Find complex solutions of a quadratic equation

Learn how to find complex solutions of a quadratic equation with these two examples.

Example #1:

Solve 8x2 + 200 = 0

Subtract 200 from each side of the equation

8x2 + 200 - 200 = 0 - 200

8x2 = -200

Divide each side of the equation by 8

(8x2)/8 = -200/8

x2 = -25

x = ±√(-25)

x = ±√[(25)×-1]

x = ±[√(25) × √(-1)]

x = ±(5 × i)

x = ±5i

x1 = 5i

x2 = -5i

Check for 5i

8(5i)2 + 200 = 0

8(25i2) + 200 = 0

8(-25) + 200 = 0

-200 + 200 = 0

0 = 0

Check for -5i

8(-5i)2 + 200 = 0

8(25i2) + 200 = 0

8(-25) + 200 = 0

-200 + 200 = 0

0 = 0

Example #2:

Solve x2 + 4x + 5 = 0

Because this time there is a linear term (4x), you must solve it either using the quadratic formula or by completing the square.

Let us solve by completing the square

Solve x2 + 4x + 5 = 0

Isolate the quadratic and the linear term.

x2 + 4x + 5 - 5 = 0 - 5

x2 + 4x  = -5

Complete the square

x2 + 4x + 22 = -5 + 22

(x + 2)2 = -5 + 4

(x + 2)2 = -1

x + 2 = ±i

Subtract 2 from each side of the equation

x + 2 - 2 = ±i - 2

x = ±i - 2

x1 = i - 2

x2 = -i - 2

Check for i - 2

(i - 2)2 + 4(i - 2) + 5 = 0

Factor out (i - 2)

(i - 2)[(i - 2) + 4] + 5 = 0

(i - 2)(i + 2) + 5 = 0

i2 - 22 + 5 = 0

-1 - 4 + 5 = 0

-5 + 5 = 0

0 = 0

Check for -i - 2

(-i - 2)2 + 4(-i - 2) + 5 = 0

Factor out (-i - 2)

(-i - 2)[(-i - 2) + 4] + 5 = 0

(-i - 2)(-i + 2) + 5 = 0

(-i)2 - 22 + 5 = 0

(i)2 - 22 + 5 = 0

-1 - 4 + 5 = 0

-5 + 5 = 0

0 = 0

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