BELIEVE IN ALGEBRA, NOT CALCULATOR
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- čas přidán 22. 08. 2024
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blackpenredpen | 曹老師
i did some mental math, but hit a wall at trying to find the square root of 63,252,753,001
that’s some impressive mental math assuming you’re telling the truth . Is there a trick or something
@@iamgroot3615 theres no trick hes probably lying
r/iamverysmart
AngryAxew there's no reason not to be able to mental math those numbers
Like 500(500+1) which is easier which is 250000+500 and it similar to the end
I mean it's OBVIOUSLY 251501.
1990: we'll have flying cars by 2019
2019: 2=1+1, wow I'm a genius
LOL
2+2 is 4, minus one that's three quick maths
Flying cars... you can't even have a sharpie that could change color. Say, red and black.
@@ghotifish1838 topical meme reference
Well it can also be 2=500-498
I just started learning English, but the explanations are clear and interesting even at my levels of English. Thanks a lot 😁👍
Юлия Охременко I am
Glad to hear!
x2
U r indian Chinese korean or ....???
@@harelavv8806 the name may seem obviously Russian to some but not all
@@blackpenredpen hii
My last words whispered in a final breath : "Don't forget the +1"
😂
LMFAOOO
He never resumed the video
He added the one wdym 🧐
Wow... this essentially proved that if you take the product of four consecutive -numbers- integers and add one to it, than it's gone be a square number.
Awesome! Good spot!
Gábor Tóth holy shit you’re right! That’s crazy man
Yup!!
The most pathological case I can think of is -1 thru 2, and yes indeed I get 1, which is a perfect square.
My professor had us prove a more general result: take the product of four numbers in arithmetic sequence, then add the fourth power of their common difference. Show that the result is a perfect square.
Instead distributing at 4:18
u = x^2 + 3x + 1
(u - 1)(u + 1) + 1 = u^2
So the root is x^2 + 3x + 1 = 251501
Bryan Lu omg that cat!!!!
Or even u=x^2+3x, then u^2+2u+1
digno de nyan cat, jajajajaj
My life is a lie. I thought u subbing was only for integrals
@@mattat3847 nah man, sub whenever it makes the problem simpler
Olympic math taught me that insanely hard problems often had elegant solutions, this is no exception.
: ))))
@@leif1075???
@@leif1075 people like these are called problem solvers...
@@leif1075 well it took me about 5min to solve it so I think is not impossible to solve. All of this types of equations where you have 4 consecutive numbers multipled are done like this
Lol,This problem is actually super easy,(Every single Olympiad contestant would have solved this question,at some point of their life) .Insanely hard problems need not have simple solutions . That's a downside of the math Olympiad .They make you expect difficult problems have simple solutions.(Although,most imo contestants don t fall for this fallacy).Real insanely hard problems have not been solved by anyone,yet.
"If you're using a calculator, why are you watching this video?"
Sanity check.
hahahahaa
I am a calculator, not a person
*laughs in Shakuntala Devi
Did you know that 2 = 1 + 1?? I bet not!
jk : )
blackpenredpen No i don’t, i need a calculator to this
That's pretty funnt, but sometimes such things are the most difficult to see, for example: we have f(x)=x^4+8x^3+18x^2+8x+17, and a question, for which x, the function f(x) is a prime. You can check infinitely many cases and never know the answer, but what makes this question easy (but on the other hand is not so obvious), is that 18=17+1. Because then we have (x^2+1)(x^2+8x+17), which has to be a prime. One of them has to be = 1, the other one has to be some prime then... We are left with only 2 cases, because we know, that 18=17+1. : )
No but I knew 2= 0.9+1.1.
I Known 2+2 = 5
What i know is 5/2 = 2 with int data type
i put it in my calculator and got 251501, that was easy
Boo!
@The Balton American calculators are beasts
@@Netherexio Agreed but Americans aren’t
@@paradox9265 What do you mean?
