quadratic equation but in terms of 5

We will solve sqrt(5-x)=5-x^2 but we will square both sides first! We will get a very nice quadratic equation in terms of 5 and then we will use the quadratic formula to solve it. This is a very fun math olympiad type of equation for algebra 2 and precalculus students! Special thanks to dear Cirnobyl for this impressive solution! Who would have thought to write that equation in terms of "5"?! That video from 2018, Dear Tejas: 👉 • Dear Tejas, challenge ...
See other people's takes on the same problem
by Mu Prime Math: 👉 • Square Root Problem: S...
by Syber Maths: 👉 • Solving a radical equa...
We can also solve the quadratic equation x^2-5x+4=0 in a similar way:
👉 • Quadratic Equation in "1"
Subscribe for more math for fun videos 👉 bit.ly/3o2fMNo
🏬 Shop math t-shirt & hoodies: teespring.com/...
**Thanks to ALL my lovely patrons for supporting my channel and believing in what I do**
AP-IP Ben Delo Marcelo Silva Ehud Ezra 3blue1brown Joseph DeStefano
Mark Mann Philippe Zivan Sussholz AlkanKondo89 Adam Quentin Colley
Gary Tugan Stephen Stofka Alex Dodge Gary Huntress Alison Hansel
Delton Ding Klemens Christopher Ursich buda Vincent Poirier Toma Kolev
Tibees Bob Maxell A.B.C Cristian Navarro Jan Bormans Galios Theorist
Robert Sundling Stuart Wurtman Nick S William O'Corrigan Ron Jensen
Patapom Daniel Kahn Lea Denise James Steven Ridgway Jason Bucata
Mirko Schultz xeioex Jean-Manuel Izaret Jason Clement robert huff
Julian Moik Hiu Fung Lam Ronald Bryant Jan Řehák Robert Toltowicz
Angel Marchev, Jr. Antonio Luiz Brandao SquadriWilliam Laderer Natasha Caron Yevonnael Andrew Angel Marchev Sam Padilla ScienceBro Ryan Bingham
Papa Fassi Hoang Nguyen Arun Iyengar Michael Miller Sandun Panthangi
Skorj Olafsen Riley Faison Rolf Waefler Andrew Jack Ingham P Dwag Jason Kevin Davis Franco Tejero Klasseh Khornate Richard Payne Witek Mozga Brandon Smith Jan Lukas Kiermeyer Ralph Sato Kischel Nair Carsten Milkau Keith Kevelson Christoph Hipp Witness Forest Roberts Abd-alijaleel Laraki Anthony Bruent-Bessette Samuel Gronwold Tyler Bennett christopher careta Troy R Katy Lap C Niltiac, Stealer of Souls Jon Daivd R
---------------------------------------------------------------------------------------------------
💪 If you would also like to support this channel and have your name in the video description, then you could become my patron here / blackpenredpen

[@cirnobyl9158]

Thank you for making this video, I'm glad you enjoyed the solution! I first saw this solution from Dr. Titu Andreescu.

[@blackpenredpen]

Thank you for sharing!!! We all love this solution!

[@YT-Observer]

@@blackpenredpen have you investigated the new version if how to solve quadratic equations? professor Loh kzbin.info/www/bejne/jnyliWOoaK9gb7s

[@SuparnoBhattacharyya]

@Adam Romanov I found this trick fascinating!

[@sumitnirmal4226]

I have seen this technique before ( probably in Problems in Mathematics)

[@DarkMage2k]

@@blackpenredpen kzbin.info/www/bejne/eICUhWqbl6-fhZo You did this once already lol

[@pudim3113]

When he said: "This is actually a quadratic equation...in terms of 5" I just got speechless and needed some time to process what I've just heard. It's just too amazing.

[@blackpenredpen]

I read his comment multiple times in order to really get what he was saying. My mind was blown! : )

[@sanjayvaradharajan]

@@blackpenredpen 😂😂

[@HarshSharma-dq5zl]

@@blackpenredpen actually I knew that technique , thanks to my teacher (tought me during ,how to find equation of lines in the pair of lines).

[@anneog]

Yes I had to have a bit of a lie down too 😆

[@gurk_the_magnificent9008]

100%. Literally open-mouthed in shock.

[@historybuff0393]

For all my years in math, I never realized that you could use a number as a variable and the variables as coefficients. Incredible!

