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### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 1

Lesson 17: Topic D: Lessons 38-40: Complex roots and the fundamental theorem of algebra- Solving quadratic equations: complex roots
- Solve quadratic equations: complex solutions
- Complex numbers & sum of squares factorization
- Factoring sum of squares
- Factor polynomials: complex numbers
- The Fundamental theorem of Algebra
- Quadratics & the Fundamental Theorem of Algebra
- Number of possible real roots of a polynomial

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# Quadratics & the Fundamental Theorem of Algebra

The proof of the Fundamental Theorem of Algebra for any degree of polynomial is really tough. For now, let's note that it indeed holds for polynomials of the second degree (i.e. quadratics). Created by Sal Khan.

## Want to join the conversation?

- If a real root crosses the x axis, a complex root crosses...? What does it cross? And, can it be seen?(39 votes)
- No, the complex root is not defined by crossing anything, and no, on our regular x,y graph it can't be seen. But if we make a vertical axis of the units of i (imaginary numbers), and make the real numbers run along a horizontal axis, the complex root can be seen graphed on this "complex plane." (And all sorts of fun can be had with graphing on the complex plane in precalculus)

Also, even though the complex roots don't exist on the x,y graph (since there's no such thing as a square root of a negative real number, so there's no way we can plot it on the cartesian plane of real numbers, that's why we need the imaginary plane) it can still be visualized on the x,y graph using a little trick, which is, to take our parabola and flip it upside down. This reflected parabola now has to cross the x=axis, and have 2 real roots.

If this flipped parabola has the equation y= -(x-a)^2+b, then it has the real roots a +/- sqrt(b) whereas our original parabola had the equation y- (x-a)^2+b with the complex roots a +/- sqrt(b)*i.

So, since on a complex plane its the vertical axis that has the imaginary numbers, to visualize it on our normal graph, we take the roots of this flipped parabola, make a perfect circle with the points of the two roots on the opposite sides of this circle, then rotate the points 90 degrees, as shown on this website: http://www.math.hmc.edu/funfacts/ffiles/10005.1.shtml

Hopefully this last bit wasn't confusing, it's just a way to pretend we have the complex numbers on the x,y graph(43 votes)

- What about when y = x^2. Isn't there only one root(0).(8 votes)
- Yes, but only because +0 and -0 are not two different numbers. 0 is unique in that it is its own opposite.(8 votes)

- Assuming you have two complex roots and a real root, how do you get the function from those roots?(5 votes)
- If the roots of a polynomial are at x = a, x = b and x = c (whether a, b and c are real or complex) then an equation with those roots will be (x - a)(x - b)(x - c) = 0. Then you can multiply out the brackets if you wish. Beware, though. This may not be the original function, because you can also multiply the whole thing by ANY non-zero number (real or complex) and the resulting polynomial will still have the same roots.(5 votes)

- At4:07, couldn't he just simplify and write
`1/5(i) instead of -3/5 + 4/5 (i)`

(3 votes)- In the expression:
`-3/5 + 4/5 (i)`

, the 2 terms are unlike terms. We can only combine like terms. Since the first term has no "i", it is "unlike" the 2nd term.

If the expression had been:`-3/5(i) + 4/5 (i)`

, then you could combine them into`1/5 (i)`

Hope this helps.(6 votes)

- is it possible for a quadratic equation to not touch either the x or y axis? How would you solve this?(3 votes)
- it's not possible. all quadratic equations have a y intercept at their c value, although some will not touch the x axis.(4 votes)

- What does the complex root that is found mean? For example, the root that was found was x = -(3/5) + -(4/5)i . What does that tell us about the graph or equation?(1 vote)
- There isn't anything too important about nonreal complex roots that shows on the standard 2-D graph because we are not plotting the nonreal complex values of x, but only the real complex values. However, if you graphed the function in the complex plan, with the imaginary portion being one of three axes, then you would see the place where the f(x) or y value of the function reaches 0.

As far was what it might mean in a real-world application, there are numerous applications of nonreal complex numbers, especially in electrical engineering, weather forecasting, and even in computer science.(6 votes)

- But it may only have one real root, if it touches the x-axis once like y=(x-1)^2 would(2 votes)
- In this case, there are two roots, but they are both equal to each other. y = (x - 1)(x - 1). We call this a root with a multiplicity of 2, since it appears twice in the function's factored form. Roots with multiplicity are counted however many times they need to be for the fundamental theorem of algebra. You can see if a root is like this (even multiplicity) visually if the line of the function "bounces back" from the x-axis after touching it, like how the parabola approaches the x-axis as you get to the vertex, and then it goes back in the opposite direction. Roots with a multiplicity of 1 will just go straight through. If you have a root with an odd multiplicity that isn't 1, such as in y = x^3, the graph will lessen its curve before it goes through the x-axis and resumes its behavior on the other side.(4 votes)

- what happens when the discriminant equals 0 for 2nd degree polynomials ?(2 votes)
- When the discriminant = 0, then the quadratic equation has one root (or x-intercept) that is a real number.(4 votes)

- This is a question from the course challenge (precalculus):

z^3 = -64

It states that one root is (-4) and asks us to find the other 2 complex roots.

