Algebra Know-It-ALL

386 Quadratic Equations with Complex Roots

Here’s a challenge!

Prove the second of the above statements.

Solution

Consider once again the general polynomial equation

ax^2 +bx+c= 0

The discriminant is

d=b^2 − 4 ac

Suppose that d< 0 and b≠ 0. We can use the “abbreviated discriminant” version of the quadratic formula

for complex roots:

x= [−b±j(|d|1/2)] / (2a)

Applying the right-hand distributive rule for addition and subtraction “in reverse” to the right side, we can

split that expression into a sum or difference of two ratios with a common denominator, like this:

x=−b/(2a)±j(|d|1/2)/(2a)

Therefore, we can write the two roots as

x=−b/(2a)+j(|d|1/2)/(2a) or x=−b/(2a)−j(|d|1/2)/(2a)

The fact that these are complex conjugates is obscure because the expressions are messy. To bring things

into focus, we can change a couple of names. Let

p=−b/(2a)

and

q= (|d|1/2)/(2a)

Now we can rewrite the roots as

x=p+jq or x=p−jq

These are complex conjugates (but not pure imaginary numbers) for all nonzero real numbers p and q.

You might ask, “Are p and q really both nonzero?” The answer is “Yes.” We can be sure that p≠ 0 because

b≠ 0 and a≠ 0, so −b/(2a) can’t be 0. We can be sure that q≠ 0 because d≠ 0 and a≠ 0, so (|d|1/2)/(2a)

can’t be 0.