Imaginary Roots in Factors
Now that we’ve found a way to solve quadratics that have real coefficients, a real constant, and
a negative real discriminant, let’s see what the factors of such equations look like. We’ll start
with situations where the coefficient of x is 0, so the roots are pure imaginary.
A specific case
Consider again the quadratic equation we solved earlier, in which the roots are pure imaginary
and additive inverses:
3 x^2 + 75 = 0
We simplified this equation in the second version of the solution when we divided both sides
by 3, obtaining
x^2 + 25 = 0
We found that the roots are j5 and −j5. Knowing these roots, it’s reasonable to think that we
should be able to figure out what this equation looks like in binomial factor form. In one case,
we have x=j5. If we take that equation and subtract j5 from each side, we get
x−j 5 = 0
In the other case, we have x=−j5. We can add j5 to each side of that, getting
x+j 5 = 0
This suggests that the binomial factor form of the quadratic is
(x−j5)(x+j5)= 0
The left side of this equation is 0 if and only if x=j5 or x=−j5. Let’s multiply it out and see
what we get. To avoid confusion with the signs, we can rewrite the first term as a sum:
[x+ (−j5)](x+j5)= 0
Applying the product of sums rule gives us
x^2 +x j 5 + (−j 5 x)+ (−j5)(j5)= 0
Using the commutative law for multiplication in the third and fourth terms, we obtain
x^2 +x j 5 + (−x j 5)+ (−j×j)× 25 = 0
Note that −j×j= 1. Also, the second and third terms of the polynomial add up to 0. We can
therefore simplify to get
x^2 + 25 = 0
Imaginary Roots in Factors 387
