ALL solution to a triple exponential equation 8^x+4^x+2^x=14

[@blackpenredpen]

Solving 9^x+3^x=1 👉 kzbin.info/www/bejne/bYHUdY2anr-MncU (calculus tutorials channel)

[@축복-l1l]

0 x2 0= take isit

[@sudirosudiro4218]

Proof of Laplace Transform of Dirac Delta kzbin.info/www/bejne/aHisiXyunLZ4mJI

[@SuperYoonHo]

@@축복-l1l hi n sorry bwt ur jokes aren't funny

[@b.m.7033]

I nailed x=1.

[@YTBRSosyalEmre]

I hammered x=1

[@paolobasarini9586]

I pinned x=1

[@claws811]

I smashed x=1.

[@nathanielb3510]

I also teach pre-calculus, my favourite part is making bonus questions for quizzes. We're actually covering synthetic division right now! Love your videos.

[@blackpenredpen]

Thank you.

[@PicklePercyy]

make this a bonus question for your next test

[@MUJAHID96414]

Can you teach me calculas .I'm excited 😄

[@fuji_films]

@@MUJAHID96414 Go to school!

[@alberteinstein3612]

You’re one of those math teachers that post KZbin videos that directly relate to the lessons you’re teaching :) That’s what my Calculus teacher does

[@michapodlaszuk9025]

I thought you dropped out at the age of 15

[@pezasus]

This reminds me of a mind your decisions question, 4^x + 6^x = 9^x I’m not sure whether you’ve covered it before but it’s really satisfying to solve

[@blackpenredpen]

Oh yea that’s a really fun one.

[@sharpnova2]

that was a nice problem

[@lesterdavidoff9869]

How did you solve it

[@pezasus]

@@lesterdavidoff9869 I'll give a hint first; try dividing by each of the terms to find something that looks familiar.

[@Maths_3.1415]

Integer solution Solve for x over the integers: 2^x + 4^x + 8^x = 14 Simplify and substitute y = 2^x. 2^x + 4^x + 8^x = 2^x + (2^x)^2 + (2^x)^3 = y^3 + y^2 + y: y^3 + y^2 + y = 14 Subtract 14 from both sides: y^3 + y^2 + y - 14 = 0 The possible rational roots of y^3 + y^2 + y - 14 are y = ± 1, y = ± 2, y = ± 7, y = ± 14. Of these, y = 2 is a root: y = 2 Substitute back for y = 2^x: 2^x = 2 2 = 2^1: 2^x = 2^1 Equate exponents of 2 on both sides: Answer: | | x = 1

[@sr.tarsaimsingh9294]

"Blackpenredpen Yay!!!! " I miss that kind of initial startup. 🧐

[@andrewlees3946]

That polynomial division is so much easier than the method we were taught! Any chance we can get further explanation about how that works?

[@blackpenredpen]

Here’s a video on the comparison between synthetic division and the polynomial long division kzbin.info/www/bejne/nYaze52Ihb50a8k

[@alro3553]

It is called ruffini's method if you want to check it out!

[@DrDeuteron]

I agree, I was never satisfied with my polynomial division skills, and I blame the method. It has only come up a few times in my work (QCD phenomenology), though.

[@raytheboss4650]

nothing better than homemade baked triple exponetial equation for breakfast!

[@blackpenredpen]

Hahaha

[@janami-dharmam]

constipation or loose motion; you choose!

[@raytheboss4650]

@@janami-dharmam i chose sextic polynomial equation

[@kavyanshtyagi2563]

why dont you start teaching mathematics of olympiad level on youtube , you would be the best !!!!

[@mastershooter64]

I know right, that'd be awesome, and the putnam too!

[@justinpark939]

well, he does have an IMO problem walkthrough at the very least. I asked if he could take in some problems, but he said he was busy for this semester but feel free to give requests via his buisness email (I think I used).

[@thomasblackwell9507]

What do you mean by that. He is the BEST!

