College Algebra & Trigonometry, 2018a

50 CHAPTER 1. ALGEBRA REVIEW
If we add an additional three units to the constant term of this quadratic equation,
we encounter a third possibility.
x 2 − 6x +12=0
a =1,b= −6,c=12
x = −(−6) ± √ (−6) 2 − 4(1)(12)
2 ∗ 1
= 6 ± √ 36 − 48
2
= 6 ± √ −12
2
= 6 ± i√ 12
2
≈ 6 ± 3.464i
2
= 6 2 ± 3.464i
2
≈ 3 ± 1.732i
Here the discriminant is negative, which leads to two complex-valued answers.
If the equation has real-valued coefficients, the complex roots will always come
in conjugate pairs. Complex conjugates share the same real-valued part and have
opposite signs in their complex-valued (or imaginary) parts: a ± bi
Graphically, the previous problem was one step away from not intersecting the
x-axis at all and the additional three units that we added on to get y = x 2 −
6x +12moves the graph entirely away from the x-axis. Because the roots are
complex-valued, we don’t see any roots on the x-axis. The x-axis contains only
real numbers.