Can the quadratic formula still work if a, b, c are not constants? 

I created an incredible example based on this: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-j6ri-2S-hxU.html

I mean what else are you supposed to think about in the car

The black and red pens at the back are the most underrated thing on the frame though

The quadratic formula is essentially just the completing the square method, which is just algebraic manipulation, so it definitely works for any equation. Whether or not completing the square is actually useful in that case, though, is a different story

Somebody stop him, he's getting too powerful

The first thing i thought after seeing the thumbnail is "What!!! is this man going insane?"

I took numerical analysis back in my undergrad. One technique to solve non-standard equations, was to find an initial guess, and then somehow find a way to isolate x, and just repeat the process. Similar to Newton but more basic

This gives me similar vibes to that equation you did long time ago about a quadratic equation in terms of 5.

" a, b, c Qua-, dra-, tic" 😂

At this point, bro is waiting for math 2 to release.

Now plug a quadratic equation into a quadratic equation

Hey bprp I was scrolling through your older videos and I saw that you made two promises 1. Sing the quadratic formula song! 2. Solve an easy integral in German language! :-) Can you please do them? Maybe just a Instagram reel but please fulfill your promise!

After getting the first solution, we can just check the original equation is even, so with that we know x_2 = - x_1.

Newton's method (introduction & example) ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-iVOsU4tnouk.html

My first reaction. It should produce a valid "value" for x for any a(x),b(x),c(x) as long as a(x)0. After-all it's just completeing the square. However, the solution (as you point out) may not produce anything useful or simpler. But your sin(x) example is pretty cool as you get a different equation that has the same roots - and that is definitlely very cool. The formula is sort of acting like a transformation - well in my mind anyway.

This is really interesting as the quadratic formula can be derived just from algebraic manipulation so obviously it can work with non-constant terms. But I’ve never seen anyone use it that way and never considered doing that before either! The only condition I can think of is a=/=0 but you can just check the a=0 case before doing the quadratic. Very fascinating stuff!

Another cool example, where we would use the Lambert W function is b=3exp(x) and c=-4exp(2x), we'd get 25exp(2x) under the square root which is 5exp(x)

hi so u can actually solve the previous cubic in this way using the quadratic formula the way u showed u get the eq x=(-1+sqrt(25+24x))/x if u rationalize this u get (2x^2+5)^2=25+24x if simplified u get 2x^3+5x-12=0 if u subtract both eq u get x^3=6 as one solution then using x-6 as a divisor u can find the other roots

If I was given that equation with sin(x) and cos(x)^2 , I would have just used cos(x)^2 = 1 - sin(x)^2 and then I have a quadratic equation in sin(x). Just use the quadratic formula on that. That seems more straightforward to me than using the formula on x.

An interesting application would be for finding roots for equations of the form ax^3+bx+c/x, doing the trick from this video, then multiplying by x and taking the square root for a final solution.

You just can't get enough of abusing the quadratic equation every time you upload :)

Just a small doubt on my part. while we derive the quadratic formula, it is usually established that a is not zero(since it wouldn't be a quadratic if it was). However if a, b and c can be functions in x(How does this generalize to all functions? Piecewise functions too?) zero(or any root of a) can be a perfectly legible root for the equation but the denominator for the quadratic formula can become undefined. Eg. (x-2)*x^2 + x - 2. I am really curious about this idea, but I don't think we will be able to derive powerful conclusions without rigorously answering these questions. I highly enjoyed your video nonetheless. Thank You.

Blackpenredpen is back again!!!

Where did you buy the euler on the wall?

Hey you should go over that Terrance Howard thing where (sqrt(2))^3/2=sqrt(2) and explain it in your own words.

Easier example than the ones in the video: x³ - 5x + 4/x = 0. You can take a = x, b = -5, c = 4/x and get x = (5 +- 3)/(2x), and multiplying by x gives the two equations x² = 1 or x² = 4, yielding the four solutions x1,2 = +- 1 and x3,4 = +- 2. (Obviously, you could also get these solutions by multiplying the original equation with x and then solving that bi-quadratic equation by standard methods.)

For the end, there's an easier way to think about the second root. It actually make sense that the values are opposites as the original function is even. x^2 is even, xsinx is even as it's the product of two odd functions, and -1/4cos^2(x) is even because cosine is even. The sum of even functions is even. f(x) = f(-x), so f(0.335418) = 0 = f(-0.335418)

Bros taking quadratic to whole another level 💀

Recently, I was looking at the equation ta^2+xa+y=0, where t is a function of x and y and assuming a is constant. I'm pretty sure, by just taking for granted the applicability of the quadratic formula to solve for a, you can eventually solve for t and y as functions of x. For example, for t(x) I got: t(x)=-2x/a-2/a^2+-[(x/a)^2+(8/a^2)(x/a)+4a^-4]^1/2 and I'm pretty sure that putting in my y(x) and t(x) into the original equation does yield a valid solution. However, tbh I don't understand why my approach was justified, and I'd love to understand it better. Edit: if it wasn't clear, I didn't start assuming t and y were variables of x. I assumed t, x, and y were implicitly related and tried to solve for t in terms of x and y, and then eventually you get another quadratic equation involving x and y and solving for y in terms of x. Plug that y(x) into t(x,y) to get t(x)=t(x,y(x))

