Complex Zeros and Quadratic Applications
October 22, 2026

[00:00:05] If the discriminant is negative, then the zeros are going to be complex zeros, they're going to be imaginary zeros.
[00:00:16] So speaking of that, let this be any quadratic function. How many zeros are we guaranteed to have?
[00:00:41] Two. What kind? Two complex zeros. They might be real or they might be imaginary, but we definitely have two complex zeros. Agreed?
[00:00:54] What are the two zeros? To find zeros, what do we do? If we plug in x = 0, we get f(0) = a times 0 squared plus b times 0…
[00:01:25] That gives the y-intercept. We want zeros, so plug 0 into f(x), or y. Then how do we solve?
[00:01:52] By factoring, completing the square, or the quadratic formula. This is how we derived the formula: the two zeros are [−b ± √(b² − 4ac)]/(2a).
[00:02:18] Does that make sense? It is abstract because these are variables, but they are all numbers. These are the two zeros. I could write it this way.
[00:02:37] We agree? Okay.
[00:02:44] Let me put an absolute value inside the square root to make this correct.
[00:02:51] If b² − 4ac is negative, these zeros are imaginary.
[00:03:09] We get a square root of a negative number: the square root of its magnitude times √(−1). And √(−1) is i.
[00:03:26] The absolute value makes that negative discriminant positive inside the radical; the i comes from √(−1).
[00:03:41] These are the two zeros: −b/(2a) plus the square-root term. I will keep the absolute value there.
[00:04:07] And −b/(2a) minus the square-root term over 2a.
[00:04:22] These are two complex numbers.
[00:04:32] What's the only difference between these two?
[00:04:38] Plus versus minus. What do we call that? Conjugates.
[00:04:46] This is an important fact: if b² − 4ac is less than zero—negative—
[00:05:03] the two zeros are complex conjugates of one another.
[00:05:27] 2i is a known zero of f(x).
[00:05:36] The leading coefficient is −3. Find the standard form of f(x).
[00:06:05] We know one complex zero. It's imaginary. And we know the leading coefficient.
[00:06:16] We know one complex zero. It's imaginary.
[00:06:20] Okay.
[00:06:24] So what does that tell us?
[00:06:26] Its complex conjugate is also a zero. The complex conjugate of 2i is also a zero.
[00:06:45] And what is the complex conjugate of two i?
[00:06:51] 0 + 2i. Its conjugate is 0 − 2i, so −2i is also a zero.
[00:07:09] And now we're going to write it in what form? Factored form. When we know the zeros, we can write factored form.
[00:07:17] So it's going to be
[00:07:21] f(x) is the leading coefficient, −3, times (x − 2i)(x − (−2i)), which is (x − 2i)(x + 2i).
[00:07:46] Now we just check. Okay. So that is the function, but it's written in factor form. And how are we going to get that into standard form?
[00:07:55] Distribute.
[00:07:57] Minus 4i².
[00:08:06] Those two terms add to zero, and i² is −1.
[00:08:16] So the product is x² + 4.
[00:08:20] And that's not quite in standard form.
[00:08:24] f(x) = −3x² − 12.
[00:08:29] And there it is.
[00:08:31] I could ask other questions: find the vertex, graph it, or what else?
[00:08:47] The axis of symmetry. That one is easy.
[00:08:51] I could have done no work. What's the axis of symmetry at the very beginning? Zero. Zero, because it's this number over two a.
[00:09:01] Let's try another one slightly more complicated. What was that first one I did? What was it?
[00:09:09] 3 + 2i is a known zero.
[00:09:22] And
[00:09:27] the
[00:09:33] vertex.
[00:09:39] Let’s use the axis of symmetry.
[00:09:42] What do we know is also a zero? 3 − 2i is also a zero.
[00:09:53] So we can write factored form. What is the question asking? Good question.
[00:09:59] Write the standard form
[00:10:09] of f(x).
[00:10:15] We know the zeros, so we can almost write the factored form of f(x).
[00:10:23] We don’t have a leading coefficient, so leave an a there for now.
[00:10:32] It is x − (3 + 2i)
[00:10:39] times x − (3 − 2i).
[00:10:46] Let's multiply this out.
[00:10:49] We will figure out a. For now: a times (x − 3 − 2i) times (x − 3 + 2i).
[00:11:07] Then multiply it all out. There is a lot of distributing. How many terms? Nine.