[@PubicGore]

Because he's not good enough for it. I'm not trying to be rude, he's said it himself several times. You should check out Michael Penn, he's very good at explaining and he does tons of math contest problems.

[@qsykip]

Olympiad questions are a very different beast. They’re not necessarily hard per se, but there are a lot of clever tricks that you need to learn. Being good at Olympiad questions doesn’t necessarily make you good at uni maths, and the same is true vice versa. Not saying there aren’t a ton of people that are good at both though, just that they are very different skills.

[@derwolf7810]

1:58 "Okay, how do we solve this cubic equation? Hmm, we cannot factor this; not even by grouping, right?" I would suggest to use the Cardano formula to solve any cubic equation of the form A t^3 + B t^2 + t y + D = 0; although i should admit that the simplification of u and v easily might become hard to find/see, so the results might look awkward. Define a := B/A, b := C/A, c := D/A, p = (-1/3 a^2 + b) q = (2/27 a^3 - 1/3 a b + c) z := u + v = t + a/3. ==> t^3 + a t^2 + b t y + c = z^3 + p z + q (Edit: Corrected this line) If z is of the form u+v (u + v == z) then z^3 + p z + q = z^3 + (-3 u v) z + (- u^3 - v^3) and with delta = (q/2)^2 + (p/3)^2 the following values of u and v are fulfilling the above condition: u = cbrt(-q/2 + sqrt(delta)) and v = cbrt(-q/2 - sqrt(delta)) Then the solutions for t are: t = (u + v) - B/(3 A), t = -(u + v)/2 - B/(3A) + i sqrt(3)/2 (u - v) and t = -(u + v)/2 - B/(3A) - i sqrt(3)/2 (u - v) The values here are: a := B/A = 1, b := C/A = 1, c := D/A = -14, p = (-1/3 a^2 + b) = (-1/3 1^2 + 1) = (-1 + 3)/3 = 2/3, q = (2/27 a^3 - 1/3 a b + c) = (2/27 1^3 - 1/3 1 1 - 14) = (2 - 9 - 378)/27 = -385/27, delta := (q/2)^2 + (p/3)^3 = 148257/(2 3^3)^2, sqrt(delta) = sqrt(148257/(2 3^3)^2) = sqrt(2372112)/6^3, u = cbrt(-q/2 + sqrt(delta)) = cbrt(-(-385/27)/2 + sqrt(2372112)/6^3) = cbrt(1540 + sqrt(2372112))/6 = (7+sqrt(57))/6, v = cbrt(-q/2 - sqrt(delta)) = cbrt(1540 - sqrt(2372112))/6 = (7-sqrt(57))/6, (u + v) = (7+sqrt(57))/6 + (7-sqrt(57))/6 = 7/3, (u - v) = (7+sqrt(57))/6 - (7-sqrt(57))/6 = sqrt(57)/3 So the solutions here for t are: t = (u + v) - B/(3 A) = 7/3 - 1/(3 1) = 2, t = -(u + v)/2 - B/(3A) + i sqrt(3)/2 (u - v) = -7/3/2 - 1/(3 1) + i sqrt(3)/2 sqrt(57)/3 = -3/2 + i sqrt(19)/2 t = -(u + v)/2 - B/(3A) - i sqrt(3)/2 (u - v) = -3/2 - i sqrt(19)/2

[@blackpenredpen]

😮 this is so cool!

[@janami-dharmam]

You did not explain the steps: A,B,C and D can be reduced to three terms without loss of generality and the roots will be unchanged. It is always possible to remove the quadratic term by a suitable choice of variable t:= t+a; Then we get u^3+p*u+q=0; cf: (a+b)^3-3*a*b(a+b)-a^3-b^3; then the rest falls in place. (p=-3*a*b and q=-a^3-b^3; solve for a and b and x=a+b) Details here: sites.oxy.edu/ron/math/312/14/projects/Fernandez-Gosselin.pdf

[@derwolf7810]

@@janami-dharmam Yes, i could have explained why that works. I just wanted to give an example analogue to the formula (x_1,2 = (-B±sqrt(B^2 - 4AC))/(2A)) for quadratic equations (A x^2 + B x + C), which typically is also used without explanation. I admit, given that readers might not have heard of that method, an explanation would have been nice. On the other hand, i think the other parts also might need additional explanation and i am too lazy for a full proof (at least in a ascii-text-form), so i only gave a rough overview.

