# Proving the quadratic formula using calculus? (just my attempt)

## Метаданные

- **Канал:** blackpenredpen
- **YouTube:** https://www.youtube.com/watch?v=tPUtHKriq_4
- **Дата:** 10.04.2026
- **Длительность:** 6:41
- **Просмотры:** 34,775
- **Источник:** https://ekstraktznaniy.ru/video/51427

## Описание

I have been wondering if we could prove the quadratic formula using calculus, so here's what I came up with. It still involves mostly algebra but I will try to see if I can use Newton's method of doing so soon!

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## Транскрипт

### Segment 1 (00:00 - 05:00) []

Today, I'm going to attempt proving the quadratic formula by using calculus, and this is what I came up with. Check this out. First, we start with the quadratic equation in the standard form, and the goal is to get X by itself. First, let's take a look at the picture. Let's say our parabola looks something like this. This is y equals ax squared plus bx plus c. And the solutions to this equation will be the x-intercepts here and here. Yeah, I know. Sometimes the parabola might not have x-intercepts because it may be up here and open up or down here open down. In that case, you can just add or subtract constants and work that out, but I'll leave that to you guys. So, let's say this is what we have. Now, just be told, the next step is the only step that needs the calculus. What I'm going to do next is I'll be using calculus to find out at what X we will get the vertex. And that's precisely where the slope of the tangent line is equal to zero, and that's y prime equals zero. So, look at this and do the derivative. y prime is equal to 2ax + b, and we want this to be zero. Solving for X, you get this will be at X equals -b over 2a. And as we all know, this right here is also the vertex formula of a parabola. Good. Now, what we are going to do next is, for simplicity purpose, I'm going to shift the parabola so that it's going to be symmetric at the center, y-axis, right here. So, the picture will look something like this now. Hmm. How can we move it, though? Well, I'm going to do a substitution real quick. Let We use X already, we use Y already. You can use Z, but that's like too complex. I'm going to use T. What do I do with T? I want T to be X plus B over 2a. And the reason why is because if you look at the blue picture, this picture is in terms of X, and the vertex is at -b over 2a. Now, if you put X right here, -b over 2a + b over 2a, T is zero. That's where the vertex that one for the T axis. All right. And what I'm going to do is I'm just going to subtract this from both sides, and we get X is equal to T minus b over 2a. And then put this into all the X's, and we get a times t minus b over 2a squared plus b 2a, and then plus c. We want this to be zero. And then just go ahead and work this out. Squaring this, we get t squared minus two times this times that, and then plus this thing squared, so it's b squared. Let me actually just write it out. b over 2a squared. And then plus distribute bt, and then minus b squared over 2a, and then plus c is equal to zero. And now we just distribute the a. That will give us at t squared, and then two and two cancel. A and a cancel, so that's minus bt. And then this right here is b squared on the top over four. And this right here will be a squared, but we multiply the a right here, so it's just going to be a to the first power. And then right here we will have minus bt minus or plus bt minus b squared over 2a, and then plus c. That's equal to zero. And the beauty of doing this is that we don't have the T terms anymore. And then right here we can combine like terms. We just need to multiply this by two and two. Okay, so we get at squared, and this term and that term have the same denominator now, and that will be -b squared over 4a. And then I need a plus c. And then let's move these two terms to the other side. That's at squared equals b squared over 4a plus c, so minus c. And then right here let's get a common denominator. So, 4a 4a. This looks really good, right? So, at squared equals b squared minus 4ac all over 4a. Okay. And then divide both sides by a. Let's multiply by 1 over a on both sides

### Segment 2 (05:00 - 06:00) [5:00]

like this so that this and that cancel. So, we get t squared equals b squared minus 4ac all over 4a squared. And then do the usual thing. Take the square root both sides. Plus or minus cancel. So, we get T, which is going to be plus or minus the square root part stays on the top like this. But on the bottom, square root of four is two, and square root of a squared is Technically, the absolute value of a, but we have the plus minus already, so we don't have to worry too much. Now, T is what? T is X plus b over 2a. So, I'm going to just change that right here. And then finally, we just have to move this to the other side, and they both have 2a on the bottom. So, ladies and gentlemen, X is equal to -b plus or minus square root of b squared minus 4ac all over 2a. Yep. This is it. I know the calculus part is only like maybe like 5% of the entire proof, but I did kind of use some calculus. I have to admit that the majority is the algebra because this is still an algebra question. Yeah. And I also tried to use the Newton's method to prove the quadratic formula, but I still end up doing a lot more algebra than the actual calculus. So, I'm going to try to think about how I can really use calculus to prove the quadratic formula. If one day I can figure it out, maybe I have a better hope with Newton's method. In that case, I will let you guys know. Anyways, that's it.