[00:11:17] You could do it. X times X is X squared. X squared.
[00:11:25] −3x + 2ix.
[00:11:37] Minus three X.
[00:11:42] −3x + 9 −6i.
[00:11:57] −2ix + 6i
[00:12:06] −4i².
[00:12:14] Everything with i cancels, or i² turns into −1.
[00:12:21] 2ix and −2ix cancel. If they didn’t, we multiplied something incorrectly. We have a times x²…
[00:12:34] −3x − 3x is −6x,
[00:12:39] and 9 + 4
[00:12:44] is 13.
[00:12:45] I don't think I gave you enough information.
[00:12:52] I don't think so. I messed up.
[00:12:56] There are actually many possibilities.
[00:12:58] This is ax² − 6ax + 13a. We agree.
[00:13:11] So we know.
[00:13:14] Negative 6 A.
[00:13:19] Over.
[00:13:22] Two a. This is our negative b— Sorry negative.
[00:13:30] This is equal to 3.
[00:13:32] No I didn't. I take that back. I screwed up.
[00:13:39] We don’t have enough information to answer the question.
[00:13:45] Okay. Yeah.
[00:13:46] Let’s instead say this is the leading coefficient.
[00:13:55] Then we substitute 3 here.
[00:14:06] The function would be 3x² − 18x + 39. Okay.
[00:14:14] If you are given just one real zero, you cannot determine the other zero.
[00:14:22] But given a nonreal complex zero, its conjugate is also a zero,
[00:14:30] so we know the other zero.
[00:14:33] A quadratic function has vertex (−3, 2) and passes through
[00:15:01] (−1, 14).
[00:15:08] Find the standard form.
[00:15:13] How we can do this.
[00:15:16] Okay. They give us the vertex.
[00:15:19] We can almost write vertex form; the only missing quantity is the leading coefficient. So we got to leave this as an A.
[00:15:29] x − (−3) is x + 3, squared, plus 2.
[00:15:39] And last thing I do is find the leading coefficient. Multiply that out.
[00:15:47] f(−1) = 14. That is what this ordered pair means. We plug that in here.
[00:15:54] 14 equals a
[00:15:58] times (−1 + 3)², plus 2. And now we got an equation with just one variable.
[00:16:08] And we do a little bit of work.
[00:16:11] 14 − 2
[00:16:14] equals a times 2 squared.
[00:16:19] That is 4a.
[00:16:22] Divide by 4.
[00:16:30] So this is the.
[00:16:35] This is the equation in what form.
[00:16:39] Vertex form. How do we get standard form? Just start multiplying out.
[00:16:47] I will go quickly: x² + 3x + 3x + 9, then plus 2.
[00:17:00] Plus 27 plus 2 is 29. Yes.
[00:17:08] OK. So if you're given some information.
[00:17:14] Suppose I gave you two real zeros
[00:17:19] and one ordered pair.
[00:17:22] Substituting that point lets us find the leading coefficient.
[00:17:29] OK.
[00:17:32] Going back last year. Where does.
[00:17:39] Where do y = 2x + 3 and y = −3x − 7 intersect?
[00:17:56] You might have done problems like this last year. Yes. The others like this.
[00:18:09] And we need to find.
[00:18:12] Where they intersect. We won’t finish this example, but we will go a little further. How will we go about doing this.
[00:18:22] OK.
[00:18:25] Yeah we're not going to try to graph them because it might not need a very nice point.
[00:18:30] We could if they met at a nice point and figured out. We have y = 2x + 3 and y = −3x − 7.
[00:18:44] We want an ordered pair that makes both equations true.
[00:18:51] And we have another name for that. What do we call it.
[00:18:56] A system of equations.
[00:19:00] OK. And last year there were like three ways to solve systems equations.
[00:19:12] Elimination, substitution, and…
[00:19:15] I think it was the only way.
[00:19:21] Yeah. Yeah. Which we don't do. OK. So for this what we do.
[00:19:31] Substitute. We take this guy. Substitute in right. This is y, so substitute it for this y. OK.
[00:19:40] We get 2x + 3 = −3x + 7. [Correction on screen: the original constant is −7.] And we solve this and get some number for X right.
[00:19:50] So is that our solution. No. No, that is only the x-coordinate. Substitute x into either original equation to find y.
