3Blue1Brown Episode 1 Lockdown Math, A simpler version of the quadratic formula.: https://www.youtube.com/watch?v=MHXO86wKeDY https://www.youtube.com/watch?v=MH

3Blue1Brown Episode 1 Lockdown Math, A simpler version of the quadratic formula.: https://www.youtube.com/watch?v=MHXO86wKeDY https://www.youtube.com/watch?v=MHXO86wKeDY It's a good lesson, thanks to him I can derive the whole from my mind. It did take some time with the second part (finding deviation d), but I got there :) I recommend to watch the video, for people who prefer to read, I'll derive the whole thing from my mind (the video makes it quite intuitive) and write it down. ___DEMONSTRATING THE 3BLUE1BROWN AWESOMENESS OUT OF MY OWN MIND___ I wasn't doing my best to memorize the intuition when he gave the lecture, otherwise it wouldn't have taken me 60 minutes to write it down ;-) It starts with a sketch of a parabola that has 2 roots. Something like: x^2-4, see a sketch: https://www.desmos.com/calculator/q8mkfjmylc https://www.desmos.com/calculator/q8mkfjmylc It then shows that the intuition of the quadratic formula is that it's basically the midpoint m plus or minus the deviation d. So for root left and root right you get: l, r = m ± d Sketch a parabola yourself, take the midpoint and draw arrow -d and arrow +d to the roots to visually verify that this is true. In order to make this slightly easier, he suggests to immediately take the formula ax^2+bx+c=0 and rescale it to x^2+bx+c=0 and work from there. The bx and c are not the same symbols, you should see it as b-prime and c-prime in the second equation. ___GETTING THE MIDPOINT m___ Now you need to figure out what m is. m is the midpoint, how do we calculate the midpoint algebraically? Well, if you'd average out the roots, then you'd get the midpoint, so m = (l+r)/2. "But we don't know the roots!" That's true, but suppose we would, how would that look like? It would look like: y = (x-l)*(x-r) for roots l and r Expanding it makes: x^2 -l*x -r*x + l*r --> x^2 - x(l+r) + l*r Comparing it to x^2 + bx + c this means that: b = -(l+r) [**] and c = l*r Remember that m = (l+r)/2. So we can somehow substitute -(l+r) for some version of m. Let's rewrite to see what it is. m = (l+r)/2, 2*m = l+r, -2*m = -(l+r). -2*m = b (see [**]) therefore b = -2*m. Let's rewrite m in terms of b. -b = 2*m, -b/2 = m. So m = -b/2. We have our m in ways we understand! m = -b/2 from our rescaled x^2+bx+c=0 equation. ___GETTING THE DEVIATION d___ How do we get d? 3Blue1Brown shows a very cool thing here about midpoints geometrically and numerically. 3*5 = 15, 4^2 = 16 1337 * 1339 = 1338^2 - 1 (checking if correct: 1790243 = 1790244 - 1, yep) So the closer the roots are to the midpoint, the closer it's looking as if you're multiplying the midpoint with itself (makes sense). How does it look like you make the deviation big? Resuming with 3*5 (my own insights, 3B1B probably showed similar things). l*r = c, m^2 = q (I use c because it's related to the c from x^2+bx+c = 0, I use q because I need a letter) 3*5 = 15, 4^2 = 16 2*6 = 12, 4^2 = 16 1*7 = 7, 4^2 = 16 [*] The deviations are clear to see, they are: 3,5 = 4 ± 1 2,6 = 4 ± 2 1,7 = 4 ± 3 Observation: multiplying the roots l and r gives c. c is somehow related to the deviation as we can see that in the first case 16 - 15 = 1 which is the deviation we want. Let's try the following, and see how close we get to the deviation we want: d1 = 16 - 15 = 1 --> seems to check out d2 = 16 - 12 = 4 --> does not check out, we need to square root it d3 = 16 - 7 = 9 --> does not check out, we need to square root it What did we just calculate? m^2 - c = d^2 I find it hard to explain why this is the case. 3Blue1Brown might have an explanation for it. I simply feel the following "if you square a midpoint, and you multiply 2 points equally far away from the midpoint in opposite directions, that's kinda like weirdly squaring a midpoint, so there must be some relationship there. The further those 2 points are, the bigger the deviation must be. This deviation cannot be d since we're talking about squares here. Ah m^2 - c = d^2! Yea that seems to look alright." Square rooting it we get: d1 = 1, from m = 4, l,r = m ± d --> 3,5 = 4 ± 1 d2 = 2, from m = 4, l,r = m ± d --> 2,6 = 4 ± 2 d3 = 3, from m = 4, l,r = m ± d --> 1,7 = 4 ± 3 So let's put it into a formula. d^2 = m^2 - c --> e.g. 3^2 = 4^2 - 7 (see [*] ) So d = sqrt(m^2 - c) ___PUTTING IT TOGETHER___ We know the quadratic formula is l, r = m ± d for x^2+bx+c = 0 which was rescaled from ax^2+bx+c=0 (again the bx+c symbols scale with it since I am too lazy to write b-prime and c-prime :D). We know m. m = -b/2 We know d. d = sqrt(m^2 - c) So the whole thing for left root l and right root r: l,r = -b/2 ± sqrt(m^2 - c) for x^2+bx+c=0.