I did it geometrically. The 6 solutions will be the 6 points along the unit circle in the complex plane. Since the cosine of 60° is 1/2, you know ±1/2 are the real parts of the complex solutions, and the imaginary parts work out to be ±i√3/2.
I was looking for this comment. This method is really easy and straightforward for these kinds of solutions, especially when the degree is higher. In some cases, such as this one, you can use 30-60-90 or 45-45-90 triangles to solve if you are not so familiar with trig.
I like the kind-a-same approach. x^6 = 1 has 6 roots, meaning there is something related to 6 -- which is 360°/6 = 60°. Rest is simpe, x = 1 · cis(k·60°)
Your videos are honestly the best. I watched your video about getting all possible solutions for x³=8. And then tried getting all solutions to x³=5, x⁴=5, x⁵=5, and x⁶=5 using the same factoring method you used. I managed to solve all if then except for the fifth power one. And I'm still working on that. I'm a ninth grader btw ans the math I take it school is super simple and your videos help me to push myself to my limit in one of my favourite subjects and I learn a lot. You don't need anyone to tell you this. But you're the best.
One of my fav French movie quotes was from a cop show with a fast talking forensics guy. The one cop (visitor) turns to the other and admits only understanding half of what the forensics guy said. The other cop said "Don't worry, I'm the same way. But as long as your half and my half aren't the same, we'll be fine." Comments like yours reminds me of that scene.
I always tend to do it with moivre and polar coordinates, it is the one-size-fits-all solution, but sometimes not so time effective :) so I appreciate to get more input for other solutions.
I once had a dream where a mathematician was arguing with Bob Ross that data can be whimsical. When Bob asked how, the mathematician proudly said said i is imaginary.
p=b/a, q=c/a so (-b±√(b²-4ac))/(2a) = -b/2a ±√( (b²-4ac)/(4a²) ) = -p/2 ±√( b²/(4a²) - ac/a² ) = -p/2 ±√( (½ b/a))² - c/a ) = -p/2 ±√( (p/2)²-q ) or in the special case a=1 --> p=b, q=c: (-b±√(b²-4c))/2 = -b/2 ±√( (b²-4c)/4 ) = -p/2 ±√( b²/4 - c ) = -p/2 ±√( (p/2)² - q ) edit: I don't like the pq-formula, because it introduces a fraction inside the root, if p isn't even -- and needs additional work beforehand, if a isn't 1.
So an n-th order equation has n solutions on the complex plane and these solutions lay on the vertices of an n side regular polygon. If it has odd number of solutions, then one is real. If it has even number of solutions, then two of them is real
In general, x^n = 1 has n solutions, and one of them will always be x = +1. They are uniformly distributed angularly around the unit circle. DeMoivre's theorem covers the general case. As for x^n = -1, it will also have n solutions. If n is odd, one of them will be x = -1. If n is even, none of them will be real, and all will be complex conjugates of each other. Divide 180 degrees (or pi radians) by n, and that's the angle of one of the solutions, CCW from the +real axis. The remaining solutions are uniformly distributed angularly.
