Dear Tejas, challenge ACCOMPLISHED! (Speed run, Uncut)

[@simpletn]

You know shit got real when he pulls out the purple marker

[@astellagaming]

account hahhahaha flyyyy

[@michaelroditis1952]

Just made your likes evil

[@HasanabiOOC]

Hahaha

[@stolen2952]

😂😂😂😂

[@wjrasmussen666]

Does he ever break out the plaid marker?

[@angelmendez-rivera351]

Just in case that anyone did decide to try to square the equation instead, 5 - x = (5 - x^2)^2, which would give you a quartic equation. Add to the equation the expression 4(x^2 - 5)y^2 + 4y^4 to obtain 4y^2·x^2 - x + 4y^4 - 10y^2 + 5 = (x^2 - 5 + 2y^2)^2 = 4y^2·x^2 - x + (4y^4 - 20y^2 + 5). Then divide by 4y^2 to get (x^2/2y - 5/2y + y)^2 = x^2 - x/2y^2 + (y^2 - 5 + 5/2y^2). We want x^2 - x/2y^2 + (y^2 - 5 + 5/2y^2) to be a perfect square, which implies we want 1/4y^4 = y^2 - 5 + 5/2y^2, or 1 = 4y^6 - 20y^4 + 10y^2, or 4(y^2)^3 - 20(y^2)^2 + 10(y^2) - 1 = 0. Then, we must solve this cubic for y^2. Thus, we really want to solve 4z^3 - 20z^2 + 10z - 1 = 0 and let z = y^2. To solve this, divide by 4 to get z^3 - 5z^2 + (5/2)z - 1/4 = 0. Let z = a + 5/3, so (a + 5/3)^3 - 5(a + 5/3)^2 + (5/2)(a + 5/3) - 1/4 = a^3 + 5a^2 + 25a/3 + 125/27 - 5a^2 + 25a/3 + 125/9 + 5a/2 + 25/6 - 1/4 = a^3 + 115a/6 + 1225/54 = 0. This is a depressed cubic equation. Now, notice that (u + v)^3 + 3uv(u + v) - (u^3 + v^3) = 0. Thus, if a = u + v, then 3uv = 115/6 and u^3 + v^3 = -1225/54. u = 115/9v, so u^3 + v^3 = (115/9)^3/v^3 + v^3 = -1225/54, which implies v^6 + 1225v^3/54 + (115/9)^3 = (v^3)^2 + (1225/54)(v^3) + (115/9)^3. Then v^3 = {-(1225/54) + sqrt[(1225/54)^2 - 4(115/9)^3]}/2 or {-(1225/54) - sqrt[(1225/54)^2 - 4(115)/9)^3]}/2. Regardless of rhe choice, u^3 will be equal to its conjugate. This implies that a = cbrt[{-(1225/54) + sqrt[(1225/54)^2 - 4(115/9)^3]}/2] + cbrt[{-(1225/54) - sqrt[(1225/54)^2 - 4(115)/9)^3]}/2], which implies z = -5/3 + cbrt[{-(1225/54) + sqrt[(1225/54)^2 - 4(115/9)^3]}/2] + cbrt[{-(1225/54) - sqrt[(1225/54)^2 - 4(115)/9)^3]}/2]. Simplifying this would be useful. 54^2 = 9^2·6^2 = 3^4·3^2·2^2 = 2^2·3^6, while 9^3 = 3^6. To get the common denominator, the inner radicand would equal [(1225)^2 - (115·4)^2]/(54)^2 = [(1225)^2 - (460)^2]/54^2. (1225)^2 - (460)^2 = (1225 - 460)(1225 + 460) = (1 685)(765)/54^2 = 337·153·5^2/54^2 = 337·17·15^2/54^2 = 5 729·(15/54)^2. This leaves the outer cubic radicands -[1225 - 15·sqrt(5 729)]/108 and -[1225 + 15·sqrt(5 729)]/108 respectively. As you can see, this is a treacherous path, and we are dealing with expressions far more complex than what was shown in the original problem and its solutions. Clearly, this is not the way to solve, and anyone would have given up halfway through.

[@blackpenredpen]

Wow!!!

[@notdarkangelu]

Man, one like for your commitment even though I didn't understood a single fucking line 🤣

[@petachad8463]

I gained nothing from reading this.

