What are the solutions to the equation (z+3)^2+(z-3)^2=10 ?

The solutions to the equation (z+3)^2+(z-3)^2=10 are z=2i,z=-2i

Find the zeros of (z+3)^2+(z-3)^2=10

The zeros of (z+3)^2+(z-3)^2=10 are z=2i,z=-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(z + 3)^{2} + (z - 3)^{2} = 10

z = 2 i, z = - 2 i

Solution steps

(z + 3)^{2} + (z - 3)^{2} = 10

Expand

(z + 3)^{2} + (z - 3)^{2} : 2 z^{2} + 18

2 z^{2} + 18 = 10

Move

10

to the left side

2 z^{2} + 8 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

Simplify

0^{2} - 4 \cdot 2 \cdot 8 : 8 i

z_{1, 2} = \frac{- 0 \pm 8 i}{2 \cdot 2}

Separate the solutions

z_{1} = \frac{- 0 + 8 i}{2 \cdot 2}, z_{2} = \frac{- 0 - 8 i}{2 \cdot 2}

z = \frac{- 0 + 8 i}{2 \cdot 2} : 2 i

z = \frac{- 0 - 8 i}{2 \cdot 2} : - 2 i

The solutions to the quadratic equation are:

z = 2 i, z = - 2 i

x^{2} + x + 29 = - 9 x + 8

4 x - 7 = x^{2} - 2 x + 2

630 - 204 x + 12 x^{2} = 0

17500 x^{2} + 6140 x + 37000 = 0

6 p^{2} - 5 p - 2 = - 6 p^{2}

What are the solutions to the equation (z+3)^2+(z-3)^2=10 ?
The solutions to the equation (z+3)^2+(z-3)^2=10 are z=2i,z=-2i
Find the zeros of (z+3)^2+(z-3)^2=10
The zeros of (z+3)^2+(z-3)^2=10 are z=2i,z=-2i