[@MA-bm9jz]

Well you havent been paying attention to classes then,when you are solving a differential equation there is this method called varriation of constants(very similar to this),or when you find the minimal polynomial of (for example) sqrt(2) over Q

[@Nylspider]

@@MA-bm9jz bruh here's a bit of advice, don't tell someone "well you haven't been paying attention in class then" That's really rude and I can only imagine what the other person feels

[@colleen9493]

ItzARubix yes I agree.

[@matejdudjak9969]

@@MA-bm9jz Also not everybody has the same interests on the same levels, maybe this person does math sometimes or just for 1 lesson in college, or maybe in other things. It would be like me mocking you for not listening to some thing you didn't like in class just because I love it and know it.

[@historybuff0393]

Matisan Andrei In differential equations that method is called variation of parameters, not variation of constants.

[@vaxjoaberg9452]

"...in terms of the variable 5." ...the variable 5. Now that's some advanced maths.

[@Nylspider]

@Jeremy Edwards what the actual frick Edit: was there like a bot that was asking for subs or something or are you talking about the original commenters?

[@sesamtoast9431]

Yeah its so sick that you can imagine every number as a constant variable, like 5=1*5 and it behaves just like using x. This video teached me so much

[@sanjayvaradharajan]

@@Nylspider don't get worked up such idiots always exist jus ignore em😂

[@saebelorn]

@Jeremy Edwards how is it thinking you are smarter than people because you consume youtube 'skeptical' content, the intellectual equivalent of white bread. read a book

[@Nylspider]

@@sanjayvaradharajan ig you're right. It does make me angry that people say mean things tho

[@nathanaelmoses7977]

X^4 = I'M THE SCARIEST VARIABLE YOU CAN'T SOLVE THIS EQUATION! 5 = you are a variable?

[@kinshuksinghania4289]

5 says "But I'm the constant"

[@LabMember-lq1ul]

Crewmate ,There is 1 imposter among us

[@realbignoob1886]

@@LabMember-lq1ul yes

[@shivamchouhan5077]

@@LabMember-lq1ul you commented 79 years ago 😂

[@xninja2369]

X^4 is really easy to solve generally , and has different forms like a and d are the same etc ... Btw X^3 with no integral roots is harder and don't get me to talk about x^5 😂

[@dudurododoizi8547]

Damn that's what I love in math : the changement of perspective

[@a_llama]

*change in

[@sungodmoth]

@@a_llama how dare you insult the great language of franglais

[@a_llama]

@@sungodmoth langue de merde

[@clanc_huranku]

Broke: treating the variable as a constant Woke: treating the constant as a variable

[@LearnFirstEarnNext]

Ha ha ha

[@LearnFirstEarnNext]

👏 Great 👍

[@sarthakjain1824]

Quadratic equation in 5!!! MIND BLOWN never thought of it that way

[@sinpi314]

In 5 triple factorial? That’s a big number (!)

[@Firefly256]

@@sinpi314 5*2 is big?

[@sinpi314]

@@Firefly256 depends, 5*2 bigger than an infinitely many numbers, but is also smaller than infinitely many numbers.

[@leandroteles7857]

If anyone has trouble visualizing an equation in terms of "5", you can just replace "5" with another variable like "y", then solve for y. You'll get a solution for y in terms of x, but you know that y is 5, so you put back the 5 and solve again. But the real question is: how could you know beforehand that solving for "5" would cause the quartic term to disappear?

[@quantumsigmaqed6312]

Because the problem isn't supposed to be obvious

[@slingman1074]

Its an alternative solution..... question was not designed for this method

[@Yadobler]