It works more generally on any commutative ring of characteristic either 0 or not divisible by 2 when a is invertible. This is a particular case. Also i'm sure people have used the quadratic formula like this, unlike you claim in the discription, i think i have before when solving some differential equation.

try x^5+x^3+x=0, where a=x^3, b=x^2, and c=x. Not as impossible as it looks because we can divide out an x, but it's still a quartic. But you can get all 4 complex roots using the trick you showed in this video.

Sir, How to know the no. Of rows in DI method? For intergration

UR CHANNEL is the BEST PROF)

Nice thought!!!

Newtons method was another programming execise I did on my Sinclair ZX81 in the early 1980s. I did have to explicitly program in the x function ands its derivative . It was very satisfying. I did need the extended memory module for this though, which was all of 16k.

Good old functional analysis. It works point by point so you can define the solution.

I like your Curiosity because im the same way with numbers with different formulas

Espectacular... gracias.

This is very cool!

Could someone link me the vide about Newton thing? I dont see it in the description

Thanks for the vedio

i really like this even if i don't understand everything because i still lack some math knowledge that is used in this video... but i would love to learn it and hopefully school will teach it to me soon lol.

I completed the square in my mind I knew it's fine but still wanted to watch

thanks sir

can that be done with matrices?, at least the nxn ones?

I knew i was not the only one who thought about this stuff on weird times!!! I also wonder upon this while trying to sleep..

Why didnt you solve with the newton's method to solve the original equation directly?

pretty cool!

This channel is just epic

I’m glad to have woken up early and saw your new post 😊

Hey @blackpenredpen I have a question for you What does it 'exactly' mean by 'taking log both sides' Let's say I have an equation x-1=3 so the usually it's said that '-1' is transferred to the other side and it's sign changes but in reality we add +1 on both sides making x=4 Now let's say logx(100) =2 So by saying taking log both sides Does it mean that log(logx(100)= log(2) ??? I don't understand this concept can you explain this in a video my teachers never really write this step and jump to the answers 😅

hmmm, recursion might be neat, where a=ax^2+bx+c.

A question: I understood that you could integrate over an interval (because that's what you do in your final year), by considering the limits of this interval, but in prep school, you integrate over different domains, so what's the point of integrating over an edge, a surface or something else?

How fantastic man

Are you excepting challenge problems?

Good job now can you think what is the relationship between the roots and the vertex of the parabola ?? TAI BPRP

Please try to integrate 1/x^5+1

Great

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ax² = qua bx = dra c = tic 0 = equation

We can now discover warp drive, alcubiere drive, Dyson sphere and time travel. What a discovery

Using the fact that [f(D_x)^2 - 1] y = 0 => [f(D_x) +/- 1] y = 0, you can show that, given ' [b(D_x) / a(D_x)] y ' exists, the quadratic formula also holds for certain sorts of differential operators a, b, c. (Analytic functions evaluated at the d/dx should work.) Notably, this is _not_ useful in solving most DEs, as I am aware of no techniques for solving the sort of non-linear non-sense that is x*y = [ - b(D_x) +/- sqrt( b(D_x)^2 - 4a(D_x)c(D_x) ) / 2a(D_x) ] y Best I can see is that the value of a solution at x=0, after applying the Quadratic-Formula'd Operator, would be 0. It is nifty to be able to isolate the non-constant part of the coefficients of a (very particular) DE, I guess.

The quadratic equation has been used before in numerous ways...like finding the inverses of the hyperbolic trig functions

I need you help: Integral of ((x^2 + sin x)/(x^2 +1))dx

My answer to this question is "That depends." For instance, it would not work with x*x^2 + b*x + c = 0. It would work with x^2 + b*x*y + y^2 = 0 or x^2 * x^2 + b * x * x + c = 0 though.

Thanks for an interesting way to use the quadratic formula. I notice if you are trying to solve a linear equation, a=0, and the formula seems to break down.

How about: a=x, b=2x, c=x. Following your approach we get to x=-1. But we somehow lose the alternative x=0 solution.

I’m at the beginning of the video. It seems that it should be possible for this to work although I’ve never even considered it before.

Overall, I wonder why there was no call to the intermediate value theorem in order to talk about how much solution there is. It would be because he didn't want to bother talking about it simply enough, but I rarely saw it precised in any of his videos sadly

Is there not a 'Lambert-sine' kind of function for analytically solving equations that involve x and sin(x)?