[@lcex1649]

I assumed it can be written as the product of two trimonials: (ax^2 + bx + c)(dex^2 + ex + c). I solved it with a system of equations, letting c=-4 so that every variable is an integer. This led to (x^2 - x + 4)(x^2 + 5x - 5)=0 and solving x from there was pretty straightfoward

[@flashbang8673]

i aint readin allat🔥🔥

[@barak363363]

*and here I just thought to power 2 both sides like an innocent child..*

[@einsteingonzalez4336]

¡Que vergüenza! (What a shame!)

[@koala2587]

Same

[@justinsantos5751]

Why is it wrong? I haven't tried it yet

[@Gal_Meister]

@@justinsantos5751 it's right, u can square both sides and then use polynomial division. Answers are the same

[@SebastienPatriote]

You technically could as there actually is a formula to solve fourth degree equations. I don't recommend it however...

[@gabrielh5105]

The teacher during the whole class: stories of his life, personal thoughts, etc The teacher when I leave 4 mins to go to the bathroom:

[@sbypasser819]

best joke ever

[@kepler4192]

B-but the stories are fun to hear!

[@shambachakrabortyofficial1611]

😂😂😂😂😂

[@tumak1]

Excellent solution ...nicely explained at 314 words per second! Cheers

[@blackpenredpen]

tumak1 : )

[@mchappster3790]

Is this a pi joke lmao?

[@Rekko82]

My math teacher wrote 314 faster than him though. I still love math despite facing a speed writer in math lessons in high school. He used right hand to write and the left hand to clean at the same time.

[@ayazraza8287]

best comment with a good sarcasm

[@cheshstyles]

@@Rekko82 no way

[@cirnobyl9158]

This problem has a bit of notoriety in the math olympiad community for having a funny alternate solution: sqrt(5 - x) = 5 - x^2 Square both sides 5 - x = 5^2 - 2x^2*5 + x^4 Rewrite this as a quadratic equation. But wait, how can we make a quadratic when there's an x^4 term? The key is to not write it as a quadratic in the variable x; write it as a quadratic in the variable 5: 5^2 - (2x^2 + 1)*5 + x^4 + x = 0 Use the quadratic formula to solve for the variable 5: 5 = (2x^2 + 1 +/- sqrt(4x^2 - 4x + 1)) / 2 5 = x^2 + x or 5 = x^2 - x + 1 Finish by solving both quadratics. Remember to throw out the two extraneous solutions where 5 - x^2 is negative, due to our first step.

[@ethang8250]

Thank you for widening our mathematical perspectives

[@blackpenredpen]

Unbelievable! Thanks for this mind-blowing solution!

[@mathfincoding]

Who went straight here after watching the video inspired by this comment?

[@CauchyIntegralFormula]

I always thought this video was about this solution, since it's pretty famous, so I never checked it out. I'm surprised to see it's not

[@rabeakhatun2819]

nice man

[@DarkMage2k]

The fabled purple pen was only mentioned in the legends. This video marks the speed of its glorious power to solve maths. On the hands of this man lies the power to overthrow deities.

[@RahulSingh-zo7sm]

Cool

[@VoteScientist]

Set playback speed to .75.

[@atharvakapade]

Yup it’s that of the legend

[@dekaprimatiodeandra6679]

Teacher: you have 5 minutes before the exam begins Me: *watch this video*

[@Mistyfgdf]

OMEGA LUL you made a mistake

[@sb-hf7tw]

OMEGA LUL hilarious

[@Saifthebest01]

Misty Diablo I would've just looked at memes which would backfire cause I would be distracted while writing cause I'd keep internally laughing

[@amanpandey2714]

Don't u have to write the exam

[@danilov114]

Call in a sub.... That is better...

[@77Chester77]

black pen, red pen, blue pen, purple pen,...this is getting out of hand!!!

[@Theraot]

There is also a green pen, and am hoping the orange pen will make a return in a few days

[@leftysheppey]

there was no black pen in this video. maybe he's taking a holiday

[@77Chester77]

@@leftysheppey well, there was Mr.blackpenredpen :-)

[@sebastiancastro7382]

Sounds like Goku's hair color 😂

[@bagusamartya5325]

I might be too late but do you mean.. Getting out of pen

[@Jamesz]

i’m beginning to feel like a rap god, rap god

[@blackpenredpen]

Hahaha

[@jesusshrek1271]

All my people from front to the back knot back knot

[@avinavverma2315]

@@jesusshrek1271 back nod*

[@jesusshrek1271]

@@avinavverma2315 soori for ma engris. Nyan nyan

[@avinavverma2315]

@@jesusshrek1271 it ish awkay ma frind dunt wiry

[@mairisberzins8677]

"It's not good, it's bad it's dangerous, Infact, its a trap."