But the real question is: how could you know beforehand that solving for "5" would cause the quartic term to disappear? that's maths for you. trying to find werid ways to re think everything. because maths is really what we define it to be, as long as it plays by the rules we set. for this, id say it comes when you are stuck with a x^4 and not sure what to do, but then you see a 5^2 which you forgot to turn into 25, and thank god you didnt, so you decide to try doing that i think that's a common type of quadratic equation. where 0 = ax^2 + bx + c, and the values (a), (b) and/or (c) contains another unknown variable. the idea is thinking of in which "domain" do you want to solve the question. i think this becomes more relavant in calculus and how you can solve with either time or frequency as the domain. but for a guy like me, if you didnt substitute y = 5 for me, i dont think id realise that its a case of just solving for y, and then subbing in y again and solve for x. ------------------------------------------------------------------- have a watch at kzbin.info/www/bejne/kH7Oepx8qJhofrM I love this. matt parker tries to prove that for any prime p where p > 3, p^2 - 1 = 24 "the square of any prime is one greater than a multiple of 24 so he does the rigourous approach, of turning (p^2 - 1) = 24k into 4 different equations and showing they all are factors of 24 but his colleague asked why he didnt do it the "easier way": p^2 - 1 = (p+1) (p-1) number line: A | - (p - 1): is an even number | - (p): odd, not multiple of 2 and 3 since prime | - (p + 1) is an even number | V so (p-1) and (p+1) are even, multiples of 2 either (p-1) or (p+1) are multiple of 4, since every other EVEN number is (ie, 2 isnt, 4 is, 6 isnt, 8 is) lastly, in a group of 3 consecutive numbers, one is multiple of 3. Its not p. so it's either p+1 or p-1 so (p+1)(p-1) have factors of 2, 4 and 3. 2*4*3 = 24. ------------------------------------------------------------------- if you ask me, like what parker said, in hindsight, the 2nd method looks elegant is makes sense. Just like y=5 and solving the quad eq for y, then subbing in y, and then solving for x. if it pops in, then you can do it. Else you'll never see it. we'll just do it like parker's first method, and this man's initial video of not squaring both sides. its more rigorous and painful, but straight forward. ------------------------------------------------------------------- i think that's what olympaids try to do. They arent seeing who's the fastest but who's the most creative. Doesnt come without lots and lots of exposure, experience and practice. rote memorisation is cancer and curbs creativity but it exposes you to many different possible scenarios, it unlocks many hidden dots on the map. then your mental creativity and flexibility will connect those dots. the fewer the dots, the less ways to connect and the longer the distance between 2 adjacent dots. but ye untill those dots help pay my bills i'll keep it as an hobby

[@kshitijsharma3170]

@@Yadobler damn boy

[@lucasgameiroborges6086]

@@Yadobler great insights! Really appreciate you efforts. I was wondering some of the same points in my head

[@x0cx102]

woah! a bit more than a year ago I watched your dear tejas video on this problem and commented that you could also "solve for five" seeing the equation as a quadratic in five, ie. shifting the parameter. and now you really did do it. awesome. that gives three ways of solving the problem: - solving for five - writing as nested radical - noticing that lhs and rhs are inverses of each other and intersects at y = x

[@blackpenredpen]

I must had missed ur comment and I am sorry about that.

[@angelmendez-rivera351]

You can write the equation as a quadratic equation with respect to x, by noticing that if f(x) = 5 - x^2 and g(x) = sqrt(5 - x), then f[g(x)] = x, and since g(x) = f(x), this implies that f(x) = x, which is easy to solve, because you can rewrite it as 5 - x^2 = x, which is equivalent to x^2 + x - 5 = 0, which gives the solutions provided in the video. The method in the video is very creative, though, and it most useful for someone who is not aware of the theorem I stated above. It probably also is more applicable in more contexts than the theorem above.

[@pritivarshney2128]

WOW yes we can do that!

[@blackpenredpen]

Hey Angel, Mu Prime Math and I discussed that method a few months ago. However, in this case, I don't think we can find both the solutions with that method. Mu Prime Math has a video on when this method can actually find all the solutions.

[@danmarino900]

How the HECKAROONI did u deduce that f(x)=x?? I read this a thousand times trying to see how you jumped to that step and i can’t see it.

[@tracyh5751]

@@danmarino900 It needs to be f[f(x)]=x.

[@raptor9514]

@@tracyh5751 I did such a mistake too. What about f(x)=1/x? Or f(x)=x-a, where a is a constant? Then f(f(x))=x

[@shaharblank98]

That's some next level Algebra right there! Amazing :)

[@blackpenredpen]

Glad that you like it! : )

[@dmorgan0628]

Can you imagine being in algebra 2 and the professor or teacher decided to toss that equation out as a mindfuck, but not really score it as a real question?

[@MuPrimeMath]

What an interesting method!

[@blackpenredpen]

I couldn’t believe it when I saw the comment!

