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@rodbenson58798 ай бұрын
Wow I'm shocked at the negative comments on this video. First there are those complaining the steps are not explained well even though he did the steps right down to basic arithmetic, along with one comment complaining the opposite (can't please all the people all the time and all that). Secondly, his solution is still a clever purely algebraic way of dealing with problem which assumes only basic maths. As for those droning on about Pascal's triangle, well why don't you make a video using this solution and see how long it takes you to first teach how the triangle comes about as a way of finding the exponents of a binomial expansion. Man there are a lot of haters out there. So let me just encourage the content creator and say well done on making a very carefully explained video of an elegant solution to the math olympiad problem.
@davidhowe69058 ай бұрын
Well said!
@BP-gn2cl8 ай бұрын
Right
@Player-pj9kt8 ай бұрын
People are complaining how he skipped a step at 7.17 without explaining it.
@rodbenson58798 ай бұрын
@@Player-pj9ktin defence of the content creator, he had already expanded (a+b)^2 showing it to be equal to a^2 + b^2 + 2ab so someone watching carefully should be able to spot the refactorisation although I guess a quick sentence explaining he is reversing the process would have been helpful for those who know very little highschool algebra. However, this slight improvement does not justify the negativity of some of the comments.
@davidhowe69058 ай бұрын
It's the same procedure as expanding (x + y)^2 = (x + y)(x + y) = x^2 + y^2 + 2xy but with x=a^2, y=b^2@@Player-pj9kt
@englishchannel51852 ай бұрын
Good teachnique for complex math solving
@AprioryRus8 ай бұрын
I assume this problem is about right grouping and then using just 2 equations: 1st (a-b)^2=a^2-2ab+b^2 2nd a^2-b^2 =(a-b)*(a+b) This way it should be possible to avoid complicated calculstions. And 111-11=100 pressumes to use those equations.
@kamilrichert84465 ай бұрын
If one wants to be extra rigorous, you need to make sure that the squared and rooted number is not negative before you cancel the power and the root. In this example it is obviously positive, but under any test-like conditions it's always better to show having considered such things
@rssl55002 ай бұрын
But a^2+b^2+ab is always positive
@kamilrichert84462 ай бұрын
reread the last sentence please@@rssl5500
@DavidVonR6 ай бұрын
I tried to do this one mentally. After awhile, I had to take the square root of 125,903,841, which I knew had to be roughly 11,xy1. Finding the digits of x and y was tough.
@jonathansobieski29626 ай бұрын
There is an algorithm for doing square roots by hand which makes the calculation very doable without guessing.
@RvVx75 ай бұрын
@@jonathansobieski2962gimme
@hajimehinata58544 ай бұрын
You got that number mentally???
@DavidVonR4 ай бұрын
@@hajimehinata5854 Yes, I am gifted at mental math and can do mental computations to many millions.
@austenmaster89812 ай бұрын
The best i got at this mentally was 100159841+111^4 Wow
@mosheshamay34758 ай бұрын
Very nice and good idea,
@vascobishop33595 ай бұрын
An interesting exercise.
@barttemolder34058 ай бұрын
The last step can be further simplified: a² + b² + ab = a² + 2ab + b² - ab = (a + b)² - ab = 111² - 11x100 = 12321 - 1100 = 11221 ...
@drinkchan48228 ай бұрын
the point was to not do 111^2 since it the whole point of this algebra was to make it possible to evaluate it in the easiest form when the algebra is converted back into numbers
@malaramesh87667 ай бұрын
Good idea
@user-qv2dd8ex8k7 ай бұрын
That works actually. I had to think about that negative number for a moment.
@rakhatthenut38154 ай бұрын
But it became even harder, how is it a simplification?
@barttemolder34054 ай бұрын
@@rakhatthenut3815 You only have to do 2 multiplications and one subtraction this way. In the general case that would be easier. For this particular set of numbers it is not necessary as it is also trivial to calculate the result one step back, but the fun is in reducing the equation to the bare minimum.