[@blackpenredpen]

Mairis ̶ʙ̶ʀ̶ɪ̶ᴇ̶ᴅ̶ɪ̶s̶ Bērziņš yup yup

[@Theraot]

captain ackbar approves

[@itsbk6192]

Straight bars

[@HPD1171]

Admiral X-bar

[@chandrabitpal9151]

@@blackpenredpen how many languages do u know u also know maths and all the other science subjects maybe at Olympiad level...😶😶😶

[@NoOne-ky1er]

'How do we do this? Let pull up the purple pen again.' Thanks, now I just need to buy a purple pen for my exams.

[@blackpenredpen]

Hahaha nice and thanks!

[@jimjam1948]

This is the reason why i subscribed. BPRP probably did it in his head in ten second but took so long just because he had to explain it to us.GREAT AS USUAL.

[@blackpenredpen]

Jamie Handitye : ). Thank you.

[@honaku95]

I think he probably fell for the trap the first time he tried it.

[@blackpenredpen]

Cuong Hoang no. I am a guy with experience. : )

[@ireallyhatemakingupnamesfo1758]

blackpenredpen the sheer top energy in the comment, wow, we stan

[@chandrabitpal9151]

@@blackpenredpen true u r the man who created world record right!!

[@shaankumar2636]

I checked twice if the video was on 2x speed

[@ElColombre27360]

How fast... Did you get MATH-ANPHETAMINE?

[@ricardobrambila1842]

Crystal Math

[@jashsamani2487]

10 seconds

[@kummer45]

It's not about the speed. It's about the accuracy that things happens. Bounded above increasing sequences comes into this. The topic is highly advanced. He explains this in another video. Nested radicals is one thing but nested alternating radicals is another subject on itself. We need more content creators like this that raises interest in real analysis and higher advanced topics in mathematics. Teachers like him are widely needed.

[@Jn-xf3tt]

When you look away for one second in class

[@vinaykumar2030]

lol underrated

[@shajahan1064]

Lmao

[@proloycodes]

2^8th like!

[@shivamchouhan5077]

(3×91)th like

[@parsecgilly1495]

hi, I have found another solution to this problem, let's say, which is based on geometric and symmetry considerations: in fact we consider the curves in the Cartesian plane represented by the left and right sides of the equation: y = 5-x ^ 2 y = sqrt(5-x) the first is a parabola, with the vertex on the y axis, the second, is the same parabola, but rotated by 90° thus having the vertex on the x axis. It is easy to verify (you can use a program that plots the curves in the Cartesian plane) that the 4 solutions of the equation are the 4 points of intersection of the two parabolas; these 4 solutions are placed on a heart-shaped figure symmetrical with respect to the straight line y = x; it is easy to verify that the first pair of solutions lies precisely on this last line. therefore, to find the first pair of solutions, we can solve the following system of two equations: y = 5-x ^ 2 y = x substituting the second in the first, we obtain: x ^ 2 + x -5 = 0 which admits the two solutions x = (-1 +/- sqrt (21)) / 2 to find the second pair of solutions, it is observed that they are symmetrical with respect to the line y = x and are found on the line y = A-x, where "A" is a constant to be determined. Therefore, the constant "A" must satisfy the two systems of equations simultaneously: 1) y = 5-x ^ 2 y = A-x 2) y = sqrt(5-x) y = A-x eliminating the "y" from the two systems of equations and rearranging, we obtain two equations of second degree in the unknown "x" and in the variable "A": 1) x^2-x+A-5=0 2) x^2+(1-2A)x +A^2-5=0 but, since the two equations must provide the same solutions, this happens, if and only if the single terms of the equations are identical and this occurs only when A = 1, therefore, the second pair of solutions is found simply by solving : x^2-x-4=0 whose solutions are: x = (1 +/- sqrt (17)) / 2

[@muhammad.2]

you wrote this so beautifully, perfect spacing, punctuation and capitalization. it was a joy reading this

[@RichardWilliams-sx5kq]

“Let’s just focus on this part right here” *proceeds to circle the entire right side of the equation*

[@Akea1243]

he didnt circ- rectangle the minus

[@ActicAnDroid]

This is the first time I've ever slowed down a video just to understand it.

