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.
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Yes, if $a$, $b$ and $c$ are allowed to be nonreal complex numbers. In fact, any complex polynomial can be written as $a(x-r)(x-r')$ (use the quadratic formula or Vieta's relations), $r$ and $r'$ being its roots. As an example, the polynomial $(x-i)(x-2)=x^2-(2+i)x+2i$ has roots $i$ and $2$.
If $a$, $b$ and $c$ are real, the roots are constrained: they are both real or nonreal complex numbers. This is caused by $z\ \text{is a root}\implies \bar{z}\ \text{is a root}$.
Ian Mateus
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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.
DanielV
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The answer is no for quadratic equations with real coefficients as others have pointed out. On the other hand it is yes for imaginary coefficients. For example:
$$(x-i)(x-1)= x^2-(i+1)x+i$$
Alex R.
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You cannot have a quadratic polynomial with only one complex root. If $$p(x)=ax^2+bx+c$$ has $$p(x+iy)=0$$, then $$p(x-iy)=0$$ (you can check this very easily).
The other two scenarios are possible. For example, $$p(x)=(x-1)(x-2)$$ has two real roots, and $$p(x)=(x-i)(x+i)$$ has two pure imaginary roots.
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This is a good question, and the answer is no, you cannot!
Suppose $a+bi$ solved $Ax^2+Bx+C=0$. The complex conjugate of $a+bi$ is $a-bi$. Would this also solve the quadratic?
Emily
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I figured I'd add an answer to note that one way to approach your question is to use something you already know about: the quadratic formula. If you have a quadratic with real coefficients, what determines whether the roots are purely real, purely imaginary, or neither? What does the quadratic formula tell you about the coefficients if the roots are purely real? What does it tell you about the coefficients if the roots are purely imaginary (note: it tells you more in this case—only very special quadratics with real coefficients have imaginary roots).
Thinking about Bryan C's answer: can you see what roots you can choose for a quadratic with complex coefficients?
dfeuer
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