What are the solutions to the equation (x-2)^2+(x+4)^2=16 ?

The solutions to the equation (x-2)^2+(x+4)^2=16 are x=-1+i,x=-1-i

Find the zeros of (x-2)^2+(x+4)^2=16

The zeros of (x-2)^2+(x+4)^2=16 are x=-1+i,x=-1-i

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(x - 2)^{2} + (x + 4)^{2} = 16

x = - 1 + i, x = - 1 - i

Solution steps

(x - 2)^{2} + (x + 4)^{2} = 16

Expand

(x - 2)^{2} + (x + 4)^{2} : 2 x^{2} + 4 x + 20

2 x^{2} + 4 x + 20 = 16

Move

16

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

Simplify

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

x_{1, 2} = \frac{- 4 \pm 4 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 4 i}{2 \cdot 2}, x_{2} = \frac{- 4 - 4 i}{2 \cdot 2}

x = \frac{- 4 + 4 i}{2 \cdot 2} : - 1 + i

x = \frac{- 4 - 4 i}{2 \cdot 2} : - 1 - i

The solutions to the quadratic equation are:

x = - 1 + i, x = - 1 - i

q^{2} + 42 q = 0

(3 x - 1) (2 x + 3) = (x + 9) (x + 8)

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

(x - 3) (x + 3) = 4, x^{2} - 9 = 4 x^{2} - 36

3 - 4 x^{2} = 0

What are the solutions to the equation (x-2)^2+(x+4)^2=16 ?
The solutions to the equation (x-2)^2+(x+4)^2=16 are x=-1+i,x=-1-i
Find the zeros of (x-2)^2+(x+4)^2=16
The zeros of (x-2)^2+(x+4)^2=16 are x=-1+i,x=-1-i