 In this video we're going to solve the equation x to the fourth minus 2 root 3x squared plus 4 equals 0. We want to solve this polynomial equation and find all complex roots. Now you might think at first this 2 root 3 like what the heck's going on with that coefficient that's actually going to be a blessing for us but we'll see that a little bit later here. You'll also notice that this this equation's missing some terms right we have an x to the fourth we have an x squared and we have a constant there's no x cube there's no x term and so this actually forms what we call it by quadratic equation it's like a quadratic equation but instead of being like a x squared plus bx plus c it's actually x to the fourth and x squared right because x to the fourth is just x squared squared so we can still solve this like a quadratic equation but instead of getting linear factors we would get quadratic factors. Now that 2 root 3 is kind of concerning me in terms of factoring so actually might instead choose the quadratic formula right you know usually goes like x equals negative b plus or minus the square root of b squared minus 4ac all over 2a but instead of being x equals it's actually going to be x squared equals because again that's that's the fundamental term in that situation x squared equals the negative b so we get 2 root 3 plus or minus the square root of b squared minus 4ac all over 2a like so let's see if we can simplify this a little bit before we go forward so we get 2 root 3 mostly it's discriminant we have to simplify here so when you take 2 root 3 when you square that 2 squared is 4 the square root of 3 squared is 3 so 4 times 3 is a 12 and then with the other one you get this negative 4 times 1 times 4 that's a 16 this is all over 2 so we can actually see now our discriminant's going to turn out to be negative so we get the square root of negative 4 right here which the square root of that turns out to be plus or minus 2i over 2 you can factor out a 2 from the numerator so you get root 3 plus or minus just i here over 2 that's 2's cancel and so we get summarizing here that x squared equals root 3 plus i and root 3 minus i so the quadratic formula did turn out to be super helpful in this situation but we didn't solve the equation yet we have x squared equals the square root of 3 plus or minus i so we have to then take the square root of these things so we get x equals the square root of root 3 plus i like so plus or minus and then we have plus or minus the square root of 3 minus i again so these these are excuse me the square root of that so those are the solutions technically speaking but it's like what number is that right we have to compute these square roots of the complex numbers and so let's investigate those if we take the modulus of the square root of 3 plus i well you have to take the square root of root 3 squared plus 1 squared this gives you the square root of 3 plus 1 which is the square root of 4 which is equal to 2 i want you to notice here that this actually doesn't matter on this middle term right here because it would change everything it wouldn't change anything in the in the remaining process so both of these numbers have a modulus of 2 what are their arguments their arguments will of course be a little bit different if you do the argument R root 3 plus i like so we have to compute compute arc tangent of 1 over the square root of 3 this happens at pi 6 we are in the first quadrant notice how it's positive positive so this is going to give us pi 6 if we do the argument of root 3 minus i this does change things a little bit mostly the quadrant right you're going to get now a negative 1 over root 3 we're going to get an angle that references it's going to reference to pi 6 but which quadrant are we in now we're in positive negative so this is actually giving me the fourth quadrant right there so we need an angle which references to pi 6 in the fourth quadrant that's 11 pi 6 like so and so then let's let's revisit our numbers from above we get that our solutions are going to look like x equals plus or minus the square root of well these numbers 2 e to the pi i 6 and then plus or minus the square root of 2 e to the 11 pi i over 6 like so all right so taking the square root it's the same thing as taking the one half power so we need to times the denominator of that x 1 and by 2 so our solutions are going to be x equals plus or minus the square root of 2 times e to the pi i over 12 and then likewise plus or minus the square root of 2 e to the pi i uh the 11 pi i over 12 like so so these are going to be our solutions so what do you do with like what do you do with pi i over 12 let's do this to the side here e pi i over 12 so we have to compute cosine of pi over 12 plus i sine of pi over 12 pi over 12 of course it's the same thing as 15 degrees you might remember that one we did that previously that's going to give us root 6 plus root 2 over 4 and then here we're going to get i times the square root of 6 minus square root of 2 over 4 like so that's e to the pi i over 12 the other one is similar when you do when you do the 11 because notice that 11 pi twelfths is the angle in the second quadrant that references to pi twelfths aka 15 degrees so that one's going to look like e to the 11 pi i over 12 you get of course cosine of 11 pi over 12 plus i don't forget the i sine of 11 pi over 12 notice of course that pi minus pi twelfths is 11 pi twelfths that's how i got the reference angle so in this in the second quadrant this is going to be negative cosine of pi twelfths plus i sine of pi twelfths so you would end up with in this case a negative root 6 minus root 2 over 4 in this case then you're going to get plus i root 6 minus root 2 all over 4 you get these numbers here and so then let's consider then the solutions the four solutions now that we've taken care of these exponential expressions using Euler's identity there so we get x equals the square root of 2 times e to the pi i 12 uh and so we're going to take this number here and times it by the square root of 2 um square root of 2 there that's going to give you a factor of 2 because notice notice that whenever you see these things here like the square root of 6 plus square root of 2 that's just the same thing the square root of 2 times the square root of 3 plus 1 and a similar statement can be said for these ones right here so when you times things by the square root of 2 you're going to get 2 over 4 you can just factor the 4 out of everything there 2 over 4 times what's left behind you get root 3 plus 1 plus 2i over 4 times root 3 minus 1 like so the next one very similar uh you're going to end up with 2 over 4 times and but by the next one i mean we're going to choose the negative case this time right so really it's just let's just put the plus or minus in there for the moment and that that's kind of messy never mind i will put a negative right there root 3 plus 1 and then a negative right there so now let's move on to this case over here using the 11 pi i over 12 that we were considering down here uh and so in that situation again take out the 2 over 4 leaving behind a negative root 3 minus 1 and then for the next one you're going to get a 2 4 i times root 3 minus 1 like so and then if you times that thing by negative 1 you get negative 2 4 negative root 3 minus 1 minus 2 4 i root 3 minus 1 and then let's simplify those expressions to some degree so x equals the first one in the end we end up with root 3 plus 1 over 2 and then we're going to get plus i root 3 minus 1 over 2 uh next we will just get that same number times by negative 1 so you get negative root 3 plus 1 over 2 minus i root 3 minus 1 over 2 then we'll do our third one right here uh we're going to get a negative root 3 plus 1 over 2 and then we're going to get a i root 3 minus 1 over 2 and then the last one i kind of out of space i'll just put it down below it's just the it's just the third one times negative 1 so that's going to give me root 3 plus 1 over 2 and then a minus i root 3 minus 1 over 2 and so we've now found the four complex roots to this polynomial yeah at first it started out pretty good just how you use the quadratic formula then you know we have to take the complex square root it's got a little bit messy but we work through it using the trigonometric forms we're able to do these things uh we're yeah we're able to do them but i guess there's not much more to say than that and so that brings us to the end of lecture 34 and also brings to the end of our lecture our chapter i should say about complex numbers and how we can use trigonometry to actually help us work through complex algebraic problems like how we're able to solve this polynomial equation using trigonometry a technique that's too advanced for even a college algebra class you would very much struggle in a place like that it's actually a pretty cool application if you agree with me please give these videos a like subscribe if you want to see more videos like this in the future and as always if you have any questions on any of the videos you're watching any of any of my math videos please post your questions in the comments and i'll be glad to answer them bye everyone