@jpl5694 ай бұрын
Well, the problem stands mainly upon the property : a^4 + b^4 + (a + b)^4 = 2 (a^2 + b^2 + ab)^2. Let us prove it with a far simpler calculation : We define P(X) = (X^2 + aX + a^2)^2 Then P’(X) = 2 (2X + a) (X^2 + aX + a^2) We define Q(X) = X^4 + (X + a)^4 + a^4 Then Q’(X) = 4 (X + a)^3 + 4 X^3 Using the well-known x^3 + y^3 = (x + y) (x^2 - xy + y^2), we obtain : Q’(X) = 4 (2X + a) [(X+a)^2 - X(X+a) + X^2) That is : Q’(X) = 4 (2X + a) (X^2 + aX + a^2). Obviously, Q’(X) = 2 P’(X), then Q(X) = 2 P(X) + k. Noticing that Q(0) = 2 P(0), we have k = 0 (qed). Thanks for your interesting videos ! 🙂
@gonzoz18 ай бұрын
Use Pascal's triangle , for gods sake.
@142536898 ай бұрын
Or just the binomial theorem
@HoSza18 ай бұрын
@@14253689which has 1:1 connection to Pascal's triangle :/ (ong use potatoe! - or use spud.)
@ChristelleHilaire-lb6pu8 ай бұрын
Delta it is sayed in my country
@brownie34548 ай бұрын
@@14253689that is what he used in the video
@lorenzolombardi12116 ай бұрын
Tartaglia's Triangle
@user-dg3ji3ju3i4 ай бұрын
It is amazing how such a complex math problem after a few steps can equal a whole number and not a decimal.
@zvonimirkujundzic68673 ай бұрын
You are the best!🎉
@YualChiek4 ай бұрын
Excellent video! Clear explanation using basic principles.
@learncommunolizer4 ай бұрын
Glad it was helpful! Thank you very much
@thekennethofoz35944 ай бұрын
When I was at school, decades ago, we configured "let" statements this way: "let a=b+1". In this solution, it's done like that. But many other maths problems here on KZbin configure it as "let b+1=a". Is there a correct way, or does it just not matter? To my mind, the first way makes more sense, but I'm happy to be corrected if I'm wrong.
@hmwndp3 ай бұрын
If 2+2 = 4, then 4 = 2+2 as well. It is the same for variables.
@AlaiMacErc2 ай бұрын
Curious! I don't recall seeing the second style at all. I'd agree the first is much clearer, as it's saying which the new variable is, and it's giving a closed-form expression for its value. But one could argue any statement of the form "let P(x)" where P is some proposition that determines a value for its parameter is valid... if harder to follow.
@knotwilg3596Ай бұрын
this is of the form V(A/2) where A = x^4+y^4+(x+y)^4 expand the latter term and add corresponding terms so that A = 2x^4 +2y^4 + 4x³y + 4xy³ + 6x²y² hence A/2 = x^4 +y^4 + 2x³y+2xy³+3x²y² regroup this as (x^4 +y^4 + 2x²y²) + (2x³y+2xy³) + x²y² = (x²+y²)² +2xy (x²+y²)+ x²y² = [(x²+y²)+xy]² hence V(A/Z) = (x²+y²)+xy fill out x,y to find 10 000 + 121 + 1100 = 11221
@VolkGreg8 ай бұрын
Let s = 5.5, t = 105.5 = 100+s, and let x = the answer. 2x² = 11⁴ + 100⁴ + 111⁴ = 16s⁴ + (t-s)⁴ + (t+s)⁴ = 2[8s⁴ + t⁴ + 6s²t² + s⁴] = 2(t² + 3s²)² x = (100+s)² + 3s² = 100² + 200s + 4s² = 10,000 + 1100 + 121 = 11,221
@travisporco8 ай бұрын
nice
@evertvanderhik57748 ай бұрын
The important part is to see 111 = 100 + 11. The rest is a matter of "a little algebra yields..."
@straightpool19646 ай бұрын
Yes
@adrianhorsnell89008 ай бұрын
Great, very clear!
