + 16 x - 96 = 0

- 10 t^{2} - 12 t = - 9982

3 x^{2} + 12 x - 4

3 x^{2} + 12 x = 63

What are the solutions to the equation z^2=4+4r^2 ?
The solutions to the equation z^2=4+4r^2 are z=2sqrt(1+r^2),z=-2sqrt(1+r^2)
Find the zeros of z^2=4+4r^2
The zeros of z^2=4+4r^2 are z=2sqrt(1+r^2),z=-2sqrt(1+r^2)