It was easy, but not so beautiful like this)
Jeez thats smart
*proceeds to use the calculator to prove that 251501 is the right answer*
Nice. I did it this way:
Assume that the expression is a square number so:
x(x+1)(x+2)(x+3)+1 = n^2
x(x+1)(x+2)(x+3) = n^2 - 1
x(x+1)(x+2)(x+3) = (n+1)(n-1)
What I did then is realise that the factors of the product on the right differ by 2. Playing around you can find that:
x(x+3) = x^2+3x = n-1
(x+1)(x+2)=x^2+3x+2 = n+1
So n = x^2 + 3x + 1
Not as neat as your method though!
Thanks for the video
OMG ive almost done it completely. i just stopped at (n+1)(n-1) lol WD! i mean 'not that almost' lmao
How do I play around with it
@@joshuamason2227 Well you have the product of 3 binomials and a monomial for which we can multiply in any order. If you try a few cases, or think about it you spot that x(x+3) and (x+1)(x+2) have a difference of 2.
@@dr3w199 okie
Damn... It's a great method. Neat work. 💯
no 2=1+1/2+1/4+1/8+1/16...
you have got many misconceptions blackpenredpen!!!
When he writes 1, he's obviously just abbreviating 1/2+1/4+1/8+1/16+...
@@iabervon and when he writes 1/2 he's abbreviating for 1/4+1/8+1/16+...
@@InDstructR And when he writes 1/4 he's abbreviating 1/8 + 1/16 + 1/32+...
@ki kus won't stop me,
And when he writes 1/8 he's abbreviating 1/16+1/32+1/64+1/128+...
Whoa that converged quickly
I remember solving this exact question in my JEE ( Mains ) exam.
@Sanat R mains usually has easy questions
@Sanat R - Study Vlogs Sure, yeah, "easy question" 😬
@Sanat R - Study Vlogs Woah, really? What kinda questions do they ask? Could you send me a link?
Dafuq did i jusf watch.i lost it when the 2=1+1
You are not nerdy enough
5:48 “Back in my day kids would use *ALGEBRA* but now their brains are rotting from these darn *CALCULATORS* ”
1 + 1 = 3
And sinx/n=six=6.
You stupid, 1 + 1 ≠ 3
3 = 2 + 1
π = 2 + 1
π - 1 = 2
@@rio_agustian_ so π = 3 lol nice discovery
@@CookieJar2025 e = 3 = π
@@rio_agustian_ noob
@@Kevin-14 lool nooob
Not only 2 = 1+1, but also 0 = 1-1.
From the second row:
(x^2+3x+1-1)(x^2+3x+1+1)+1 and per the third binomial equation
= (x^2+3x+1)^2 -1^2 +1
= (x^2+3x+1)^2
I didn't understand. Where did you use binomial expansion
0 = 1-1
1 = 1*1
2= 1+1
I didn't see any binomial here... But what i see is that you used the form (a+1)(a-1) + 1 = a^2 - 1 +1= a^2
@@yogeshpathak73 I think Tibor Grün is from Germany. In German school curriculum the formula (a+b)*(a-b) = a^2 - b^2 is known as 3. binomial formula. b=1 is a special case.
Oh ok... Didn't know that. Thanks.
This is beautiful. I've been looking at it for five hours now
Continueing from: sqrt( (x^2 + 3x) (x^2 + 3x + 2) + 1 )
let y = x^2 + 3x
sqrt( y * (y + 1) + 1 )
= sqrt( y^2 + y + 1 )
= sqrt( (y+1)^2 )
= y + 1
= x^2 + 3x + 1
= (x + 1) (x + 2) - 1
= 501 * 502 - 1
= 251501
much easier to multiply :p
that is what i wanted ti type nice 👍
Take x ^2 + 3x = a
Then in step 2
a(a+2) + 1
a^2 + 2a + 1
= (a+1)^2
Yes much easier, and you see it imediatly too.
was about to say this, i think it's a lot more intuitive
Isn't it beautiful how one problem can be solved in diffrent ways, even if the idea and the method are nearly the same. That's why we love maths.
@@milanmitreski7657 Yes
You extra smart boy the time required here will be same
0:10 is this a pewdiepie reference? 😂😂
He liked it!!!