@learncommunolizer8 ай бұрын
Glad it was helpful! Thank you
@user-sl9xz6ix9e8 ай бұрын
GREAT
@learncommunolizer8 ай бұрын
Thanks ❤️🙏❤️
@shivnathbanerjee58684 ай бұрын
To find (a+b)^n expansion we can draw n+1 lines of pascal's triangle and the (n+1)th line gives the coefficients of all terms in the expansion. Then we can write the terms by decreasing the powers of a from n to 0 and increasing the powers of b from 0 to n along with the coefficients. And add all the terms. That is the expansion.
@josiahgibbs56974 ай бұрын
I wondered why he didn't do that also.
@richardleveson64675 ай бұрын
Thank you - this is a clever solution!
@marioperic97095 ай бұрын
I have started in general approach; to solve (a^4+b^4+(a+b)^4)/2. which is actually: (a^2+b^2+ab)^2 and solution in general is a^2+b^2+ab...
@wayneyadams8 ай бұрын
Fortunately, those of us not having to participate in the Math Olympiad can just punch it into our calculators. LOL It is still interesting to see how the answer is worked out using just pencil and paper. It is fascinating how many times a problem that looks complex as all get out can be simplified with the judicious use of substitutions.
@squirming_squirrels8 ай бұрын
This is defenitely 100% not a math Olympiad question.
@andreibratosin11997 ай бұрын
Lol this is not olympiad level. Where I'm from we used to do these in 7th grade .. and much faster cus this method is archaic and lengthy
@tigistafine2026 ай бұрын
I agree with your statement.
@aaronhansen7066 ай бұрын
I am in my 50s. I got my GED at 17 and I think I aced it although it is nothing but pass/fail. I did a couple years at a community college and decided I would rather labor for a living. I do some algebra in my head as I walk around sometimes. Usually to figure out what the mindset of the civil engineer was thinking. I can only do it in my head when I am distracted by music. I need to be distracted to concentrate. My fav college course was actually a lifesaving class at De Anza college in Cupertino, CA. It covered some very... bad situations and gave a ton more info than the Boy Scout stuff I learned. I do love training my brain
@YorubaMathematicsClass-Y-kp5cp4 ай бұрын
You have done well in this long process
@eskaykim54188 ай бұрын
Anyone would think of letting a=11 and b=100 and then using FOIL to simplify. I believe those who left negative comments on this video thought there should be more creative and easier way to solve this since he mentioned that there would be a 'trick'
@yapadek30982 ай бұрын
Using the "Pascal triangle" (in french) : 1, 1 1, 1 2 1, 1 3 3 1, 1 4 6 4 1, ... the developpement of (a+b)^4 is faster ;-) (NB This triangle is also used for Cn,p) Sorry for my english and thank you for the video;
@2bleubird8 ай бұрын
I think it would just be easier to multiply everything out
@malaramesh87667 ай бұрын
No not at all you need to multiply 11 six times. Then need to workout square root for huge number.
@barneyronnie8 ай бұрын
Great job; crystal clear!!
@learncommunolizer8 ай бұрын
Thank you! Thanks 🙏❤️🙏
@leftofcenter48 ай бұрын
This looks familiar from my school days what level of math is this? Algebra 3 and 4 ? Or pre calculus. Understanding those sub categories is where I lost it years ago..
@user-kr1zj6lm2u2 ай бұрын
I love the many ways to go about the solution. It is a shame that modern math ed does not encourage other processes for a soltion.
@willdejong77638 ай бұрын
Nice.
@learncommunolizer8 ай бұрын
Thanks!🙏❤️🙏
@blacktamizans78476 ай бұрын
For getting the value of a² + b²+ab You just conver into (a+b)² -ab = (100+11)² -(100 x 11) =12321 -1100 =11221 OK.
@jmich78 ай бұрын
Really really cool
@learncommunolizer8 ай бұрын
Thanks 🙏❤️🙏
@konchady18 ай бұрын
Looks like you expanded it. Here is a trick: a^4 + b^4 +(a+b)^4 = a^4 + b^4 + 2a^2b^2 +(a+b)^4 - a^2b^2 -a^2b^2 =(a^2 +b^2)^2 - a^2b^2 +((a+b)^2 - ab)((a+b)^2 + ab) = (a^2 + b^2 + ab)(a^2+b^2-ab) + (a^2 +b^2+ab)(a^2+b^2+3ab) =2(a^2+b^2+ab)^2
@rakos16 ай бұрын
1000101
@johnyoung66373 ай бұрын
I can't wait to use this.
