The quadratic formula: a tutorial
3Blue1Brown Episode 1 Lockdown Math, A simpler version of the quadratic formula.: <a href="https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=MHXO86wKeDY" rel="nofollow">https:&#x2F;&#x2F;www.youtube.com&#x2F;watch?v=MHXO86wKeDY</a><p>It&#x27;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 :)<p>I recommend to watch the video, for people who prefer to read, I&#x27;ll derive the whole thing from my mind (the video makes it quite intuitive) and write it down.<p>___DEMONSTRATING THE 3BLUE1BROWN AWESOMENESS OUT OF MY OWN MIND___<p>I wasn&#x27;t doing my best to memorize the intuition when he gave the lecture, otherwise it wouldn&#x27;t have taken me 60 minutes to write it down ;-)<p>It starts with a sketch of a parabola that has 2 roots. Something like: x^2-4, see a sketch: <a href="https:&#x2F;&#x2F;www.desmos.com&#x2F;calculator&#x2F;q8mkfjmylc" rel="nofollow">https:&#x2F;&#x2F;www.desmos.com&#x2F;calculator&#x2F;q8mkfjmylc</a><p>It then shows that the intuition of the quadratic formula is that it&#x27;s basically the midpoint m plus or minus the deviation d.<p>So for root left and root right you get:<p>l, r = m ± d<p>Sketch a parabola yourself, take the midpoint and draw arrow -d and arrow +d to the roots to visually verify that this is true.<p>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.<p>___GETTING THE MIDPOINT m___<p>Now you need to figure out what m is.<p>m is the midpoint, how do we calculate the midpoint algebraically?<p>Well, if you&#x27;d average out the roots, then you&#x27;d get the midpoint, so m = (l+r)&#x2F;2. &quot;But we don&#x27;t know the roots!&quot;<p>That&#x27;s true, but suppose we would, how would that look like? It would look like:<p>y = (x-l)*(x-r) for roots l and r<p>Expanding it makes: x^2 -l*x -r*x + l*r --&gt; x^2 - x(l+r) + l*r<p>Comparing it to x^2 + bx + c this means that: b = -(l+r) [**] and c = l*r<p>Remember that m = (l+r)&#x2F;2. So we can somehow substitute -(l+r) for some version of m. Let&#x27;s rewrite to see what it is.<p>m = (l+r)&#x2F;2, 2*m = l+r, -2*m = -(l+r). -2*m = b (see [**])<p>therefore b = -2*m. Let&#x27;s rewrite m in terms of b.<p>-b = 2*m, -b&#x2F;2 = m. So m = -b&#x2F;2.<p>We have our m in ways we understand! m = -b&#x2F;2 from our rescaled x^2+bx+c=0 equation.<p>___GETTING THE DEVIATION d___<p>How do we get d?<p>3Blue1Brown shows a very cool thing here about midpoints geometrically and numerically.<p>3*5 = 15, 4^2 = 16<p>1337 * 1339 = 1338^2 - 1 (checking if correct: 1790243 = 1790244 - 1, yep)<p>So the closer the roots are to the midpoint, the closer it&#x27;s looking as if you&#x27;re multiplying the midpoint with itself (makes sense). How does it look like you make the deviation big?<p>Resuming with 3*5 (my own insights, 3B1B probably showed similar things).<p>l*r = c, m^2 = q (I use c because it&#x27;s related to the c from x^2+bx+c = 0, I use q because I need a letter)<p>3*5 = 15, 4^2 = 16<p>2*6 = 12, 4^2 = 16<p>1*7 = 7, 4^2 = 16 [*]<p>The deviations are clear to see, they are:<p>3,5 = 4 ± 1<p>2,6 = 4 ± 2<p>1,7 = 4 ± 3<p>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.<p>Let&#x27;s try the following, and see how close we get to the deviation we want:<p>d1 = 16 - 15 = 1 --&gt; seems to check out<p>d2 = 16 - 12 = 4 --&gt; does not check out, we need to square root it<p>d3 = 16 - 7 = 9 --&gt; does not check out, we need to square root it<p>What did we just calculate?<p>m^2 - c = d^2<p>I find it hard to explain why this is the case. 3Blue1Brown might have an explanation for it. I simply feel the following &quot;if you square a midpoint, and you multiply 2 points equally far away from the midpoint in opposite directions, that&#x27;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&#x27;re talking about squares here. Ah m^2 - c = d^2! Yea that seems to look alright.&quot;<p>Square rooting it we get:<p>d1 = 1, from m = 4, l,r = m ± d --&gt; 3,5 = 4 ± 1<p>d2 = 2, from m = 4, l,r = m ± d --&gt; 2,6 = 4 ± 2<p>d3 = 3, from m = 4, l,r = m ± d --&gt; 1,7 = 4 ± 3<p>So let&#x27;s put it into a formula.<p>d^2 = m^2 - c --&gt; e.g. 3^2 = 4^2 - 7 (see [*] )<p>So d = sqrt(m^2 - c)<p>___PUTTING IT TOGETHER___<p>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).<p>We know m. m = -b&#x2F;2<p>We know d. d = sqrt(m^2 - c)<p>So the whole thing for left root l and right root r: l,r = -b&#x2F;2 ± sqrt(m^2 - c) for x^2+bx+c=0.