@@albel2094 yupp
Yupp
I'm only in 8th grade Algebra 1 but I was using variables to find how some of your factorizations works.
You went from (x^2+3x)(x^2+3x+2)+1 to (x^2+3x)(x^2+3x+1)+(x^2+3x+1).
What I did was set (x^2+3x) to a variable (a).
(a)(a+2)+1
a^2+2a+1
(a+1)(a+1)
Now substitute back in.
(x^2+3x+1)(x^2+3x+1)
When in doubt use variables..
Yes, that's using even more algebra than BPRP did.
Well yes because using the variables is actually the logic behind the solution, it's just that it was invisible throughout the process :D
Where r u from?
Agreed! Variables always help to proceed the solution.
I put x=500 but multiplied everything. In the end i got to sqrt((x+y)^2) with x=500 and y=501^2
3:20 I solved it differently.
Let y= x^2+3x. Then substitute y into the expression making y(y+2)+1, distribute so y^2+2y+1 and that is a perfect square of (y+1)^2.
Here, the square root and exponent cancel each other leaving y+1, sub back in x and then easily find the answer :)
this is also what I did and I think that this is a bit better because you don't have to split 2 into 1 + 1 and do the rest
That's what I did as well
0:09 That was PowerFul
You can also put a +1-1 inside the x^2+3x bracket and it'll be in the form of (a+b)(a-b).
that's what i thought he was gonna do as well but what he did was cool as well.
why?
Yeah, (x-1)(x+1)+1=x^2-1^2+1 seems easier to find than multiplying out exactly the right portion of the big expression.
I chose to make x = 502, which ends up yielding a nice difference of squares and a two term quadratic, which is much easier to distribute. The quartic you get has a palindromic pattern reminiscent of pure binomial coefficients, making it tempting to say the golden ratio is a root. It is, in fact, a root, so synthetically divide the quartic by the golden ratio identifying polynomial, x² - x - 1. You end up with the golden ratio identifying polynomial again, meaning that the original quartic in that square root is (x² - x - 1)², so cancel the power and the root. Plug 502 back in for x, some quick multiplying and subtracting by hand and you've got 251501.
Sorry...Time over! give me your exam!
0:01 that's my life philosophy now
i remember my math teacher asking me to prove that n(n+1)(n+2)(n+3) + 1 is always a perfect square given that n is an integer
I expressed it as sqrt((501.5-1.5)(501.5-0.5)(501.5+0.5)(501.5+1.5)+1)
You get two a^2-b^2 expressions that you can multiply out, add the 1, and then factor into a squared quadratic expression
Very neat and, as someone mentioned elsewhere, it generalizes to “1 plus the product of any four consecutive integers is a perfect square”
Seemingly elementary problems can have wonderfully elegant solutions! All we need is to substitute a number with x, and the magic begins.
now do it with CALCULUS
Why do you think this is funny?
@@davidappell3105 because suffering is funny
@@davidappell3105 Its FUNI
3:20 just put y = x^2 + 3x, then you have y(y+2) + 1 = y^2 + 2y + 1 = (y+1)^2. So the answer is y + 1, or x^2 + 3x + 1.
I also did it in this way...but that way was also fine...its all about which method comes in your head first
Dang that’s genius
Your explanation is awesome . I like your teaching very much. Thanks
I am so impressed with myself, I actually used the same method you did before watching the video =D
Jan Wrobel nice!!!!
Every body knows 1+1=2 but i know 1+1 =/= 3
♫♪Ludwig van Beethoven♪♫ Hahahhaha
I’ve got you all beat with 1+1 > 0
@@jgsh8062 nah mine's better 1+1≠1+1
The world needs more teachers like you. I'm more impressed by your teaching skills than any math. Much respect.
i watched this video this video right before my math competition and the same type of question came up on the task sheet. Thank you very much!
for those wondering the question was
202120212019(202120212021)(202120212023) all over 100010001 x (202120212021 squared +4)
a very good perspective and a very good solution. thank you!!!
