Not sure where to start? These are the lessons and games we recommend most to students, parents and teachers on Basic-mathematics.com. Each lesson explains the topic in plain language, walks you through worked examples step by step, and ends with practice so you can check your understanding. Pick one below, or use the menu to find lessons on fractions, algebra, geometry, statistics and more.
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