[@vannchen9422]

I just took a math competition and will appreciate if you went over this problem: what is the sum of (1/2) + (1/3 + 2/3) + (1/4 +2/4 + 3/4) … + (1/100 + 2/100 + 3/100 +4/100 … + 99/100). Thanks and have a great day!

[@hannahjamaj531]

Hmm

[@ahotbanana]

The more of your videos I watch, the more I see math as an art form. There's more creativity in coming up with and solving these problems than there is in the average pop song

[@AliKhanMaths]

Another great video! Channels like these inspire me to share my own maths content!

[@Ironmonk036]

I didn't know about the 2nd channel. Subscribed!

[@professorpoke]

00:10 laughs in minecraft villager

[@blackpenredpen]

😆

[@TommyLandon]

I am not even a math major, or need complex math to make a living, but I always liked watching your videos!!

[@blackpenredpen]

Thanks. I appreciate that. 😃

[@TommyLandon]

@@blackpenredpen Thank YOU for keeping me curious and intrigued! :)

[@Maths_3.1415]

My solution Solve for x: 2^x + 4^x + 8^x = 14 Simplify and substitute y = 2^x. 2^x + 4^x + 8^x = 2^x + (2^x)^2 + (2^x)^3 = y^3 + y^2 + y: y^3 + y^2 + y = 14 Subtract 14 from both sides: y^3 + y^2 + y - 14 = 0 The left hand side factors into a product with two terms: (y - 2) (y^2 + 3 y + 7) = 0 Split into two equations: y - 2 = 0 or y^2 + 3 y + 7 = 0 Add 2 to both sides: y = 2 or y^2 + 3 y + 7 = 0 Substitute back for y = 2^x: 2^x = 2 or y^2 + 3 y + 7 = 0 Take the logarithm base 2 of both sides: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or y^2 + 3 y + 7 = 0 Subtract 7 from both sides: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or y^2 + 3 y = -7 Add 9/4 to both sides: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or y^2 + 3 y + 9/4 = -19/4 Write the left hand side as a square: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or (y + 3/2)^2 = -19/4 Take the square root of both sides: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or y + 3/2 = (i sqrt(19))/2 or y + 3/2 = -(i sqrt(19))/2 Subtract 3/2 from both sides: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or y = (i sqrt(19))/2 - 3/2 or y + 3/2 = -(i sqrt(19))/2 Substitute back for y = 2^x: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or 2^x = (i sqrt(19))/2 - 3/2 or y + 3/2 = -(i sqrt(19))/2 Take the logarithm base 2 of both sides: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or x = log((i sqrt(19))/2 - 3/2)/log(2) + (2 i π n_2)/log(2) for n_2 element Z or y + 3/2 = -(i sqrt(19))/2 Subtract 3/2 from both sides: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or x = log((i sqrt(19))/2 - 3/2)/log(2) + (2 i π n_2)/log(2) for n_2 element Z or y = -(i sqrt(19))/2 - 3/2 Substitute back for y = 2^x: x = (2 i π n_1)/log(2) + 1 for n_1 element Z or x = log((i sqrt(19))/2 - 3/2)/log(2) + (2 i π n_2)/log(2) for n_2 element Z or 2^x = -(i sqrt(19))/2 - 3/2 Take the logarithm base 2 of both sides: Answer: | | x = (2 i π n_1)/log(2) + 1 for n_1 element Z or x = log((i sqrt(19))/2 - 3/2)/log(2) + (2 i π n_2)/log(2) for n_2 element Z or x = log(-(i sqrt(19))/2 - 3/2)/log(2) + (2 i π n_3)/log(2) for n_3 element Z

[@nidhiagrawal3354]

In the case where t = 2, you have to put x = 1 + 2nπi/ln(2) because we are no longer only working with reals where n is in the set of integers. For the case where t ≠ 2, we don't need the "+ 2πi" because it's already included in the log. You can also simplify the log by taking the ½ out and making it a - 1.☺️ Edit: I changed the 2πi to 2nπi and said that n is in the set of integers.

