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