[@Nylspider]

Yo Mu prime

[@blackpenredpen]

@@sanyamgarg9584 Appreciated! : )

[@treanungkurmal5718]

@@blackpenredpen Love You sir ! Your methods are always interesting be it the integration by parts method, or this one ! Superb method Sir...... Regards to Mu Prime Sir

[@PrismofScience]

This man is truly on a different level of mathematics

[@blackpenredpen]

Thanks to that viewer tho! : )

[@victoralvarez3519]

Ok i need a minute. Where are My math books

[@BigDBrian]

I remember there was one story of a student who'd solved a problem and ended up with the correct answer, after taking the derivative of a function with respect to "2". It amazes me, though I can see the legitimacy of the case of this video better.

[@washieman2445]

How do you take the derivative with respect to 2?

[@minamagdy4126]

probably like y=2, d/dy. This is based on an interpretation of the question in a constant-value-variable way, which this video demonstrates. (to be honest, I took the idea from another comment explaining the validity of bprp's steps in this way)

[@BigDBrian]

I've tried to find the thread which had many fun examples of false proofs for correct answers (such as anomalous cancellation) but I didn't manage to find it, unfortunately.

[@danielnelson6131]

This was honestly a really bizarre solution method to this equation. I've never seen a polynomial being solved that way. Pretty incredible!

[@duggydo]

Freaking ingenious! The person who thought of this is a Boss!

[@particleonazock2246]

@Chandler Bing Frankly, we are not all taught to appreciate lexicon in the English language, I see.

[@ramazantaikocusu]

I was someone hating math with all my heart. You're the one making me start doing "math for fun". Thanks a lot, teacher!

[@blackpenredpen]

Happy to hear that! : )

[@michaelz2270]

The quartic polynomial you have there x^4 -10x^2 + x + 20 does factor as (x^2 + x - 5)(x^2 - x - 4) so there's a connection between your solution and the quartic polynomial... but I have to say I've never seen this neat trick to factor quartics, very nice :)

[@sueyibaslanli3519]

I want to say just : " wow! "

[@sueyibaslanli3519]

It takes just a second

[@ethancheung1676]

First time seeing this, was amazed and opened my eyes. In retrospect, the 5 is not so different than a radical eg Sqrt(3) or i. We just get too used to the ease of evaluating the numbers like 5 we don’t see the duality or changing perspective between the number and the variable

[@Stelios2711]

Holy mother of Maths!... That was spectacular! I once saw a little video of Flammable who did such a thing for fun, but I've never expected to see this trick (solve the quadratic with respect to something that is... known) in action to a more challenging problem! A lot of people say that it is a rather standard trick that it is taught in Math Camps around the globe. Good to know that.

[@astraestus8828]

The beauty of mathematics. It seems so obvious when you understand it and it blows your mind.

[@Penndennis]

Thank you so much for that. Man! did that ever put a smile on my face. What a brilliant insight to pass on; thanks again to you and Cirnobyl.

[@nguyenthai3140]

The guy that discover the quadratic folmula: "This is gonna very useful for solve equation in term of f(x)" blackpenredpen: "It's not just x"

[@theimmux3034]

That's impressive, really shows understanding of maths. You can't end up with this just by learning all sorts of formulae and methods by heart.

[@TheOddsMustBeCrazy]

Regarding the quadratic formula, I didn't know you were allowed to do that, replacing x with 5. Very interesting.

[@BV-mg1ek]

5 to x^4: you may have outsmarted me, but I outsmarted your outsmarting! (I love jojo memes, and I'm not afraid to admit it)

[@tanishphopalkar5864]

That marker transition is very smooooth

[@hbarudi]

Nice math trick to solve this... Most student struggle to figure it out under pressure...

[@Meilo0110]

I've just gained brain cells.

[@MathrillSohamJoshi]

Earlier 'don't square both sides' And 'watch in 0.75x' Now 'Square both sides' And 'don't watch in 0.75x' Haha

[@ianmoseley9910]

clever trick - seems obvious when you explain it but would never have thought of it myself

[@thevastuniverse246]

2:15 "EvERybOdY DiVidEd by..." Love that... XD

[@blackpenredpen]

Hahaha, thanks

[@leadnitrate2194]

Sounds like "erybaady divaadid baa" Love it though.

[@aniruddhvasishta8334]

For those of you who are having trouble understanding what he did: If you just replace 5 with another variable then it's just a quadratic in that variable.

[@weonlygoupfromhere7369]

I hope we can meet in the future

[@ethandole2218]

I would be exctatic to meet bprp!

[@prithujsarkar2010]

Yea

[@phiilw2068]

I would have loved to see that video 1 year ago, and then use that method of solving in my math test, my teacher wouldve been so astonished ::::

[@-Me-yl2ou]

Wtf i just watched 😲😲😲😲😲 Completed my graduation in maths but never had anything like this.