[@salimahmad3533]

I can't understand your method 🙄. How you use x = 1 +2 πi?. Please explain it in better way .

[@nidhiagrawal3354]

@@salimahmad3533 Oh sorry it shouldn't be 2πi, it should be 2nπi/ln(2) because we're working with base 2 not base e. Where n is the set of integers. I was used to working with base e. Here was my thought process at that time. Since ln(x) = y is the same as x = e^y and we know from the Euler's formula that we can add 2nπi to y and e^y won't change. Therefore, ln(x) has infinitely many solutions. Btw ln(x) means natural log of x, which means log base e of x. I just forgot to add the n part. Now there is actually a way to change the base from 2 to e. lb(x) = y ln(x)/ln(2) = y x = e^(ln(2)y) We know from the Euler's formula that we can add 2nπi/ln(2) to y and e^(ln(2)y) won't change. Btw lb(x) means binary log of x, which means log base 2 of x. I hope this makes sense.☺️

[@salimahmad3533]

Thank you.

[@nidhiagrawal3354]

@@salimahmad3533 No problem.☺️

[@chowring]

Nice video! Since we are extending the field of solutions to all complex numbers I believe we should add a 2*Pi*k*i where k is any integer to any solution.

[@ДенисКосько-н9и]

didn't know u had another channel now i do. subscribed.

[@playdead1255]

Finally one of these math videos that I can actually understand. Although it's probably only because I am currently in precalc honors.

[@inotmark]

Munch's painting is the perfect background to some of those equations.

[@hassanalihusseini1717]

Me in my first maths class.... 😃

[@elementalgamer9879]

I'm not from an English speaking country but heard of precalculus and that stuff in tv shows and movies. Now I realise that here in sweden it's just normal maths so found this useful.

[@fabriziosantin7420]

Shouldn't all the logs be considered complex logs? So, with infinite solutions?

[@blackpenredpen]

Yes. And you can see my ln(a+bi) video in the description : )

[@cadaver123]

6:52 Yeah! This is just so cool 🦒❤

[@SirboOfficial]

I did it in a different method. I separated 14 into 2^2... and the other side, as well. Then, you can easily see that x=1

[@guilhermerocha2832]

I live for the "as always that's it" Thank you for your great work!

[@julienelson6506]

YAY!! You DID become a teacher!

[@leun6768]

I feel so proud for solving it but I used the natural logarithm instead

[@TH--NguyenMinhTu]

I think that method of factoring out cubic equations like that is called Horner’s method in Vietnamese and it’s pronounced the French way

[@gregoryzelevinsky9837]

Why do you not feel the need to add 2*pi*i*log_2(e)*n to all the solutions? :)

[@benjaminhoeckel7590]

I never knew you could factor like that. I was always taught to use long division, but this is way faster!!!

[@ΑΝΤΩΝΗΣΠΑΠΑΔΟΠΟΥΛΟΣ-ρ4τ]

I mean, if we're talking only about real solution, we could argue that f(x) = 8^x + 4^x + 2^x is constantly increasing, thus it will either intersect with g(x)=14 at one point, ot not intersect with the graph of g(x) at all. By inspection, x=1 is a solution, therefore, it is also the only real solution.

[@domanicmarcus2176]

At time 4:49, you said "3 minus 28;" shouldn't it of have been 9 minus 28? Either way, the answer is negative 19.

[@debblez]

is it possible to solve something like 2^x + 3^x + 5^x = 10 ?

[@dimitrispapadimitriou9330]

Yes it is

[@debblez]

@@dimitrispapadimitriou9330 how?