[@blackpenredpen]

: )

[@thomaswilliams5320]

Another way to solve this is to notice that two sides are inverse functions of eachother. The intersection of a function and an inverse function will lie on the line y = x. Therefore this can be solved simply by setting either side equal to y = x.

[@MichaelRothwell1]

Excellent observation. Algebraically, this is f(x) = f^(-1)(x) => f(f(x)) = x. For a solution we can solve f(x) = x. This is what BPRP did. But there could in theory be more solutions, depending on f. E.g. if f is self inverse, such as 1/x, then any x in the domain is a solution.

[@MichaelRothwell1]

So your argument is not water tight. If the graph of f meets y=x, this also solves f equals its inverse. But for f equals its inverse at x, we just require both (x, y) and (y, x) lie on the graph of f for some y, not that y=x.

[@kaimm8900]

@@MichaelRothwell1 could we use his argument if we do have more information regarding the solution?, in diaphontine equations for example?

[@NVDAbets]

This observation is extraordinary. But how does it find the second solution? I can't seem to find it.

[@thomaswilliams5320]

Jifu Wen I couldn't either, since technically 5-x^2 is only the inverse function of root(5-x) for x > 0 and the other solution lies in the negative x.

[@justin-7887]

Not only fast but also fairly well explained. Good job.

[@blackpenredpen]

Justin - thanks!

[@Treegrower]

Changing the x = sqrt(5 - sqrt(5 - ....)) to simply x = sqrt(5-x) was such a clever move, I never would have thought of that! Who would of thought you could simplify it to a quadratic equation. Well done.

[@kali3828]

Would have*

[@user-zb8tq5pr4x]

actually once you've done a couple of infinite sums like these it is really easy to notice this. Nothing you would have easily thought of by yourself, but hey, that's what learning is for

[@brendanwoods4773]

I don’t get why, can someone explain?

[@stevefrei2588]

@@brendanwoods4773 x= one half the square root of 5

[@peacewalker991]

@@stevefrei2588 suuuper old post I know - but what is the reason for this? Why can you do the substitution, is this just something you need to know?

[@Tassdo]

Well, once you applied your first trick to get x^2+x-5 = 0, you could also square the original equation and factor this polynomial out, getting (x^2+x-5)(x^2-x-4). You can then easily find all roots (and discard the irrelelevant ones) (Neat trick btw)

[@ogorangeduck]

we need more of this type of speedrun

[@blackpenredpen]

OK!

[@maxsch.6555]

Agreed

[@ISoldßinLadensViagraOnEbayఔ]

@@blackpenredpen Fun fact: x=1. The end.

[@xenon1308]

This man can now reedem his "top 10 rapper Eminem is afraid to diss" reward

[@Taterzz]

playing it at 2x for even faster math. there is no limit to this man, he diverges.

[@MrPetoria33]

Never noticed you could solve algebraic equations recursively before. Neat.

[@littlebigphil]

Fixed point iteration and certain infinite continued fractions are similar. You do have to be careful you don't get a divergent limit when you do this though.

[@MarcoMate87]

That irrational equation is equivalent to the system formed by the following: 5-x >= 0 5-x^2 >= 0 5-x = (5-x^2)^2 The first two inequalities are solved by -sqrt(5)

[@manuelrojas9547]

Why x_1 and ×_4 aren't acceptable? >.

[@valeriobertoncello1809]

@@manuelrojas9547 because if you plug them in you get negative values under the radical. They are indeed solutions but they're on the imaginary plane

[@ashishpradhan9606]

Thanks bro

[@think_logically_]

Factorization wasn't really obvious. However this solution doesn't leave an open question for the alternating case. This is why I prefer it.

[@Liamdhall]

sqrt(5-x) and 5-x^2 are inverse functions of one another. A function and its inverse will always meet one another on the line y=x, therefore setting either sqrt(5-x) =x or 5-x^2 = x will produce the same solutions as the original equation to be solved.

[@JM-sq3ic]

This is a very pretty argument and avoids the risk of divergent series. Top thinking!

