Algebra Know-It-ALL

392 Quadratic Equations with Complex Roots

The roots are therefore

x= [−b±j(|d|1/2)] / (2a)

= (− 6 ±j8) / (2 × 5)

= (− 6 ±j8) / 10

=−6/10±j(8/10)

=−3/5±j(4/5)

Stated individually,

x=−3/5+j(4/5) or x=−3/5−j(4/5)

The binomial factor form of the original quadratic must therefore look like this:

{x− [−3/5+j(4/5)]}{x− [−3/5−j(4/5)]}= 0

We should multiply this out to be sure we have found the correct binomial factors. First, let’s

convert the subtractions into negative additions, obtaining

{x+ (−1)[−3/5+j(4/5)]}{x+ (−1)[−3/5+ (−j)(4/5)]}= 0

The distributive law for multiplication over addition lets us rewrite this as

{x+ 3/5 + [−j(4/5)]}[x+ 3/5 +j(4/5)]= 0

We now have two trinomial factors on the left side! We can multiply them out to get the equation

x^2 + (3/5)x+j(4/5)x

+ (3/5)x+ (3/5)^2 +j(3/5)(4/5)

+ (−j)(4/5)x+ (−j)(4/5)(3/5)+ (−j)(j)(4/5)^2

= 0

Fortunately, we have some pairs of terms that cancel out, so we can simplify to

x^2 + (6/5)x+ 9/25 + 16/25 = 0

and further to

x^2 + (6/5)x+ 1 = 0

Multiplying both sides of this equation by 5 gives us the original quadratic in polynomial

standard form:

5 x^2 + 6 x+ 5 = 0