[@dimitrispapadimitriou9330]

@@debblez You can solve this equation by bringing 10 on the left side of the equation. So you have: 2^x+3^x+5^x-10=0 Then let f(x)=2^x+3^x+5^x-10 And then you find f(x) monotonicity, which is nondecreasing By finding monotonicity you prove that your function is 1 to 1 (or Injective function) and that your function could have maximum 1 solution for f(x)=0 By looking at the function it's clear that x=1 is a solution for f(x)=0 and thus the only because f(x) is a 1 to 1 function. So the solution to the first equation is x=1 Sorry if there are any mistakes in English I am not a native speaker.

[@lucaslucas191202]

It's really hard to find all answers (complex as well) since you can't just substitute 2^x=t. Then you end up with t + t^(ln3/ln2) + t^(ln5/ln2). I wouldn't even know where to begin. Wolframalpha also just shows the real solution x=1, but that's not really interesting.

[@lucaslucas191202]

Actually this really intrigued me and I continued looking for complex answers. You can prove that the real part of the complex number must be between 1 and about 1.922. The explanation is rather long, so I'll only write it if you want it. I'm also pretty sure there are an infinite amount of complex solutions, but the imaginary components can get extremely high really quick. The reason the question is so hard is the same that there might be an infinite amount of solutions. 2^z and 3^z rotate with different frequencies as you change the imaginary part, and since ln(2) and ln(3) are irrational you will never be able to find a value for the imaginary part that makes them reach the same degree again. So they're wildly unpredictable. I did not discover a way to find the exact solutions, but you can use an algorithm like the one shown in this video kzbin.info/www/bejne/mGipqYOppsuehbs to find approximations.

[@fahdfarachi6232]

Here is my method : By putting "f" the function in the video. I simply noticed that 14 = 8 + 4 + 2 We have then f(x) = f(1) And since f is increasing strictly on IR+, then it's injective. Meaning that x=1.

[@Risu0chan]

Don't forget the non principal branches of the logarithm. There is an infinite number of solutions of the form log(blah)/log(2) + k 2πi/log(2), as well as 1 + k 2πi/log(2), for all integers k

[@zaprazvan]

love ur content man , respect!!

[@TreeCube]

There are actually more solutions! Homework for you, what are those secret solutions?

[@TreeCube]

HINT eˣ is not injective for complex x.

[@girlgaming1993]

yes imaginary plane, just throw in 2pi(n) where n is any integer

[@girlgaming1993]

oh wait maybe not

[@MikehMike01]

6

[@TreeCube]

SOLUTION Let's look at the equations we got earlier (using approximations so it's not messy): 2ˣ = 2 2ˣ = -1.5+2.179…·i 2ˣ = -1.5-2.179…·i Let's focus on the first equation: 2ˣ = 2 Recall: e^(2πk·i) = 1 for integers k Using that, we can rewrite the first equation into this (a bit similar to solving sin(y)=sin(x)): e^(x(Log(2)+2πn·i)) = e^(Log(2)+2πm·i), where m and n are integers and Log is the *principal natural logarithm* Then cross the e's: x(Log(2)+2πn·i)) = Log(2)+2πm·i Solve for x: x = (Log(2)+2πm·i)/(Log(2)+2πn·i) *x = (0.693…+2πm·i)/(0.693…+2πn·i)* We can solve the second and third equations similar to how we did the first equation: x = (Log(-1.5+2.179…·i)+2πm·i)/(Log(2)+2πn·i) x = (Log(-1.5-2.179…·i)+2πm·i)/(Log(2)+2πn·i) *x = (0.972…+2.173…·i+2πm·i)/(0.693…+2πn·i)* *x = (0.972…-2.173…·i+2πm·i)/(0.693…+2πn·i)* Turns out there are infinitely many solutions huh? Note: To compute Log(z): Log(z) = ln(r)+iθ You can prove this by writing z as polar coordinates.

[@mr.inhuman7932]

Okay, so this is unrelated to the video, but I have a question: In the Video about the limit as n-->inf. For (n!/n^n)^1/n you said that 0⁰ is unfefined. You have actually said this multiple times. But everywhere I look it clearly says that 0⁰=1. I'd love a video about that! Also I love this channel! Thanks for all the conent you make for us!