[@goldfing5898]

This was understandable to me, in contrast to the original video. I would need a slow motion video. Is there the original version around, without speeding up?

[@thecrazyeagle9674]

That's not the case though? Solving 5-x^2 = x gets us the wrong answer.

[@Liamdhall]

It gets the first solution where x>0, but you're right that it doesn't find the second one where x

[@thecrazyeagle9674]

@@Liamdhall Haha, awesome you responded 3 years later 😀

[@qmzp2]

Pro tip: Play at x0.75 speed

[@itsalongday]

Pro tip: Play at x2 speed

[@danielangulo2119]

Protip: MAKE SURE THE SYNTH AND THE VOCALS ARE IN THE SAME KEY. Anyone?

[@sigvelandsem8669]

You can also find the solution for the positive intersection of the right and the left side, by noticing that the left side is the inverse of the right side, therfore the intersesction of the right side and the left side must lie on the line y=x. so just put either the right side or the left side equal to x and solve.

[@daroncoal2945]

i guess this is the reason why the 0.5 speed exists on youtube

[@allyourcode]

What he's doing with ellipsis substitution is kind hand wavy tho. At no point does it actually extend out to infinity. At every step along the way, you have only ever done a FINITE number of substitutions... The alternating/negative is even more sketch, because x somehow magically disappears from the right side entirely (via this "continued radical" identity, which is pretty neat). I think what's going on here is that we are implicitly relying on the fact that as we do more and more substitutions, the influence of x goes to zero, because it comes under more and more radicals as we do more and more substitutions. Therefore, the right hand side can be replaced with lim as radical_count -> inf of F^radical_count(x) where F(x) = sqrt(5 - sqrt(5 - x)). So, one thing that we are missing is a proof that this limit even exists in the first place. If so, I think the rest is ok. Or maybe we can just proceed based on the assumption that the limit exists (which is what is implicitly going on in the video), and then double check at the end that the "solution" that we "dervied" actually works. I think you can easily imagine a similar problem where "proceed based on the assumption that the limit exists" blows up in your face, and then, you'll be left wondering what went wrong. For example, let's change F to be F(x) = 2x. Then, F^n(x) = 2^n * x. Well, that does not converge except in the special case of x = 0. It only works for special values of F (such as F(x) = sqrt(5 - sqrt(5 - x))).

[@DarkMage2k]

You're a math rapper man

[@sc-ek6qz]

Hmm

[@tejasv.g5339]

when your name is Tejas and you see this on top of recommended

[@tejaskulkarni6041]

T3MPURR lol my name is Tejas as well

[@sarkar_ma]

And you're like "bhai ye kaunsa scene kardia maine?"

[@jongyon7192p]

As a speedrunner I love what you did. ...Will you speedrun a 4 part contour integral?

[@MG-wj5bn]

This is crazy to see you here, I am super interested in the 0xA community and saw your HMC video a while back, cool to see other people in that community in other places I visit as well.

[@ヤマナカシンヤ]

First of all, i'm foreigner so i can't understand his English,but i realized that i can understand what he tryed to saying Mathematics is "universal language"

[@DiegoTuzzolo]

Please proof formula on 3:36 !!!!

[@x-lightsfs5681]

I would love to see it!

[@zzz942]

There is a recursion, so that may help you

[@niltonsilveira4199]

@@zzz942 Successive Square Roots with Alternating Sign - Bong Soriano

[@kaimm8900]

@niraj panakhaniya thanks!!!!

[@mohammadzuhairkhan2096]

Expand sqrt (a-sqrt (a-x))=x. You will get x^4-2ax^2-x+a^2-a=0 which can be simplified to (x^-x-1)(x^2+x+1-a). Also note that x^2+x+1-a has two roots, and the positive one equals (sqrt (4a-3)-1)/2.

[@taikaherra8937]

First the Minecraft, and now the math itself. These speedruns are getting wild.

[@OleJoe]

My idea is to let 5=a. Solve for a, then replace a with 5 and solve for x. Then throw out the extraneous x solutions.

[@LD-dt1sk]

Sometimes I forget that math is very serious and professional

[@blitz6588]

I CAN SWEAR BY JUST LOOKING AT YOU THAT YOU WERE DESPERATE TO BREATHE OUT THE SOLUTION...... NEVERTHELESS AWESOME SKILLS DUDE!!!

