# Video: Solving Quadratic Equations over the Set of Complex Numbers

Determine the solution set of 𝑥² − 8𝑥 + 185 = 0 over the set of complex numbers.

03:47

### Video Transcript

Determine the solution set of 𝑥 squared minus eight 𝑥 plus 185 equals zero over the set of complex numbers.

So what we’re gonna do here is solve this quadratic equation. And the way I’m gonna approach that is I’m gonna subtract 185 from each side of that equation first. And that leaves me with 𝑥 squared minus eight 𝑥 on the left-hand side and negative 185 on the right-hand side. What it leaves me with is one 𝑥 squared on the left-hand side minus eight 𝑥.

Now, we’re gonna use our experience of completing the square. To have an initial guess so that can be expressed as 𝑥 minus four all squared or 𝑥 minus four times 𝑥 minus four. Because we had one 𝑥 squared, we just got one 𝑥 and one 𝑥 here. And because we had negative eight 𝑥 here, we’re gonna take half of negative eight, which is negative four.

Now, when I multiply that out, I get 𝑥 times 𝑥 which is 𝑥 squared, 𝑥 times negative four which is negative four 𝑥, negative four times 𝑥 which is another negative four 𝑥, and negative four times negative four which is positive 16. So 𝑥 squared minus four 𝑥 minus another four 𝑥 plus 16. Well, negative four 𝑥 take away another four 𝑥 is negative eight 𝑥.

So it turns out that I guess we’re slightly wrong. This here, 𝑥 minus four all squared, isn’t quite the same as 𝑥 squared minus eight 𝑥. Yes, it has got 𝑥 squared minus eight 𝑥 in it, but it’s also got an extra plus 16 — an extra add 16. So if I took that 16 away from 𝑥 minus four times 𝑥 minus four, then I would just be left with 𝑥 squared minus eight 𝑥. So 𝑥 minus four all squared minus 16 is the same as 𝑥 squared minus eight 𝑥 and that’s equal to negative 185. Well, I’m trying to solve this for 𝑥. I’m gonna add 16 to both sides, and this gives me 𝑥 minus four all squared on the left-hand side and negative 169 on the right-hand side.

And to solve this for 𝑥, we are going to have to take square roots of both sides so that when it’s here 𝑥 minus four all squared, we’ve just got 𝑥 minus four. But we encounter a slight problem because there are no real solutions to the square root of negative 169. So I’m gonna re-express negative 169 as 169 times negative one.

So taking the square root of the left-hand side gives me just 𝑥 minus four and over on the right-hand side, the square root of 169 is 13. So really there are two solutions here. Negative 13 would also give us 169 if we squared it.

And then we can use a definition of imaginary numbers; 𝑖 squared is equal to negative one. So 𝑖 is the square root of negative one, to say that the square root of negative one is 𝑖. So 𝑥 minus four is equal to positive or negative 13𝑖. Now if I add four to each side of the equation, I get 𝑥 on the left-hand side and plus or minus 13𝑖 plus four or four plus or minus 13𝑖 on the right-hand side.

Now if we look back at the question, it asked us to determine the solution set. So I need to write this out in solution set notation. The solution set contains two values four plus 13𝑖 and four minus 13𝑖.