[@nikhilnagaria2672]

This works because in the derivation of the Quadratic formula, you assume LITERALLY NOTHING except a≠0. So, whenever you have for some non-zero a, ax^2+bx+c=0 immediately implies x = 1/2a (-b ± sqrt(b^2-4ac)) I always tell this whenever I teach this. Here's another simpler version of solving a more simpler problem: 2x + 4 = 20 (2)^2 + x(2) - 20 = 0 (2) = 1/2 (-x ± \sqrt{x^2 -4(1)(-20)} (4 + x)^2 = x^2 + 80 x^2 + 8x + 16 = x^2 + 80 8x = 64 x = 8 Even when you have something a mess like 2x cos(x) ln(x)^2 + (4x^3+1)ln(x) + tan(x)arcsin(x) = 0 The Quadratic formula still works, although it won't lead you anywhere. (The equation is made up as an illustration, I don't know how to solve it legit.)

[@zdino9116]

That's why is IMPORTANT to check out the comments. There could be interesting things. ;)

[@skcena6543]

This is so cool.I never looked quad equation in this way.Its always x but never though it can be '5'

[@rishitgarg4222]

I'm really happy to say that our Maths teacher had introduced us to this method!!!

[@Andrei71236]

What a plot twist - one of the greatest solutions I've ever seen

[@garyhuntress6871]

"have a look! Man, seriously!" Loved that :D

[@claudemiropacheco]

Freaking amazing. Easy to understand after you explained, but I would never think about it myself.

[@Craznar]

That's what happens when art and maths collide. Well done @Cimobyl

[@nisargbhavsar25]

This method is great!!

[@user-my7ki4it3s]

Mind-blowing 🤯 I'd never come up with solving quadratic equations in terms of NUMBERS. Thank you for the video

[@gravitonX_]

Nothing is more beautiful than Physics and Mathematics ! ❤️🥰

[@EpicMathTime]

This is a beautifully creative solution.

[@danyalkhan1347]

Is that really a thing ?? Well i am engineering student and never in my life i remember treating a constant as a variable while solving an equation.😂 Incredible mate !

[@PVempati]

Really? You didn't have finding the inverse of a quadratic function in 11 and 12th?

[@DrinkExchanger]

This makes me happy. Thank you

[@vinayk9528]

The coolest quadratic solving ever, and try wearing a white shirt to highlight your beard

[@blackpenredpen]

Unfortunately, if I wear a white shirt, then I will be blent in with the whiteboard. That's why I never wear white shirts in my videos.

[@vinayk9528]

blackpenredpen oh its ok i understand.

[@boogabooga4388]

WAIT WHAAAT!! OMG!! quadratic in terms of 5?! IT BLEW MY MIND!!

[@ajithxyz7950]

I can't feel my legs after watching this. Everything seems like a lie now. XD Good video. I will be sharing it with my friend.

[@nikitakipriyanov7260]

This also could be quite easily solved directly by the Ferrari method. I've used a half of a sheet of a paper to reach the same answer. The method is based on that if you have a complete square on one side, you must also have a complete square on the other. After squraring both sides we get: 5-x=(5-x²)² Let's add following to both sides: 2z(5-x²)+z², where z is still unknown variable. This wouldn't destroy the complete square in the right part. Our equation becomes: 5-x + 2z(5-x²)+z² = (5-x²)² + 2z(5-x²)+z² After expanding left part and contracting right one: -2 z x² - x + z²+5+10z = (5-x² + z)². Now, let's find z such that left part of an equation is also a complete square. Its discriminant is: (-1)² - 4 (-2z) (z²+5+10z) = 8z³+80z²+40z+1. We need that to be zero in order to have a complete square: 8z³+80z²+40z+1 = 0 This is main disadvantage of Ferrari method: you need to solve a cubic equation. Fortunately, this one has a rational root. All rational roots of such equations have the form ±a/b, where a is some divisor of a free term (1 in this case) and b is some divisor of a term with largest degree (8, i.e. b could be 1, 2, 4 or 8), so we might try all of them to find any that works. You can convince yourself that z=-1/2 is a root. We can now solve it completely, but we don't need, as we need just any z. Actually two other roots of z equation will lead to the same roots of original equation, just grouped other way. We don't lose anything by skipping them, so let's continue with this one. Putting z=-1/2 into our equation: (9/2-x²)²=(x-1/2)². This could be true either if 9/2-x²=x-1/2 or 9/2-x²=-(x-1/2), so instead of a 4th order equation we now have two quadratic ones. They could be written as: x²+x-5=0 and x²-x-4=0. First one has roots (-1±sqrt(21))/2, second one (1±sqrt(17))/2. Now we have to drop stray roots that don't really satisfy original equation and this will get us the same answer as in the video.