[@meerable]

I never cease to be surprised by these recursive methods of solutions on your channel) it's magical!)

[@meghadridebnath1413]

The term you have used to solve the first equation is called ' The formula of Sridhar Acharya'. He was an Indian and since I am an Indian too I liked your hasty process to work out this amazing equation...

[@jinhuiliao1137]

Let y=sqrt(5-x). Then 5=y^2+x. Substitution and have y=y^2+x-x^2. Then factorize: (y-x)(y+x-1)=0 y=x or y+x-1=0 Substitute y=sqrt(5-x). and solve

[@aswinibanerjee6261]

Squaring both sides would be easier. Everything you have to do that at first solving the equation wrt 5(wrt 5 the equation is quadratic) then wrt X

[@gnpar]

Nevermind, got it. I had never seen that before. Neat!

[@Ivan-Matematyk]

This equation equals to the system: 5-x^2=y, 5-y^2=x, y>=0. From here follows that y=x or y+x=1. Then 5-x^2=x or 5-x^2=1-x. Next all is clear.

[@kimisun8315]

Two sides are inverse functions of each other => intersect at y=x

[@flipperpluto_BG]

This is so interesting. GOOD JOB CONGRATULATIONS!!

[@JohnDoe-wb2ci]

that was kinda cool math seems to be even more interesting than i thought

[@Wild4lon]

BPRP is a gateway drug

[@fergalmdaly]

Squaring both sides and then look for 2 quadratic factors works out pretty nicely. The fact that x^3 term is missing means that the factors are (x^2 + ax + b)(x^2 - ax + c) and a(c-b)=1 forces a=+/- 1 so this just becomes (x^2 + x + b)(x^2 - x + c) and quickly you get quadratics in the video.

[@damianmatma708]

03:50 - Seeing this formula, I get three very important questions in my head: 1) Can you do the video with proof of this formula? 2) And what is the formula for the second alternating series of infinitely nested radicals? I mean if the formula for the FIRST alternating series of infinitely nested radicals is: *√{ a - √[ a + √( a - √[ a + √{ … } ] ) ] } = [√(4*a - 3) - 1] / 2* then what is the formula for the SECOND (shown below) alternating series of infinitely nested radicals: √{ a + √[ a - √( a + √[ a - √{ … } ] ) ] } = ??? I know it will be: *√{ a + √[ a - √( a + √[ a - √{ … } ] ) ] } = [√(4*a - 3) + 1] / 2* (Note *"+ 1"* - not "- 1") but *how to prove it?* 3) And what are the general formulas: *√{ a + b*√[ a - b*√( a + b*√[ a - b*√{ … } ] ) ] } = ???* *√{ a - b*√[ a + b*√( a - b*√[ a + b*√{ … } ] ) ] } = ???* and *how to prove them?*

[@Nylspider]

I think he proved the formula in his infinitly nested Michael Jordan video

[@linzong-e3o]

hey,are you still there? i think i prove it myself,please let me know if you would like to see the proving process

[@ashwinraj2033]

@@linzong-e3o Sure! Please.

[@linzong-e3o]

@@ashwinraj2033 sorry, i have a lot work to do , so it was 5 days later till i saw your comment

[@linzong-e3o]

@@ashwinraj2033 please click into my channel , it has the explaining video, i don't know why i couldn't paste the link

[@sayanacharya2967]

I think a better solution will be adding -x on both sides. Like Root(5-x)-x=(5-x)-x^2 Or, Root(5-x)-x=( Root(5-x)-x).(Root(5-x)+x) So, Root(5-x)-x=0 or Root(5-x)+x=1 From there solve it and take out the extraneous solutions.

[@shmuelzehavi4940]

Very nice way of solving, however the solutions are still the same two ones. The 2 extraneous solutions do not satisfy the original equation. A very elegant solution.

[@gruk3683]

Him: How to solve without squaring both sides Also him at 1:55

[@nikolaskhf]

Speed run... but you still explain it.. what a great teacher...

[@blackpenredpen]

: )))))

[@mariomario-ih6mn]

0:10 the way he smiled.

[@erickwat3216]

I like how he makes a video later by squaring both sides, then sets up the quadratic formula as 5=... absolutely brilliant

[@AndroidGamingrepublic555]

I love his fast calculation abilities. Wait "super fast"

[@tejasappana4097]

Thanks for letting me know!