The answer: 4(cos60,isin60)

But I cannot figure out how to get it. Please help(2 votes)- First, find the absolute value (modulus) and angle of -64, so we can rewrite it in polar form. The absolute value of -64 is 64, and the angle it creates from the positive x-axis is 180 degrees. So, in polar form -64=64[cos(180)+isin(180)]. Now rewrite the entire equation. z^3=64[cos(180)+isin(180)]. Take the cube root of both sides. z=(64[cos(180)+isin(180)])^(1/3). When we raise complex numbers to a power in polar form, we multiply the angles by that power and raise the modulus to that power. So, z=64^(1/3)*[cos(180*1/3)+isin(180*1/3)]. Simplify to get 4[cos(60)+isin(60)].(2 votes)

- But f(x) = (x+9)^2 only has one root..(2 votes)
- 𝑓 actually has only one
*unique*root, namely -9. However the function still has two roots total (as it can be written as the product of two binomials equated to 0. We call such a root a*double root*which is a root with multiplicity 2. Therefore, 𝑓 still satisfies the Fundamental Theorem of Algebra.(2 votes)

## Video transcript

Voiceover:Let's say that
we have the function F of X being defined by the
second degree polynomial, 5 X squared plus 6 X plus 5. The fundamental theorem of algebra tells us that because this
is a second degree polynomial we are going to have exactly 2 roots. Or another way of thinking about it, there's exactly 2 values for X that will make F of X equal 0. So I encourage you to pause this video and try to figure out what
those 2 values of X are. So when I, if we're
looking for the X values, that would make this
expression equal to 0. That's essentially trying
to solve this equation. At 5 X squared plus 6
X plus 5 is equal to 0. And there's no obvious way I can think of factoring this right
over here so I'll just resort to the quadratic formula. So the quadratic formula tells us that X, where X is a solution to this equation, is going to be equal to negative B. I'll try to color code it. Negative B, this is right,
this is B right over here. Negative B, so negative 6, plus or minus, plus or minus the square
root of B squared, of B squared, minus 4 times A, times A, times C, times A times C. All of that, all of that over
all of that over 2 times A. All of that over 2 times A. So what does this simplify to? This is going to be equal
to so I'll go back to, this is going to be equal to negative 6. I'll try to keep it color coded still. Negative 6 plus or minus,
plus or minus the square root. Now, what is this going to be equal to? This is 36 minus 100. So negative 64 all of that over 2 times 5. All of that over 10. Now this is interesting. We are trying to take the square
root of a negative number. Or another way of thinking about it, B squared minus 4 A C is less than 0. B squared minus 4 A C which is often, often times called the
discriminant of this quadratic. This is less than 0. So this is less than 0. This part of the quadratic
formula we're going to try to take the square
root of a negative number. This is going to be a negative number. So we're going to result
in an imaginary number. So what that gives us
are not 2 real roots, but 2 non real complex roots. So this is going to be equal to negative 6 plus or minus 8 i, the square root of negative 64 is 8 i. If we extend the principle
square root function to imaginary numbers or complex numbers. All of that over 10. Or we could say X is equal to, let's see if we find a
greatest common divisor. If we try to reduce this,
so we're going to have let's see they're all divisible by 2 so it's negative 3 over 5,
that's the same thing as negative 6 over 10, plus or
minus 4 over 5, 4 over 5 i. Or you could say that the 2 roots, which are non real complex roots, so it's X is equal to
negative 3/5 plus 4/5 i, that's one of the roots. And then the other root is X is equal to negative 3/5 minus 4/5 i. Notice, we satisfied the
fundamental theorem of algebra. We have 2 roots, they're non real, but the fundamental theorem of algebra says, "hey look we're just going to have "at least if we're Nth degree polynomial, "we're going to have n complex roots." They could be real or
they could be non real and we see that right over there. And we also see that they are conjugates. And the quadratic formula kinda sets up a situation where we are, especially if this ends
up being less than 0 and so this when you take the square root you get an imaginary number you see where those conjugates appear. Now let's verify graphically
that this is indeed the case. That this does not have any real roots. So let's get a calculator out. So let me go into graph mode. So Y is equal to. Let me clear what I must have
been doing out here before. So Y1 is equal to 5 times X squared, plus 6 X plus 5 and then let me set a reasonable range here. So I don't know. Actually, I know very little
about this function right over. So I'm going to set minimum
range at a negative 10, X max I don't know positive 10. X scale is 1, let's see Y max let's see, this thing actually does
get pretty big pretty fast. So let's say Y max is and the Y scale, I don't know I'll say 100. 100, the scale I'll make it 10. Each of those notches are going to be 10. Y minimum, we want to see the X axis to make sure it doesn't cross
it so let's go at negative, I don't know, negative 20. And now let's graph this. I'm hoping that I've actually
captured it and I have. So there you have it, you see that this thing does not intersect the X-axis. And in fact, we could,
we could zoom in on it. Let's see it's a little bit
hard to, it's a little actually, let me just change the range a little bit. So let me just change the range. So let's make our X min, let's make it 5, let's make our X max, whoops
not 50, let's make that 5. And let's make, negative 5 and positive 5. And let's see, let's make our Y max, let's make our Y max equal
to 20 and then our Y scale, I don't know we could make it 2. Okay I think this will
get us much closer in, there we see, it does
not intersect the X axis. This does not have any real roots but it has 2 non real complex roots.