[@lukaradulovic7904]

finally! someone mentioned the ferrari method!

[@prashantshukla6018]

I wish I had u as a teacher in my coaching u teach the world's best maths . I'm nimisha from india , love from india sir . I wanted to ask, where am I supposed to ask a doubt from u sir ?.

[@mehuljain6703]

Salute to the man who figured out this way if doing this problem

[@srpenguinbr]

ax+bx^{cos²(1)}+cx^{cos(2)}=0 can be solved using this I actually proved you can always use it to solve equations in the form ax^A+bx^B+cx^C=0, with 2B=A+C (in order to get a nice square root), A>B>C. However, you can always factor out x^C and use the substitution x²=u. So the approach I took to make more interesting was using some trig identity

[@chenmarkson7413]

absolutely MIND BLOWN.

[@050138]

Awesome way to reimagine a polynomial.... Variable becomes constant and constant becomes variable....

[@noakes3067]

Wow. Big fan of outside the box thinking. Great job teacher. Where could I get one of those "Integrals for you" posters?

[@blackpenredpen]

Thanks! I made it myself and you can find it in the description. : )

[@MultiRRR123]

That's... Not something that crossed my mind, okay. You always deliver!

[@bonkdragon5504]

I had to take a deep breath for a quadratic formula solution! BRILLIANT

[@masonbarker7841]

You are a better teacher then my last 3 teachers. Take it how you want it’s the truth

[@leoneschle3112]

My mind was literally blown when I saw that!

[@frozenmoon998]

Great to be so early, definitely worth it for this video :) BTW yes you can write an even equation as a quadratic (such as sixth power or fourth power), probably with some restrictions though, just as letting x^2=y and doing it in y, substituting back in and etc. :D

[@MA-bm9jz]

Or just notice the left hand side is the inverse of the right hand side,in terms of inverse functions,so the equation is rewritten like f=f^(-1) ,as you know the graph of the inverse function is simetric to the first bisector to the graph of the function, so if they do intersect(like in the equation above) they intersect on the line y=x so the equation is equivalent to f(x)=x or f^(-1)=x so 5-x^2=x

[@angelmendez-rivera351]

Matisan Andrei x |-> sqrt(5 - x) is only the right-inverse of x |-> 5 - x^2, because x |-> 5 - x^2 is not injective in R

[@MA-bm9jz]

@@angelmendez-rivera351 yeah but its bijective on those intervals taken seperatly,set an x there and so on

[@MA-bm9jz]

@@angelmendez-rivera351 solving 3 second degree polynomials is computationally easiee than solving a 4 degree one

[@angelmendez-rivera351]

Matisan Andrei I am aware of that, but the statement that you wrote on your comment is wrong, so I corrected it, so that a person that reads it and tries to use the theorem uses it correctly. You cannot go around correcting people if you are going to make mistakes yourself.

[@coolmangame4141]

it's big brain time. tbh this method is very amazing

[@blackpenredpen]

It really is! I was amazed when I saw the comment.

[@SeeingDreams11]

Very nice. The process of solving may make you think about being a math major.

[@vfnikster]

Much simpler, if we want only one root. We've got inverse functions on both sides, at least on positive ray x>0, they intersect on the line y=x, so change the left hand side to x, and get x=5-x^2, calculate the first positive root. The second root could be trickier. The method presented is interesting but works only if there are inverse functions with squares or square roots.

[@endwayphysicist]

bro... in terms of 5... I wouldn´t have tried that even after hours looking at it, it was just mindblowing

[@mohamedmoustache258]

i donT have any words just ...WOW

[@kunal_chand]

This is pure mathematical beauty.

[@asjadraza3826]

I am a jee aspirant in class 12 in India The same method was taught to me by my maths teacher in class 11th in quadratic equations....