@isabellanievesthegaminggir65589 ай бұрын
👍🏻
@learncommunolizer9 ай бұрын
👍👍👍
@ShinyStarOfDeath6 ай бұрын
The issue with those videos is that they are made with the solution and tactic already in mind, rather than experience it first hand. Although I understand your tactic , as others have already stated, you rushed it as the most critical point (which renders this whole process useless). Probably because you forgot the next step, but had the answer in mind.
@trondarnepettersen51968 ай бұрын
The expression under the last rootsign should have been written [(a^2+b^2)+ab]^2. Would have been more clear.
@navi27104 ай бұрын
I just read it as : X = a2 + b2 Y = ab So (X + Y)2 = x2 + y2 +2xy = (a2 ×b2)2 + (ab)2 + 2ab(a2 + b2) Therefor (X + Y)2 is also (a2 + b2 + 2ab)2
@lukaskamin7555 ай бұрын
Are binomial theorem (of Newton) and squaring a sum of 3 terms such unheard things in Japan?
@xaashi20226 ай бұрын
Wow
@learncommunolizer6 ай бұрын
Thanks 👍☺️
@mirzatayerejepbayev83677 ай бұрын
Very good !!!
@learncommunolizer7 ай бұрын
Thanks a lot! 😊👍
@gavintillman18847 ай бұрын
Better to expand (a+b)^4 all in one go, binomially?
@xyz.ijk.9 ай бұрын
Usually you show us every step. This time for some reason you didn't show us the critical step. Why did you do that? It's extremely difficult to understand how to combine all of those terms into SQR of (a^2+b^2+ab)^2
@serbanudrea94299 ай бұрын
Just expand it according to (x+y)^2 = x^2 + y^2 + 2xy by letting x = a^2+b^2 and y = ab.
@xyz.ijk.9 ай бұрын
@@serbanudrea9429 Excellent; thank you for taking the time to respond. This was very helpful.
@hybridaccounts8 ай бұрын
Not difficult. Just the basics of (a + b)² = a² + b² + 2ab
@huyminhha6588 ай бұрын
why didnt you recordnize the (A+B)^2 equality
@xyz.ijk.8 ай бұрын
@@huyminhha658 obviously because I need a lot more work. That's why I value this channel.
@user-fb1qo8go1b3 ай бұрын
excellently !!
@learncommunolizer3 ай бұрын
Thank you very much!!
@bobajaj42243 ай бұрын
you could just use Pascal's triangle to find (a+b)^4 instead of going through 2 steps...
@alperbykgln69554 ай бұрын
❤
@David-Dash-IBA5 ай бұрын
It feels like a puzzle. You are stumped until you know the trick.
@paulorodriguesbarros74182 ай бұрын
Solução elegante!!!
@canr7726 ай бұрын
6:50 u said 2a^2b^2 but it was 3a^2b^2
@gyrlyninja5 ай бұрын
THANK YOU!!!! I thought I was going crazy!
@LilCletus5 ай бұрын
How to spend all your time on one problem and fail the test.
@jonathansobieski29626 ай бұрын
If you know how to calculate square roots by hand then this question is relatively easy to just do the calculation.
@Abhaykk19948 ай бұрын
Good explanation as always. We can just use the binomial theorem. Could be solved in a jiffy. This is way too long and max time allowed for such a questions is 2 mins Max.
@anirudhabanerjee68484 ай бұрын
Most easy method i am telling you. If possible then note it down.... We know that in the language of exponent we define ^nroot a as a^1/n So root 11^4 becomes 11^4×^(1/2) root 100^4 becomes 100^4×^(1/2) And root 111^4×^(1/2) Now by fraction we know that a+b+c/2 = a/2 + b/2 + c/2 Thus, 11^4×^(1/2)/2 + 100^4×^(1/2)/2 + 111^4×^(1/2)/2 After cancellation we get =11^2/2 + 100^2/2 + 111^2/2 = 11×11/2 + 100×100/2 = 50 × 100 + 111×111/2 = 121/2 + 5000 + 12321/2 Using Associative law here Then, 5000 + 121/2 + 12321/2 = 5000 + 121 + 12321/2 = 5000 + 12442/2 = 5000 + 6221 = 11221. SIMPLE!
