The Quadratic Formula, Part 2

In the previous section, we limited ourselves to only using real numbers for our solutions. This caused us to run into a challenge when the number under the square root in the Quadratic Formula became negative since we couldn't simplify that to a real number. If we expand our range a little and allow our answers to be complex numbers rather than just real numbers then every quadratic equation will have a solution.

Working with imaginary numbers is a key skill to working this kind of question. Remember that the basic rule is that sqrt(-1) = i. This means, for example, that we could simplify something like sqrt(-51) this way:

$$x = \frac {-b \pm \sqrt {b^2 - 4ac}} {2a}$$

$$\sqrt {-51} = \sqrt {-1 \cdot 51} = \sqrt {-1} \cdot \sqrt {51} = i \sqrt {51}$$

The important thing to remember is that when you see a negative under a square root, you can "pull the negative sign" outside of the square root as i.

With the exception of handling the conversion of the negative square root to the imaginary number i, this type of question proceeds exactly the same way as the ones in the previous section.

Example 1

Solve the equation x² + 2x + 5 = 0.

This is Example 4 from the previous section where we had to stop because the part of the formula under the square root became negative. I'll repeat the process here and then keep going with our new "complex number" approach.

Since the equation is already set equal to zero we can start by picking out the values we'll need for the formula.

a = 1b = 2c = 5

If we substitute those values into the formula, we get:

$$x = \frac {-2 \pm \sqrt {2^2 - 4 \cdot 1 \cdot 5}} {2 \cdot 1}$$ $$x = \frac {-2 \pm \sqrt {4 - 20}} {2}$$ $$x = \frac {-2 \pm \sqrt {-16}} {2}$$

This is where we ran into trouble before. If, however, we allow complex numbers then we can simplify $\sqrt {-16}$ to 4i and keep on going.

$$x = \frac {-2 \pm 4i} {2}$$ $$x = \frac {-2 + 4i} {2}, \frac {-2 - 4i} {2}$$ $$x = -1 + 2i, -1 - 2i$$

Example 2

Solve the equation 2x² - 5x + 10 = 0.

First, we need to pick out the a, b and c values that we'll substitute into the formula. Remember that the a value is the number in front of the x² term, the b value is the number in front of the x term and the c value is the constant at the end. This means that, for our equation, we have:

a = 2b = -5c = 10

If we substitute those values into the formula, we get:

$$x = \frac {-(-5) \pm \sqrt {(-5)^2 - 4 \cdot 2 \cdot 10}} {2 \cdot 2}$$ $$x = \frac {5 \pm \sqrt {25 - 80}} {4}$$ $$x = \frac {5 \pm \sqrt {-55}} {4}$$ $$x = \frac {5 \pm i\sqrt {55}} {4}$$

Since $\sqrt {55}$ can't be reduced or simplified any further. That's our final answer. It's a little messy with both a square root and a fraction left in it but these problems are often going to come out that way.

Video Lectures

Solving Equations with the Quadratic Formula, 2

In this lecture, we're goint to take the next step with the Quadratic Equations and see how it can be used in the case where the "discriminant" part is negative.

Dynamic Pratice - Using the Quadratic Formula