Lol me too, i guess we Indians are taught to go through problems fast. When I saw quad equation going in higher degrees, i went back to the problem again and solved it instantly using Pythagorean triplets. 😂
I did the same thing. By 3:40 the process devolved into a glorified guess-and-check anyway on the assumption that the solution was a rational number, which is no different than what we did. He might as well have applied that reasoning much earlier in the process and done this far faster, and without needing the rational root theorem. He technically assumed it was merely rational, not a whole number, but it was an assumption nonetheless and much more work.
it's really nice to see how you solve it differently than I was taught in Brazil (where I live). for example, we don't "subtract/divide both sides", we essentially do the same things but we say we are swapping the term side (and changing the operation, so obviously 2x = 8, and we bring the 2 that is multiplying to the right side, where it now divides, so x = 8/2 = 4). we also call the quadratic formula Bhaskara's formula for some reason. anyways nice video!!
The bhaskara thing is an mistake that someone made more than a hundred years ago that became popular, probably cause it’s way better to say bhaskara than “fórmula do polinômio de segundo grau”
Red! Awesome video, any polynomial above a degree of two just makes me want to cry. As an aside, I was working on this problem that asked how many factors a certain number had (it was some highly composite number, I want to say 600?) My immediate thought was to get the prime factors and count from there, as you did in the video. But I was having some trouble with the combinations. Maybe I'm just a bit rusty and need to relearn my combinatorics, but do you have a suggestion on how to go about it?
First find the prime factors. 600 has a prime factorization of 2^3 * 3 * 5^2. Every factor will be a combination of these prime factors. Each prime factor has 1 plus its exponent as the number of ways you can add it into a factor (0 copies, 1 copies, 2 copies, etc.), so you can use the fundamental counting principal. In this case you would get (3+1)(1+1)(2+1)=4*2*3=24 factors. Hope this helped. I could provide some clarification if this was not clear.
Aqua Green. I'm flabbergasted how this channel doesn't have a million+ subscribers. Andy you're such a smart and intelligent guy. Sorry if my english was bad, but you did a great job!
Orange. Because of your videos I'm starting to realize that mathematics is about have more and more weapons (formulas) in your toolkit and being creative with where and how to apply them. A 3,000ft view that I never recognized when studying math in school. Thank you for this window into the next level of understanding mathematics.
I have discovered your videos recently, and I noticed you always detail how you get from (a+b)² to a²+2ab+b². I don't know in your country, but in France we are taught what we call "identités remarquables" (remarkable identities) that we are supposed to know: (a+b)² = a²+2ab+b² (a-b)² = a²-2ab+b² (a+b)(a-b) = a²-b²
Purple, this was really good. It would be cool to have a test that's like one massive question like this. except I realized there's a big issue if I'm wrong
Amazing Videos Andy, thn a lot for awakening a long time sleeping knowledge in my brain :-) btw. that is the application you use to move around all the shapes and formulas so easily?
mint! at that point where there are too many x inside the equation i would probably just look at the multiple option and try the answer one by one see which one is correct
It could be shortened if you find the C for the big triangle and substract with x and then put it into the c^2=a^2+b^2 formula and it wil be like this 100=(x-3)^2 + (( surd 2x^2+2x+25)-x) and you will get x=11
Oh my God, literally today I was working on a problem and realized I could use the Rational Zero Theorem but I didn't remember it, and couldn't find anything in writing that I could understand. However, seeing it actively performed in front of me really helps with that! I also didn't realize, but I guess I needed a refresher on polynomial long division! Oh and if you're curious, the math problem was: "A+B+C=1 A²+B²+C²=2 A³+B³+C³=3 What is (A,B,C)?" I'm only using algebra and so far that's been cruel, I'm seven pages into it and probably inefficiently solving it, but I'm doing it until I got the answer without a calculator for some reason!
Red, though I first worked through all the ways to use quadrilateral formulas. The other missing side length is non integer and made a hash of things. Doing the quadrilateral area formula against the sum of triangles became a mess. Discarding that I did as you dud in a more convoluted way. Rather than struggling through finding the roots of the quartic, I threw it into a spreadsheet, plotted and quickly solved it that way. Though not proper for this sort of solution, 50 years of engineering on - I use them a lot to bash through problems. But then too, I tend to solve for all of the missing elements and nit just the one sought in the problem statement. Force of habit.
At this type questions i first check if there is a special triangle. (3-4-5, 5-12-13, 8-15-17, 7-24-25) Difference of big triangle's edges is 7. so i tried 8 15 17 triangle and it fit perfectly.
Purple! I hated math in school (back in the 50's and 60's - crappy teachers made math boring/confusing) -- but I love solving puzzles and I've always been good with all the sciences (except for the math part that I have just always avoided). Anyway - watching your videos makes math seem fun and maybe it's not too late for me to start learning math at 70 years old?