@yula005real4 ай бұрын
Молодец, очень подробно изложено.
@user-nc8dy7sd6f8 ай бұрын
Очень красивый пример! Моё решение немного отличается от вашего, но в общем всё одинаково.
@laogui24254 ай бұрын
PAINFULLY slow - for the inner expression anyone doing this level maths can go straight to a^4 + b^4 + 2a^3b +2ab^3 +3 a^2 b^2 in a single step! Then (a^2+b^2)^2 + a^2b^2 +2 (...) then (a^2+b^2 +ab)^2 - 3 steps, and without the need for substitution with a,b
@lucksys4 ай бұрын
Be carefull when you simpify the square root of the square (this is a module definition).
@TMoDDD8 ай бұрын
I like when you say powa xd
@mariats28653 ай бұрын
Красивое решение, понятное объяснение, Вы лучше всех объясняете на Ютубе
@nagamanib653916 күн бұрын
The technique explained is very lengthy and can be solved easier
@mathpro9264 ай бұрын
I like maths
@Tirelesswarrior7 ай бұрын
Beyond basic algebras 1 and 2, geometry, high school statistics, and precalc, math becomes so abstract. 😅😅😅 I suspect that being well grounded in the courses I've mentioned is enough to survive the business world. The exception is for those pursuing engineering and specific science endeavors
@CraigH9996 ай бұрын
Still don't understand how to solve this. It seems like just random things get written down out of nowhere. So frustrating to be so bad at math and I try to learn with videos like this but I just get more confused and frustrated.
@logica_19896 ай бұрын
You should give up. Stop banging your head on something you were not made for. You're never going to be good at this. Pursue something else.
@funprog2 ай бұрын
You need to remember the algebra identities (a+b)^2 = a^2+b^2+2ab etc
@nickkunst9526 ай бұрын
9:37 It's plot twist that you didn't do mental arithmatic for simple addings after all of these complex calculations..
@doghousedon16 ай бұрын
Easy peasy. 42
@terry_willis5 ай бұрын
Excellent explanation until about 8:16 when that step eluded me.
@michaeltieber35507 ай бұрын
Only saw the thumbnail and calculated it in my head so I might have made a mistake, but my solution is 11221.
@omerhamzasacan735 ай бұрын
Are you allowed to change (a^2)^2 + (b^2)^2 TO (a^2 + b^2)^2
@GaryBricaultLive2 ай бұрын
At least this madness resulted in a single actual number and not some messy and unresolved quadratic equation solution!
@tamarshahverdyan27234 сағат бұрын
# 45 #
@manda3dprojects9668 ай бұрын
The question is "How did you know that the square root will cancel a square in the end?", did you saw the future, or just believing that every math question is always like that?
@thothorleboiteux99007 ай бұрын
That's easy : you don't know. You just try and cross your fingers. And if it doesn't work, you try something else!
@funprog2 ай бұрын
These kinds of problems are designed to be simplified like this
@r.awilliams98156 ай бұрын
Um, yeah...I'll just go over here and hit this rock with another rock.
@waldro497 ай бұрын
Why those steps? Amazing
@swapnilmaurya77676 ай бұрын
I calculate all the problem in just 50 second Calculation is easy than trick😂😂😂
@danielsteiner70884 ай бұрын
Empirical thinking
@TWJRPGGamming9 ай бұрын
let a = 100,b=11 , =((2a^4+4a^3b+6a^2b^2+4ab^3+b^4)/2)^(1/2)=(a^4+2a^3b+3a^2b^2+2ab^3+b^4)^(1/2) =((a^2+b^2)^2+2ab(a^2+b^2)+(ab)^2)^(1/2) = (((a^2+b^2)+ab)^2)^(1/2)=(a^2+b^2)+ab = 10000+121+1100 = 11221
@TWJRPGGamming9 ай бұрын
I am almost the same as your video until "6:49" :p
@DiscoCatsMeow6 ай бұрын
This stresses me out so much. Maths has always stressed me out. It just 3 numbers 121+10000+12321 then divided in half. I don't understand all of this. No one ever explained to me why or what all the steps are for. It's literally 4 steps why is all of this necessary? Like what is the purpose? Even in school no one would ever explain to me why that was necessary. I always got in trouble because I couldn't show my work. But they never told me what work I was supposed to show, I just freaking gave up. Even on this why are they writing 47000 different numbers/letters? WHY are their letters? And where do the squares come from?😭😭😭
@rogelioroldan9527Ай бұрын
Ecotú querido. Como te gusta complicarte.