[@asjadraza3826]

I am not lying... The same ques was indeed taught to me in kota last year

[@asjadraza3826]

@@akashbanik2947 yeah

[@asjadraza3826]

@@akashbanik2947 it was exactly the same ques

[@maxwellgrossman]

I think even another way to solve this one is that you can notice that they each side is the opposite sides inverse and since they’re symmetrical over y=x if one of the sides crosses it then they both have to cross it at the same point since a point on y=x reflected over that line is itself. So you can solve 5-x^2=x which is a quadratic and then use synthetic division or polynomial long division to find the other roots that don’t cross y=x

[@sumitnirmal4226]

@blackpenredpen Sir I have a very different and easy solution to this. First we square both sides, then we arrange the terms as x^4 - 9x^2 - (x-5)(x+4)=0 If we observe carefully we can arrange x^2(x^2-9) as x^2[ x^2 + (x-5) - (x+4)] thus our final equation becomes x^2[x^2 +( x - 5 )-(x+4)] - (x-5)(x+4)=0 I hope u can clearly see the factors...... Its really difficult to write all these stuffs😅😅😅 Thank u sir (Love from India)

[@leviszhou711]

这个解题法在我初高中时候没有教过，但是当你这么写的时候 能猜到你的想法了。最后通过画图法确定两个解在-2和2之间是很ok的。

[@hassanniaz7583]

This is insane!

[@PegasusTenma1]

This looks very similar to variation of constants used to solve ODEs

[@shivangitripathi7931]

Mind blown by the fact that it was a quadratic in 5🤯🤯🤯

[@陈若明]

Suppose that f(x) = sqrt(5 - x) (for x = 0). Easy to see that f(x) = g^ -1(x), and therefore, they are symmetric about the y=x axis (see 5:30). So one of the intersection between f and g must be on that axis, which means x = 5 - x^2, or x^2 + x - 5 = 0. then the quartic equation x^4 - 10x^2 + x + 20 = 0 can be written as (x^2 + x - 5)(x^2 + Ax + B) = 0. the following are trivial.

[@giovanni1946]

That's absolutely impressive !

[@rohamyaghoubisabet1650]

A different creative solution! That's amazing! Thank you

[@NonTwinBrothers]

Damn, that 5 knows where it's at

[@blackpenredpen]

lol yea!

[@pkmkb]

The amount of problems you can solve with 5 (or any constant) taken as a variable is insane. The dude who came up with this is a genius. 👏👏

[@weonlygoupfromhere7369]

Keep up the great work man!

[@stuartyeo5354]

An absolutely nifty technique :)

[@blackpenredpen]

Indeed it is!

[@justanothergirl__]

This is such a cool way of rethinking of how to approach a problem that seems a bit out of reach. I can't wait to share this with my algebra class this semester :D

[@j.r.8176]

blew my mind completely

[@gloystar]

WoW! That's a very smart workaround. I liked it .. If you can avoid quartic equation by any means then do it, LOL!

[@afo0]

Amazing trick. I guess that you could use always the same trick but taking 1 instead of 5, and splitting the polinomial elements as you want (1^2 = 1, bla bla). You have many options to select what term goes with 1^0, 1^1 and 1^2. With luck in at least one of them you can reduce the biggest X exponents inside the square root (like in the video). And probably a bunch of solutions are no valids. At least you can end with a bunch of weird things equal to each other and to 1 XD.

[@blackpenredpen]

Very nice! I thought of the same too and was trying to come up with more examples like this. I am making a animated video on solving x^2-5x+4=0 like this : )

[@TikeMyson69]

You could also change the 5 to any other number as long as you readjust the "c" term to make up for the change. Probably not very useful in this particular case but this can be done with any polinomial equation. Very cool trick, love it!

[@superbus8102]

beautiful and unusual approach :)

[@anitagofradump5195]

Wow its amazing how "obvious" this seems after being shown that its possible. Like it makes complete sense because once you solve for 'x' it is simply a number (or set of numbers) that satisfy the equation so whenever we use the quadratic formula to solve for 'x' were really just solving a number in terms of other numbers so there was no reason to ever think it wouldnt be applicable the other way around

[@fahimfa4600]

This man is underrated genius

[@ChadTanker]

took me a minute to see the quadratic in terms of 5 cuz im absolutly not used to ignore the variable when solving equations

[@solarisone1082]

Definitely bookmarking this video.

[@skit555]

You're insane man! I subscribed xD

[@internationalfatherinlaw5585]

This is a question from an Indian examination called Gmat Another way of solving it by using the concept of range