[@fedem8229]

0⁰=1 by convenience (it makes many results such as series and other calculations much simpler), but as a limit lim x→a f(x)^g(x), where both f and g converge to 0 as x approaches a, it is an indeterminate form

[@jessehammer123]

“Everywhere you look” is wrong. We cannot define 0^0 as its own construct, without more information about its context. Generally, if forced to give a value to 0^0, we choose 1 because saying that 0^0 “equals” one breaks the least stuff and is the most useful in the most contexts. But it’s not necessary to do so, and there are examples of other things you could define 0^0 as. There are contexts in which people define 0^0 as 1 because it’s convenient (for example, in polynomials) and contexts in which people don’t define 0^0 (for example, in analysis-type subjects like calculus).

[@mahdiomidiyan3444]

Could you make a video on how to solve b differential equations like below one ? They're interesting! Especially for our university professors. Thank you! 👌🙏 y' = (2x + y + 4)/(x - y - 7) y - (x + x*y^3) y' = 0

[@GaryFerrao]

here you're explicit about the log base 2. what was the result of that poll you had a few months earlier?

[@stickmanbattle997]

I'll spend my 6 years and do maths everyday because of this man

[@HamStar_]

why did I make it to calc 3 and have never seen synthetic division before

[@myuu22]

How did this video come out on the same day that Numberphile made a video about the number 14?

[@lily_littleangel]

This is great, but what about x^8 + x^4 + x^2 = 14?

[@blackpenredpen]

😮!

[@ДенисКосько-н9и]

let t=x^2, so t^4+t^2+t=14. then use cubic resolvent. and Cardano's formula

[@ДенисКосько-н9и]

@@blackpenredpen why not

[@jayparekh8636]

Sir I'm from India & I am big fan of yours. Love You SIR..🤗 Your Teaching is Great..

[@chazzbunn7811]

I know you didn't tell us to pause the video and try this on our own, but I did anyways. I solved it the same way as you, I tutor a lot of pre cal so it came naturally to me.

[@blackpenredpen]

Nice 👍

[@scottwitoff8932]

Nice video. Lots of good algebra. And I like The Scream on the wall 😉

[@blackpenredpen]

Thanks.

[@mariosamson1716]

u forgot that u can add 2kpii/ln2 where k is an integer to any of these solutions

[@Ba2sik]

Thanks for the timestamps!

[@danmeyer0552]

I love your videos, but in this video I think you forgot the solutions x=1+(2πni/Ln(2)), where n is an integer.

[@alvaromedina1119]

Wow this was beautiful to watch

[@chessematics]

The method of division is ingenious. Can you teach these nice tricks for normal arithmetic some day?

[@ΑΝΤΩΝΗΣΠΑΠΑΔΟΠΟΥΛΟΣ-ρ4τ]

This method is called Horner division. Search it up, it's really practical when it comes to factoring polynomials (especially when it comes to the ones whose degree is greater than 2)

[@chessematics]

@@ΑΝΤΩΝΗΣΠΑΠΑΔΟΠΟΥΛΟΣ-ρ4τ thanks for giving the name! I'll be checking it out

[@yoav613]

Wolfram also knows,if you ask him nicely

[@blackpenredpen]

😂

[@TheFutureDothDream]

so this equation has two solutions: 1 and the complex solution..?

[@Pixics]

You are a math wizard, I liked learning about math that applies to real life scenarios, but in general I'm not that great at it.

[@ZipplyZane]

I don't think we need to pretend we don't know that x=1. It helps us factor the cubic polynomial. If x = 1, then t = 2, and thus we know that (t-2) is one of the factors. (I actually did it a bit differently and got (t-1) as one of the factors. I had already divided the whole equation through by 2 as I though that might give me better coefficients on the cubic polynomial. )

[@paradox_07]

Sir u are briliant.. ☺️✨

[@violintegral]

Substitutions aren't just for integrals!

[@blackpenredpen]

😆 yup!

