=(x+2i)(x+2i)

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Solution for =(x+2i)(x+2i) equation:


Simplifying
0 = (x + 2i)(x + 2i)

Reorder the terms:
0 = (2i + x)(x + 2i)

Reorder the terms:
0 = (2i + x)(2i + x)

Multiply (2i + x) * (2i + x)
0 = (2i * (2i + x) + x(2i + x))
0 = ((2i * 2i + x * 2i) + x(2i + x))

Reorder the terms:
0 = ((2ix + 4i2) + x(2i + x))
0 = ((2ix + 4i2) + x(2i + x))
0 = (2ix + 4i2 + (2i * x + x * x))
0 = (2ix + 4i2 + (2ix + x2))

Reorder the terms:
0 = (2ix + 2ix + 4i2 + x2)

Combine like terms: 2ix + 2ix = 4ix
0 = (4ix + 4i2 + x2)

Solving
0 = 4ix + 4i2 + x2

Solving for variable 'i'.
Remove the zero:
-4ix + -4i2 + -1x2 = 4ix + 4i2 + x2 + -4ix + -4i2 + -1x2

Reorder the terms:
-4ix + -4i2 + -1x2 = 4ix + -4ix + 4i2 + -4i2 + x2 + -1x2

Combine like terms: 4ix + -4ix = 0
-4ix + -4i2 + -1x2 = 0 + 4i2 + -4i2 + x2 + -1x2
-4ix + -4i2 + -1x2 = 4i2 + -4i2 + x2 + -1x2

Combine like terms: 4i2 + -4i2 = 0
-4ix + -4i2 + -1x2 = 0 + x2 + -1x2
-4ix + -4i2 + -1x2 = x2 + -1x2

Combine like terms: x2 + -1x2 = 0
-4ix + -4i2 + -1x2 = 0

Factor out the Greatest Common Factor (GCF), '-1'.
-1(4ix + 4i2 + x2) = 0

Factor a trinomial.
-1((2i + x)(2i + x)) = 0

Ignore the factor -1.

Subproblem 1

Set the factor '(2i + x)' equal to zero and attempt to solve: Simplifying 2i + x = 0 Solving 2i + x = 0 Move all terms containing i to the left, all other terms to the right. Add '-1x' to each side of the equation. 2i + x + -1x = 0 + -1x Combine like terms: x + -1x = 0 2i + 0 = 0 + -1x 2i = 0 + -1x Remove the zero: 2i = -1x Divide each side by '2'. i = -0.5x Simplifying i = -0.5x

Subproblem 2

Set the factor '(2i + x)' equal to zero and attempt to solve: Simplifying 2i + x = 0 Solving 2i + x = 0 Move all terms containing i to the left, all other terms to the right. Add '-1x' to each side of the equation. 2i + x + -1x = 0 + -1x Combine like terms: x + -1x = 0 2i + 0 = 0 + -1x 2i = 0 + -1x Remove the zero: 2i = -1x Divide each side by '2'. i = -0.5x Simplifying i = -0.5x

Solution

i = {-0.5x, -0.5x}

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