[00:20:00] So we're going to do something similar but now with quadratics.
[00:20:03] We want to find where.
[00:20:09] They intersect.
[00:20:19] First off. How many possible solutions. How many possible.
[00:20:29] Times can intersect.
[00:20:36] Does this one open up or down.
[00:20:43] So it opens up. This one open up or open down down.
[00:20:49] With one opening up and one opening down, how many intersection points could there be? Two. Can we have just one.
[00:20:56] They can meet at a single point. Well.
[00:21:00] They could be at the vertex.
[00:21:03] OK. And could there be no solutions or no intersections. Yes. Yeah. If this guy was down here.
[00:21:09] So how are we going to do this.
[00:21:12] So. OK. Same thing as. Right. y = x² + 2x − 2 and y = −x²
[00:21:23] + 4x + 10.
[00:21:27] And this is a system we're going to solve.
[00:21:30] OK.
[00:21:33] Systems are we are looking for solutions that are common to both of them. So the solution is going to be ordered pairs.
[00:21:39] And we're solve this by.
[00:21:41] Substitution. We're going to substitute this. Which is equal to Y into this Y or you could do it vice versa.
[00:21:52] x² + 2x − 2 = −x²
[00:22:01] + 4x + 10.
[00:22:07] Oh now we have an equation in only one variable. And now we're going to solve this.
[00:22:15] It's just a quadratic equation. We get everything on one side.
[00:22:19] Add x², subtract 4x, and subtract 10.
[00:22:26] 2x² − 2x − 12 = 0.
[00:22:38] And then you could solve by factoring completing the square quadratic formula. These have a common factor. Which should we do first?
[00:22:47] The two. Yeah get rid of that two. It will make it easier.
[00:22:56] And does this factor easily?
[00:23:03] It does. What's the factor?
[00:23:06] x − 3 times x + 2. The solutions are 3 and −2.
[00:23:18] Is that my answer? 3, −2? Those are X values.
[00:23:27] So to find the ordered pairs we plug them into. Either one of them. Let's plug it in here. What was the first one?
[00:23:39] f(3). 3² + 2 times 3 − 2.
[00:23:51] 9 + 6 − 2
[00:23:55] is 13.
[00:23:59] And g of… What was the other one? Negative one? Negative two. Negative two.
[00:24:24] Is that negative two? No. So they intersect.
[00:24:34] (3, 13)
[00:24:40] and (−2,
[00:24:44] −2).
[00:24:46] Okay.
[00:24:47] We could have a quadratic and a linear. Do the same thing. Substitute the linear expression into the quadratic, or vice versa.
[00:24:55] And we solve that quadratic. If you end up with the irrational solutions.
[00:25:02] Substitution into the functions may look less tidy.
[00:25:07] Find the equation of the line
[00:25:20] That.
[00:25:26] Passes.
[00:25:32] through
[00:25:41] Both.
[00:25:43] vertices.
[00:25:45] The quadratic has a vertex, and the absolute-value function has a vertex.
[00:25:56] What is the Vertex of this one? This one? easy.
[00:26:04] (3, 1). This is in vertex form. Take the opposite of this number, paired with that number. How do we find the vertex of this one?
[00:26:25] −2. −2. And now we plug it in to find the y value.
[00:26:42] 4 − 8 + 1. Is that −3? Yes.
[00:26:48] Okay, so the vertex here is
[00:26:53] (−2, −3). And then what did I ask for?
[00:27:02] The equation of the line through both vertices.
[00:27:09] I don't want to graph it. I want to find the equation.
[00:27:15] Okay, we need to find the slope first.
[00:27:19] Which is going to be right. (y₂ − y₁)/(x₂ − x₁).
[00:27:27] 1 minus negative 3, over 3 minus negative 2. That is 4/5.
[00:27:38] Now we need the y value. We can then find the y intercept or
[00:27:45] The better way is point-slope form. Oh, yeah.
[00:27:54] Which point should I use? This point or this point?
[00:28:00] I think that was the point. It's nice.
[00:28:05] It doesn't matter.
[00:28:10] y − 1 = (4/5) times ( x
[00:28:19] minus 3).
[00:28:24] And that's it.
[00:28:27] Do I need to put it in slope-intercept form? No, it is already the equation of the line. No matter what form it's in.
[00:28:36] So we can answer all kinds of crazy questions now.