Algebra Know-It-ALL

When we add j[(c/a)1/2] to each side, we get

x+j[(c/a)1/2]= 0

The binomial factor form of the original quadratic is therefore

{x−j[(c/a)1/2]}{x+j[(c/a)1/2]}= 0

To be sure this is the correct equation in binomial factor form, (that is, to be sure we haven’t
made any mistakes!), we had better multiply out the left side. To avoid confusion with the
signs, let’s rewrite the first term as a sum:

{x+ (−j)[(c/a)1/2]}{x+j[(c/a)1/2]}= 0

Using the product of sums rule, we can expand to get

x^2 +x j [(c/a)1/2]+ (−j)[(c/a)1/2]x+ {−j[(c/a)1/2]}{j[(c/a)1/2]}= 0

Taking advantage of the commutative law for multiplication in the third and fourth terms,
we can rewrite it as

x^2 +x j [(c/a)1/2]+ {−x j [(c/a)1/2]}+ (−j×j)(c/a)= 0

The second and third terms are additive inverses and −j×j= 1, so we can simplify this to

x^2 +c/a= 0

Multiplying through by a, we get

ax^2 +c= 0

which is the original quadratic.

Are you confused?

We’ve seen what happens in a quadratic when the coefficient of x^2 and the stand-alone constant are posi-
tive while the coefficient of x is equal to 0. You might ask, “What happens if they are both negative?” The
answer is simple. If a< 0 and c< 0, then we can multiply the equation through by −1, and we’ll have the
case where a> 0 and c> 0. For example, if we see

− 2 x^2 − 79 = 0

we can multiply each side by −1, getting

(−1)× (− 2 x^2 − 79) =− 1 × 0

Imaginary Roots in Factors 389