[@ΓιωργοςΓουργιωτης-ι8ρ]

I don't comment a lot but this video was actually amazing

[@blackpenredpen]

Γιωργος Γουργιωτης thanks!!!!!!

[@justinlewtp]

Let y = f(x) the= 5-x^2 Thus x^2 = 5-y x = +-√(5-y) Therefore, f^-1 (x) = +-√(5-x) f^-1 is the mirror image of f along the line y=x, hence f^-1 intersects f at the line y=x Therefore since the question is basically the form f^-1 = f , by considering f(x) = x and solving, gets the same answer

[@thechannelofeandmx4784]

Man...I want whatever kind of coffee you had before this video😂

[@pitachipenthusiast]

Nothing but respect for our favorite math legend!

[@urironen250]

Please make a video of the formula looks cool

[@plexirc]

yes please!

[@x0cx102]

Another beautiful way to solve this is actually square the equation, but insfor tead of solving for x, "solve for 5" because you get a quadratic in terms of 5 as the parameter. Upon simplifying, you get a 5 = two possible quadratic in x, which you can then directly solve.

[@KevinS47]

That is a genius thumbnail right there haha

[@CauchyIntegralFormula]

As a bit of an aside, we can set y = sqrt(5-x) = 5 - x^2. Then, y must be positive, and both y^2 = 5 - x and y = 5 - x^2, or: x + y^2 = 5 and x^2 + y = 5. We'll consider y needing to be positive at the end; otherwise, there are at least two solutions to this system of equations, namely those that satisfy x = y, and x^2 + x - 5 = 0. The quadratic equation gives us those easily. Returning to the main equation, squaring both sides gives us 5 - x = x^4 - 10x^2 + 25, or x^4 - 10x^2 + x + 20 = 0. This is a quartic, which is normally troublesome to factor. However! We already know two solutions, namely the solutions to x^2 + x - 5 = 0. That means that x^2 + x - 5 must be a factor of the left-hand side, and indeed we see through division that x^4 - 10x^2 + x + 20 = (x^2 + x - 5)(x^2 - x - 4). So the other two solutions to the quartic satisfy x^2 - x - 4 = 0, which we can again find via the quadratic formula. That's all four solutions. Just throw out the extraneous ones (which are the ones where 5-x^2 is negative) and we're done!

[@vlliss]

And I'm here doing my GCSEs. I wish I understood as much as these guys on KZbin. Are there any masters who want to give me some first hand tips? I like maths but I'm not too great at it. I wish I was so badly.

[@satwiksortur7814]

I'm an year late but bear with me, The most important part is being able to simplify every step to make it easier to calculate. Factorization will help you a lot in every step. Its also good to practice mental math to speed up your calculations.

[@tryphonunzouave8384]

That's so cool, I love seeing people rush things (well as long as they are still done correctly)

[@victoirevim9698]

You need to do maths speedruns at the AGDQ. I'd donate money for that.

[@klausg1843]

Another method: let f(x)=sqr(5-x) and g(x)=5-x^2. Then f(g(x))=g(f(x))=x. Let a=g(a). Then f(a)=f(g(a))=a. So f(a)=g(a) which means that a is a solution. And a solves a=5-a^2. from where you get the two solutions.

[@LAMG059]

Teacher : The test wont be soo hard it's only from what we studied in class The Test:

[@BenSmith-xs1yi]

I think I found a better solution. Identify that the left is the inverse of the right and so x=5-x^2, gives one solution (take the positive solution). Now comes a slightly harder part. You want the intersection of y = - x + c such that the two solutions are equidistant to c/2 ( the intersection of y = x and y = - x +c ). So first solve, - x + c = 5 - x^2, to get x = (1+- sqrt(1- 4(c- 5)))/2, the midpoint of these solutions at a given c must land at c/2, so (x_1 + x_2)/2 =1/2 = c/2, so c = 1, then take negative solution.

[@Wabbelpaddel]

Lol that's efficient. I substituted u = sqrt(5-x) Then had -u⁴ + 10u² - u = 0, and then... I used Ferrari's 4th degree polynomial formula 🤣

[@unemployed756]

0:01 Doraemon theme song.