[@jomaryambrosio]

@Blackpenredpen do you have a playlist where you teach math from the algebras to calculus?

[@alanarmstrong3186]

I made it 2^(3x)+2^(2x)+2^(x)=14 then lnbase2'd all of it. Added the x's together and somehow once I simplify I get some decimal. Just when I started thinking "finally, an easy one" ....I hear "everyone knows its 1"

[@ryanjabell]

You and Eddie woo are carrying me through early college math lol

[@DimaVaulin]

Brilliant begining of the video 👍😆

[@Mathematician6124]

If you don't mind please take an alike equation. 4^x + 6^x =9^x.

[@sohffso490]

My teacher would rage seeing a negative number inside square root

[@20icosahedron20]

Can't we abbreviate it like this? log₂((-3±i√19)÷2) = log₂(-3±i√19)+log₂(1÷2) = log₂(-3±i√19)-1

[@tithyyamnahata4944]

How to apply to be in your classes?

[@СтилянПетров-к5в]

Can you do one of those shorts you used to do about what i! (i factorial) is

[@aidenmunro6142]

Love your videos! :)

[@joshuaanoruo973]

The laugh at the beginning tho 😂😂😂😂😂😂

[@Mathematician6124]

It's really worth it.

[@chinchang5117]

To solve the equation t^3 + t^2 + t - 14 = 0, just press the calculator, Casio 96 sgplus. Mode 4,3.

[@kohwenxu]

Lol

[@kohwenxu]

Understand reference

[@RajVerma-vp1rp]

Imagine the Happiness of those who put x=1 & fortunately got it Correct!! 🤞

[@blackpenredpen]

😆

[@kingarth0r]

What did you think of the problem I sent you?

[@shawkathasan2211]

Sometimes i call you a legend of mathematics 🌚🌚

[@SuperYoonHo]

thanks

[@ZipplyZane]

I find synthetic division to be more annoying, in the same way that using matrices to solve systems of equations are. I find it easier conceptually if I can keep the variables around. I actually wound up just doing the long division in my head, term by term.

[@richardgratton7557]

Cool!😎 but I also kinda feel like the guy in the picture next to the white board!

[@navzarzar]

3 strictly increasing functions = constant => only one solution = 1

[@badman1882]

now i know, why i only got 1 mark out of 3 from that question...

[@JasonYJS_xoxo]

X is 1, isn't it?

[@DevGola-tn8yk]

When he said "but I know " My phone crashed

[@lukezeng8734]

When you factor polynomials it is much easier to do it by inspection

[@mohamedsamir9527]

Man you're great I really love ur videos keep up the great work ❤❤

[@Aivo382]

8+4+2=14 If 8^x+4^x+2^x=14 Then we can say: 8^x+4^x+2^x=8+4+2 So x=1..... This is my solution to this problem trying to not get too complex I think it's not matemathicly valid enough even when the actual X result may be 1

[@theophonchana5025]

t = 2^(x)

[@nohinbinu5017]

These q come in kvpy.

[@Maks-cw5zy]

I dont understand how the (t-2)(t^2+3^t+7). I dont understand where/how he got the -2

[@GSHAPIROY]

4:46 That sounded really weird...

[@shifatmat]

The new thumbnails look nice.

[@blackpenredpen]

Thanks.

[@davidramos4707]

Wolfram Alpha should be a sponsor.

[@Caturiya]

What is the name of "No division" NEVER SAW SUCH a THING. Did you discover it?

[@blackpenredpen]

This should be helpful: kzbin.info/www/bejne/nYaze52Ihb50a8k

[@funikstv7999]

Answer is 1 of course

[@Joy-be3gk]

Interesting!👏

[@PeterBarnes2]

You forgot about all the other solutions! The complex logarithm is multivalued! There are infinitely many solutions for each root of the polynomial.

[@dracokinerek]

How to solve: 8+4+2=14, x=1 done

[@naphatrungaphinya546]

Cant x = 1 ????

[@pi4313]

x = 1 easiest problem of my life