TITLE: Is it possible to have a quadratic equation having only one complex root? If so, what would a picture of it look like?
QUESTION [5 upvotes]: I'm also wondering the same questions about a quadratic function with two real roots, and a quadratic function with two pure imaginary roots. Is it possible? And if it is possible, what would it look like when graphed? I'm trying to fully understand this lesson.

REPLY [2 votes]: What does a quadratic with only 1 imaginary root look like?
If it has real root $r$ and complex root $ki$:
$$f(x) = A(x - ki)(x - r)$$
$$f(x) = Ax^2 - (r + ki)x + rki$$
Notice that the $x^1$ and $x^0$ coefficients are also complex numbers.  So only 1 imaginary coefficient implies that you will have complex coefficients to your quadratic.  So to graph it, you would need 3 axis:  $x$ axis, real $y$ axis, and imaginary $y$ axis.  Your graph will touch only the $x$ axis at $x=r$.
What if a graph has 2 pure imaginary roots, $S\cdot i$ and $T \cdot i$ ? 
$$f(x) = A(x - Si)(x - Ti)$$
$$f(x) = Ax^2 - (S + T)ix - ST$$
Notice that if $S + T \ne 0$, then once again you have an imaginary coefficient to the $x^1$ term.
What if the graph has 2 complex roots, $R + S\cdot i$, and $W + T \cdot i$?
$$f(x) = A(x - R - Si)(x - W - Ti)$$
$$f(x) = Ax^2 - (R + W)x - (S + T)ix + (RW - ST) + (SW + RT)i$$
You can see that $f$ will only have real coefficients when $S = -T$ and $R = W$.  In other words, the polynomial only has real coefficients when the roots are Complex Conjugates.
But what does this look like?  Well we know that since the coefficients are real that all the values of $f$ will be real.  We also know that since the roots are complex, that the quadratic will never intersect the $x$ axis.  So if $A$ is positive, you get a quadratic that opens upwards and is always above the $x$ axis, and if $A$ is negative, you get a quadratic that opens downwards and is always below the $x$ axis.