[@TheKnowledgeOfScience]

😢😢😢hmmm❤🎉

[@stackexchange1065]

Nice solution. Looking at it geometrically, plot the two functions y^2 = 5 - x (corresponding to LHS) and y = 5 - x^2 (corresponding to RHS)..they intersect at 4 points which are the solutions to the equation. Just subtracting the two equations gives you all the 4 solutions. (y - x)(y + x - 1) = 0 Hence 2 of them lie on on y=x and the remaining 2 of them lie on y +x = 1. Now expressing y in terms of x as per the straight lines and substituting it in the parabolas gives us the quadratic equations leading to the solutions

[@anmolbansal2604]

My secret weapon: ×0.75

[@blobropch0p]

Let L be the line of intersection of the planesx+y= 0 andy+z= 0. (a) Write the vector equation ofL, i.e., find (a,b,c) and (p,q,r) such thatL={(a,b,c) +λ(p,q,r)|λis a real number.} (b) Find the equation of a plane obtained by rotatingx+y= 0 aboutLby 45◦

[@givecamichips]

Nice any% Speedrun, I hope to so your entry in the 100% run including proving the sqrt((4a-3)/2) formula.

[@izzystephens3550]

These videos makes me fall in love math all over again ❤️

[@yuarkok6273]

My first thought was like: "Is he speaking english?"

[@shmaxg]

There is another nice way to solve the problem. Let a = 5, then you get sqrt(a-x) = a-x^2. Now square it and solve the equation with respect to a. It is very easy since the equation w.r.t. a is quadratic and has a simple determinant. Then you get two quadratic equations w.r.t. x that are easy to solve.

[@Wild4lon]

When BPRP starts saying 'is nothing but' you know he watches papa flammy's vids

[@blackpenredpen]

Wild4lon hello, new subscriber

[@vaedkamat484]

I don't know what video is more satisfying

[@AwesomepianoTURTLES]

Professors hate him! Learn how this man made thousands speedrunning math on KZbin just using this one small trick.

[@blackpenredpen]

: )

[@kevinl1080]

I love how happy he gets 3:05

[@jesusthroughmary]

Blue ped red pen purple pen YAY

[@blackpenredpen]

jesusthroughmary yay!!!

[@jw7416]

Rip black pen

[@jassskmaster7575]

I couldn't make out a single word you said but the math speaks for itself

[@rot6015]

Am i on some drugs or why is he talking so fast to me

[@Bicho04830]

Both.

[@sc-ek6qz]

Same.

[@arushdixit2792]

This concept is really useful in calculas problems.

[@thephysicistcuber175]

Oh and btw: degree 4 equations aren't terrible, you can always solve them with radicals :) although that wouldn't make it good for a speed-equation-solving

[@rezazom27]

(5-x)^1/2 = 5-x^2 changes to the Quartic equation: x^4-10x^2+x+20=0 which is factored into: (x^2 +x -5)(x^2-x-4)=0, thus we have two quadratic equations, which can be solved by quadratic formula and obtain the 4 solutions: (1+ √17)/2, (1-√17)/2, (-1+√21)/2, (-1-√21)/2

[@Sid-ix5qr]

I thought the whole video was a rap.....

[@yoavshati]

It wasn't?

[@papapapapapapageno]

Isn't it?

[@mandaglodon]

You are most cool mathematician that I have ever seen in this life.

[@mandaglodon]

After Einstein and Hawking.

[@tamirerez2547]

1:07 The red marker: Please write more slowly !! I'm not enough to spend ink !!! Hellllp!

[@TheGrailFinder]

Here is my solution. We want to find the intersection of the curves y=sqrt(5-x) and y=5-x^2. We can rewrite the first equation as x=5-y^2. So at the intersection we have x=5-y^2 and y=5-x^2. Subtracting one equation from the other, we get x-y=x^2-y^2 => (x-y)=(x-y)(x+y). One solution is when x-y=0=>x=y. So we can write x=5-x^2, which is a quadratic equation, solve in the usual way. When x!=y, we can divide through by the (x-y), giving x+y=1 => x=1-y. Substitute y=5-x^2 and we get x=x^2-4, which is another quadratic, solve in the usual way.

[@anthonyr.748]

"And of course, if we want to be cute--because we ARE cute" 😍 yall this man is a keeper we stan 💖

[@balrajtavanandi3332]

coolest intergral prof love it

[@meiz1795]

3:34 can't wait for the proof :>