Dividing through by (−a), we obtain
x^2 =c/(−a)
Because c> 0 and −a> 0, we know that c/(−a)> 0. We can take the positive-negative square root of both
sides to get
x=±[c/(−a)]1/2
which can be rewritten as
x=±(−c/a)1/2
Stated separately, the roots are
x= (−c/a)1/2 or x=−[(−c/a)1/2]
These are the same roots we get if a> 0 and c< 0. Both roots are real, and they are additive inverses of
each other.
Conjugate Roots in Factors
We’ve seen the binomial factor forms of quadratics where the roots are pure real or pure imagi-
nary. Now let’s look at the factors when the roots are complex conjugates, but are not pure
imaginary numbers.
An example
We can use the quadratic formula to find the roots of a quadratic equation that has complex
conjugate roots. Then we can generate the binomial factor form of the quadratic from those
roots. Let’s try this:
5 x^2 + 6 x+ 5 = 0
We have a= 5, b= 6, and c= 5 in the polynomial standard form. The discriminant is
d=b^2 − 4 ac
= 62 − 4 × 5 × 5
= 36 − 100
=− 64
The square root of the discriminant is pure imaginary:
±(d1/2)=±[(−8)1/2]
=±j 8
Conjugate Roots in Factors 391