@sandorMrBeen8 ай бұрын
My gush, you never heared from Pascal's triangle???
@pincopallino1558AlfiWord8 ай бұрын
111^2
@aaronhansen7069 ай бұрын
I have to admit, you completely lost me on this one. I love how you make algebra simple... usually. I watch your videos partly for fun. Partly to help my kids understand letters in math. I am going to have to rewatch this one
@aaronhansen7069 ай бұрын
After rewatching I got it! I'm going to challenge my senior citizen Dad to this.
@learncommunolizer9 ай бұрын
You can do it! Great to hear 👍👍👍
@aaronhansen7069 ай бұрын
@@learncommunolizer while I do understand the how part, I am still lost at the why you put the 111^4 to squared squared. It makes sense but I don't understand the why part.
@leif10758 ай бұрын
@learncommunolizer my alternate method is rewrote 111 as 110 + 1 and take 4th power then 110 is same as 11×10 which you have factor of in jther terms..see what I meana?
@leif10758 ай бұрын
@@learncommunolizerindont see why anyone would do the ma ovulation you do at 7:13..I don't see anyone thinking of that..would you agree?
@minhdangvlogs8 ай бұрын
It is Cardido indentity
@rehaozenc8 ай бұрын
😂😂😂 If we do the normal process, the solution takes 3-5 minutes, see video 10.
@DJF19475 ай бұрын
Rather laboured.
@JarppaGuru3 ай бұрын
we write that python and print answer. no calculator used
@botron-san44616 ай бұрын
I have test rn, wish me gl
@mikeneal89008 ай бұрын
How did I get the same answer by taking the sqrt of the original problem and getting (11^2 + 100^2+111^2)/2. Which is essentially the same as the very last equation
@uwearnold36808 ай бұрын
Thats my idea too. 121 +10000+12321/2 22442/2 11221 ... and need not so much paper😊
@witta4thewinwitta4u144 ай бұрын
Den in nes step...
@zonked12002 ай бұрын
The step before you reverted a and b back to 100 and 11, you didn't go over how you got root((a^2 + b^2 +ab)^2). I'm totally not seeing how you got to that point.
@billh5923Ай бұрын
He doesn't know what cancel means. He used the term cancel when he was combining terms and crossing them out. This is just bookkeeping, it is not a cancel.
I solve the same but using 111-11 in 100 ^4 bracket
@kokomo97647 ай бұрын
Haha, it said no calculators allowed. Bruh, a calculator wouldn't help me solve this.
@Eris1234516 ай бұрын
I really enjoyed watching that and despite my never being that quite that good at expansions even at my best I still thought every step was clear and explained perfectly well and I had no difficulty following it at all. People sometime miss the fact that sometimes math can just be fun.
@ChristelleHilaire-lb6pu8 ай бұрын
A=11cm ; B=100cm
@stlyns5 ай бұрын
11,221. I used a calculator.
@Genrih07 ай бұрын
Простая школьная задача. Непонятно, что в ней такого сложного.
@NegativebplusorminusthesquarerАй бұрын
(100+11)^4 does not equal 111^4
@robertgapatas8 ай бұрын
Elshadai, criador Elshadai, salvador Elshadai, o poderoso Elshadai, o grande eu sou
@RobiBue8 ай бұрын
Você esqueceu El Shaddai, o matemático
@user-qv2dd8ex8k7 ай бұрын
If you can do this you just passed college algebra. 😁😅