Meanwhile, in an alternative universe, BPRP wonders if he can use the quadratic formula if a, b and c are not constants, but functions of x. He realises that this could give something nice if the discriminant is a perfect square, so the square root can be eliminated. For good measure, he decides that the "quadratic" will be in sin x, rather than just x. Now for the discriminant Δ: If we let a=1 and b=4x (the coefficient 4 is to avoid fractions when solving for c), then b²-4ac=16x²-4c. If we want Δ=4 (a perfect square), then 16x²-4c=4, 4c=16x²-4, c=4x²-1 So he gets the "quadratic" equation in sin x: sin²x+4x sin x+4x²-1=0, with "solutions" given by the quadratic formula: sin x=(-4x±2)/2=-2x±1 which can be rewritten sin x+2x±1=0 He then gamely uses the Newton-Raphson method to find a numerical solution of each to 5dp. As luck would have it, each of the new equations has only one real solution and the original equation exactly 2 real solutions. As the "quadratic" is even in x, the solutions are opposites. He then notices that the "quadratic" in sin x he found is actually also a "quadratic" in x itself, so he now rewrites the equation in "standard" form, so as to be able to apply the quadratic formula again, but this time to a "quadratic" in x: sin²x+4x sin x+4x²-1=0 4x²+4x sin x+sin²x-1=0 4x²+4x sin x-cos²x=0 x²+x sin x-¼cos²x=0 For the rest of the story, see the video...

People should be glad Police can't check Maths reading like Alcohol reading

Your question seems to be highly connected to abstract algebra and Galois theory and maybe even algebraic geometry. Yes, a cubic (and quartic) equation exists, but iirc, you cannot write them down without nested roots, so your quadratic formula method wouldn’t get far. The proof for this I believe involves studying ring theory and examining field extensions. Then, once you get to 5th degree polynomials, you can no longer do this; there exist 5th degree polynomials with no roots which can be written with just addition, multiplication, and exponentiation, namely x^5-x-1. This is proven with Galois theory. As for your last example, you could have done exactly that with completing the square, and that’s a common technique used in more advanced studies over all sorts of algebraic rings and fields. My favorite example of such a thing is the derivation of the “Dirac equation” in physics, where you quite literally use the quadratic formula and get 2 solutions - one which corresponds to regular matter, and the other which corresponds to antimatter, which was the first prediction that positrons exist.

I used it on constant numbers once and it worked lol 😂

Yes it works, you are doing algebraic manipulation In every step (1) to (2), you are actually saying (for every x, if x is a solution to (2), x is a solution to (1)), the "for every x" effectively fixes a,b,c. For example: a(x)x=b(x) to x=b(x)/a(x). You are saying " For every possible x, if x' (one of the possible values of x) makes x'=b(x')/a(x'), then x' will also make a(x')x'=b(x') " Given correct assumptions, you can also say the converse. In the quadratic formula, just make sure a(x)=/=0 and every step is equivalent, you won't lose any roots. This is just rarely useful, isolating x and bundling every other functions with x together with a square root is rarely useful when you want to solve actual answers. However, this is quite useful in finding a numerical answer, if the right side with a(x),b(x),c(x) is convex around the answer, you can iterate to it

means, solving cubic or quartic is easier now

Interesting

Man… I want to Taylor expand the RHS so bad. Maybe to three terms, keeping the x^2 term… apply the quadratic formula again… 😅

❤

Degree 4 polynomial has radical solution formula. Use a(x) = mx + c. Now any degree 5 polynomials has radical solution formula. QED.

I have a question. What is the square root of -i?

Bro boutta add a chapter in the curriculum

Back when I was an undergrad in Chem Engineering (40 years ago), we didn't bother with closed form solutions. I just pile the terms on both sides of the equation, make a program in BASIC on my Sharp handheld - yes, they had computers back then - and let the program keep iterating until both sides of the equation equals each other.

cool

I guess its better to treat them as constants

Is this already proven to work? Would be interesting to see a rigorous prove that this applies for all functions of c

Did dr peyam retire ? 😢😢😢

Okay but if you need to do newtons, you could have just done it for the original functuon

I really liked ✌🏼QUA ✌🏼 DRA ✌🏼 TIC equaion

Even calculus is afraid of this man , somebody please stop him !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Finally, a worthy opponent.

"I was thinking about the quadratic formula. As one does."

Where is the beard? It makes you look wise.

Out of millions thoughts he only thought about the quadratic formula I mean he operated this formula multiple times already 😂

0:00 Here's the story Of a lovely lady Who was bringing up three very lovely girls a, b and c

that's helpful but i think that second degree equations must have 2 solutions : X1 and X2

The x’th root of the i’th root of the pi’th root of the i’th root of i = the (√(6x) + √(-x))th root of e^x . Solve for x

it should work because the process is nothing but just factoring terms.

YAY FIRST! I wuv the quadratic formula ❤

This is essentially just completing the square though isn't it?