Solving Equations with Complex Solutions
You often come across equations that have no real solutions — or equations that have the potential for many more real solutions than they actually have. For instance, the equation x^{2} + 1 = 0 has no real solutions. If you write it as x^{2} = –1 and try to take the square root of each side, you run into trouble.
Not until you have the imaginary numbers can you write that the solution of this equation is x = +/–i. The equation has two complex solutions.
An example of an equation without enough real solutions is x^{4} – 81 = 0. This equation factors into (x^{2} – 9)(x^{2} + 9) = 0. The two real solutions of this equation are 3 and –3. The two complex solutions are 3i and –3i.
To solve for the complex solutions of an equation, you use factoring, the square root property for solving quadratics, and the quadratic formula.
Sample questions

Find all the roots, real and complex, of the equation x^{3} – 2x^{2} + 25x – 50 = 0.
x= 2, 5i, –5i. First, factor the equation to get x^{2}(x – 2) + 25(x – 2) = (x – 2)(x^{2} + 25) = 0. Using the multiplication property of zero, you determine that x – 2 = 0 and x = 2. You also get x^{2} + 25 = 0 and x^{2} = –25. Take the square root of each side, and
Simplify the radical, using the equivalence for i, and the complex solutions are
The real root is 2, and the imaginary roots are 5i and –5i.

Find all the roots, real and imaginary, of the equation 5x^{2} – 8x + 5 = 0.
x = 0.4 + 0.6i, 0.4 – 0.6i. The quadratic doesn’t factor, so you use the quadratic formula:
The only two solutions are complex: 0.4 + 0.6i and 0.4 – 0.6i.
Practice questions

Find all the roots, real and imaginary, of x^{2} + 9 = 0.

Find all the roots, real and imaginary, of x^{2} + 4x + 7 = 0.

Find all the roots, real and imaginary, of 5x^{2} + 6x + 3 = 0.

Find all the roots, real and imaginary, of x^{4} + 12x^{2} – 64 = 0.
Following are answers to the practice questions:

The answer is x = 3i, –3i.
Add –9 to each side to get x^{2} = –9. Take the square root of each side. Then simplify the expression using i for the negative under the radical:

The answer is
Use the quadratic formula to solve for x. Simplify the expression using i for the negative under the radical:

The answer is
Use the quadratic formula to solve for x. Simplify the expression using i for the negative under the radical:

The answer is x= 2, –2, 4i, –4i.
Factor the left side: (x^{2} + 16)(x^{2} – 4) = (x^{2} + 16)(x – 2)(x + 2) = 0. Obtain the two real roots by setting x – 2 and x + 2 equal to 0. When x^{2} + 16 = 0, you find that x^{2} = –16. Taking the square root of each side and using i for the –1 under the radical gives you the two imaginary roots.