Algebra Know-It-ALL

394 Quadratic Equations with Complex Roots

This makes the above factored equation look like this:

[x− (p+jq)][x− (p−jq)]= 0

To avoid making mistakes with signs, we can rewrite this as

[x+ (−1)(p+jq)][x+ (−1)(p−jq)]= 0

and then as

[x+ (−p)+ (−jq)][x+ (−p)+jq]= 0

Using the multiplication rule for trinomials along with the commutative law for multiplication in some

of the terms, we can expand this to

x^2 + (−px)+jqx

+ (−px)+ (−p)^2 + (−jqp)

+ (−jqx)+jqp+ (−j)(j)q^2 = 0

which simplifies to

x^2 + (− 2 px)+p^2 +q^2 = 0

Now let’s substitute the “legitimate” names for p and q back into this equation. That gives us

x^2 + {−2[−b/(2a)]x}+ [−b/(2a)]^2 + [(|d|1/2)/(2a)]^2 = 0

which simplifies to

x^2 + (b/a)x+b^2 /(4a^2 )+ |d|/(4a^2 )= 0

Now remember that d=b^2 − 4 ac. Because d is negative, its absolute value is equal to its additive inverse.

The additive inverse of a difference (subtraction) can be obtained by switching the order of the terms on

either side of the minus sign. Therefore,

|d|= 4 ac−b^2

Let’s substitute the quantity (4ac−b^2 ) for |d| in the previous equation, getting

x^2 + (b/a)x+b^2 /(4a^2 )+ (4ac−b^2 )/(4a^2 )= 0

We can combine the last two terms of this polynomial into a single fraction because we have the common

denominator 4a^2. That gives us

x^2 + (b/a)x+ (b^2 + 4 ac−b^2 )/(4a^2 )= 0

which simplifies to

x^2 + (b/a)x+ (4ac)/(4a^2 )= 0