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## Algebra 2

### Course: Algebra 2>Unit 2

Lesson 6: Quadratic equations with complex solutions

# Solving quadratic equations: complex roots

Solving quadratic equations can lead to complex solutions. Listen to Sal break down the process using the quadratic formula and standard form. Then, learn how to verify these complex solutions, ensuring you've got the right answer. It's all about turning tricky problems into solvable equations! Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• If you graph a quadratic equation that has only complex roots, you will get a parabola that never crosses the X axis. Is there a type of representation or graph on which you can graph such an equation that will show at which points the equation crosses the X axis if, instead of a one-dimensional axis, you represent the horizontal as the complex plane?
• I think a way to do that is to make a 3D chart that has the complex coordinates on the horizontal axis. You would put the absolute value of the result on the z-axis; when x is real (complex part is 0) the absolute value is equal to the value of the polynomial at that point.
The absolute value is always non-negative, and the solutions to the polynomial are located at the points where the absolute value of the result is 0. You could make two representations, one for the real value of the result and one for the imaginary value of the result, but you would have to search for the point(s) where those 2 are both 0.

I came up with the following piece of code written in MATLAB:
``numberOfPointsOnTheAxis = 101;p = [1 -6 10]; % 1*x*x -6*x +10 ; the coefficients of the polynomial% the real part goes from 1 to 4realPart = linspace(1,4,numberOfPointsOnTheAxis);% the imaginary part goes from -2 to 2imaginaryPart = linspace(-2,2,numberOfPointsOnTheAxis);x = ones(numberOfPointsOnTheAxis,1) * realPart(:)' + ...    imaginaryPart(:) * ones(1,numberOfPointsOnTheAxis) * 1i;y = p(1)*x.^2 + p(2)*x + p(3);close allsurf(realPart, imaginaryPart, abs(y))hold onplot3(realPart, zeros(1,numberOfPointsOnTheAxis), ...    p(1)*realPart.^2 + p(2)*realPart + p(3), 'r-', 'LineWidth',3)colorbarxlabel('real part')ylabel('imaginary part')zlabel('absolute value of result')``

The red line is the parabola that you normally get when using only real coordinates
• Can someone explain to me what Sal is doing at onwards?
• Sal is using the same FOIL technique except now there are complex numbers.

`((3 + i) / 2)^2` can also be written as `((3 + i) * (3 + i)) / (2 * 2)`.

By using FOIL the numerator will become:
``F: 3 * 3 = 9O: 3 * i = 3iI: 3 * i = 3iL: i * i = -1(3 + i) * (3 + i)--> 9 + 3i + 3i - 1--> 8 + 6i``

Hopefully from here, the path forward is becoming clear. We know what `(3 + i)^2` equals so lets's plug it into `2 * [(3 * i)^2 / (2^2)]`.

``2 * [(3 * i)^2 / (2^2)]--> 2 * [(8 + 6i) / 4]--> 4 * (4 + 3i) / 4--> 4 + 3i``

I hope this helps!
• -i^2 = -1?
I understand that +i^2=-1 as i = sqrt-1, can someone show me how -i^2=-1 please?
Anyone? (I think its similar to the reason you have 2 answers in quadratics but I still dont know what value n should be)
• this confused me as well, but it was easier to get my head around it if I think of it as: -i^2 = (-1i)^2 = -1^2 * i^2 = 1 * i^2 = 1 * -1 = -1
• At , Sal says that (-i)^2 is -1. Why is this the case?

I understand the definition of i^2 = - 1, but not quite this part.
• It is the same as squaring any other negative.
`(-a)² = (-1)²(a²) = a²`
Thus
`(-𝑖)² = (-1)²(𝑖²) = (1)(𝑖²) = 𝑖² = -1`
• Is there any video lessons on De Moivre's theorem?
And how to find out the value of log(i) ? Can it be done by Euler's formula?
• I can't seem to find any videos on de Moivre's theorem, either, but I do know that the idea that `(cos x + i*sin x)^n = cos(nx) + i*sin(nx)` can be derived from Euler's formula. <https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-14/v/euler-s-formula-and-euler-s-identity>
I don't know if this is how you've seen it, but the version of de Moivre's theorem in my textbook is `r^(1/n)*(cos((x + 2pi*k)/n) + i*sin((x + 2pi*k)/n))`. Clearly a very different formula, but that doesn't mean it can't be derived from that first expression.
Technically `(cos x + i*sin x)^n = (r*(cos x + i*sin x))^n`, but the r is omitted because we're usually talking about the unit circle (r=1). Keeping that in mind, let's arbitrarily substitute n for 1/n. (In that first case, n represents any old number, but with 1/n, n is the number of roots)
Now we have `r^(1/n)*(cos(x/n) + i*sin(x/n))`.
...so what about the `+2pi*k`? Well, substitute x with x+2pi*k and voila!
`r^(1/n)*(cos((x + 2pi*k)/n) + i*sin((x + 2pi*k)/n))`
Note: x+2pi*k just refers to any angle coterminal with x, so they are equivalent
• Where did the 6i come from?
• I assume you are asking about the 6i that occurs at about @ in the video. If yes, then it comes from squaring (3+i).
(3+1)(3+i) = 9+3i+3i+i^2 = 9+6i-1 = 8+6i

Hope this helps.
• What if on a quadratic equation with no real solutions, I add an "i" in front of the "x^2" and then graph it? If I then found the roots to that quadratic equation, how would these roots be related to the roots of the first equation?
• think about it! So if you have some quadratic equation Ax^2 + Bx + C = 0, and you multiply each "x" term by i, you'd get iAx^2 + iBx C = 0. Now, let's try dividing the whole equation by "i": even though it's imaginary, it's still a number, right? iAx^2 becomes Ax^2, iBx becomes Bx, C becomes (C/i), and 0 becomes 0 (0 divided by ANY number, even imaginary ones, is always equal to 0). Put it together and we get Ax^2 + Bx + (C/i) = 0. Our equation looks the same as the original, except C is now different. Using the quadratic equation:
x = (-B +- sqrt(B^2 - 4A(C/i)^2))/2A. (C/i)^2 = C^2 / i^2, which equals C^2 / -1. Plug it in:
x = (-B +- sqrt(B^2 + 4AC))/2A (remember, minus -C^2 is the same as plus C^2)

Compare this to the solution of our original equation:
x = (-B +- sqrt(B^2 - 4AC))/2A

As long as A, B, and C are not zero (you're dealing with an actual quadratic equation), you can see that x is different.

Long story short: the roots will NOT be the same as the first equation! Hope that helped!
• at , why is ONLY the principal square root of 4 taken. shouldnt BOTH +2 & -2 be taken as the square root of 4??
(1 vote)
• You ask a good question and you are right in your thinking. By definition, the Principal root of a number is the same sign as the real number. For example, both -4 and +4 are the square roots of 16. So, to talk about just the principal root of 16 means we discuss the "n"th root of 16 that has the "same sign" as the number in question. Since 16 is positive, the principal square root is +4 (even though -4 is also a square root)

The concept of principal root of a number means we only have to talk about one possible answer, instead of working through both the positive answer and the negative answer, if both can exist.

So by Sal discussing the principal square root of 4, I think he is trying to simplify the discussion.