You can also this as x^2+3x=t and expression would become t(t+2)+1 =(t+1)^2 this que came in practice test for jee last week And guess what i solved that 😎😎😎👍👍
I love your videos about not using calculators (Like the Wolfram-Alpha video)! They're the best! Keep up the good work! It's nice going back to algebra sometimes...
Why It?
Yea me too. I try to mix things up a bit.
Everybody knows e^{iτ}=1
.
.
.
.
But I know 1=e^{iτ}
Nice!!!
@@user-bd9mu3ee1i That's e^{iπ}. τ=2π
My mans using tau! Up top!
@@user-bd9mu3ee1i he's using tau not pi
Doing a PhD in Literary Studies, but stuff like this is why I absolutely love maths ♥
Nice factoring method but it might have taken me a while to spot. Multiplying out and factoring isn't so bad
(x - 1)x(x + 1)(x + 2) + 1 = (x^2 - 1)(x^2 + 2x) + 1 = x^4 + 2x^3 - x^2 - 2x + 1 = (x^2 + bx +- 1)^2
= x^4 + 2bx^3 + (b^2 +- 2)x^2 +- 2bx + 1
We see this works if b = 1 and c = -1 so the answer is 501^2 + 501 - 1 = 500^2 + 2*500 + 1 + 500 = 251501
I did assume there was a nice solution, but expanding under the root to get x^4 + 6x^3 + 11x^2 + 6x + 1 was pretty easy, and then matching coefficients in (x^2 + ax + 1)^2 was straightforward too.
But yeah, the main thing is to replace 500 by x. I don't think I could intuitively see which two of the brackets would make it easier, and I'm not sure that's a better method than expanding the whole thing to only 4 terms (plus the one on the outside).
Just wanted to say after some work, some variable assigning and a lucky coincidence later, I found the answer!
My steps:
Let a=500
Expand a(a+1)(a+2)(a+3) to a^4+6a^3+11a^2+6a+1
Complete the square (or the fourth in this case):
a^4+4a^3+6a^2+4a+1+2a^3+5a^2+2a
=(a+1)^4+2a(a^2+2a+1+.5a)
=(a+1)^4+2a((a+1)^2+.5a)
=(a+1)^4+2a(a+1)^2+a^2
Observe this follows the perfect square structure.
Therefore:
(a+1)^4+2a(a+1)^2+a^2
=[(a+1)^2+a]^2
Square rooting gives:
(a+1)^2+a
a^2+3a+1
By substitution:
a^2+3a+1
=250000+1500+1
=251501
Blew my mind! Earned yourself a new subscriber! Keep up the good work!👍
I love the explanation, though I did it a bit differently. When I got to the second line, I substituted (x²+3x) as y and found that that worked much simpler than distributing 2 as 1+1.
Now this video makes me like algebra
czcams.com/video/XX2DI9E1zV8/video.html
3:19 The "WOW". LOL
When he drops the "Check this out", you know crazy stuff will happen on the board
There are things to learn from each of your videos 😁❤️
Well yes, that's the point.
Really good solution! GOOD Teacher👍
What a incredible content. Im a student of math (i'll be a teacher in the future) from Brazil. Thank you so much for sharing knowledge!
I have never been so hyped at 2 = 1+1 before.
What Everybody knows : 1+1=2
What BPRP knows : 2=1+1
.
.
.
.
.
What I know : 1+1=2 and 2=1+1
😇😇😇😇😇😇😇😇😇😇😇
BPRP know 2 = 1+1
I know 2 = 2
what happened to the comment button its gray
I know 2 = two
I love this. You did a great job of laying out a good challenge.
Why are your videos so entertaining? I'm so glad I came across this channel.
Everybody know e^2.pi.i = 1
.
.
But I know 1 = e^2.pi.i
Wrong it's
- (e ^ pi × i)
@@mundane3809 Nice try but thats -1
ignoring your name
@@nikolas9105 no
e ^ ( pi × i ) = -1
So if you make -1 negative, it become positive.
@@mundane3809 you are correct but the original comment was also correct. e^2πi = 1.
@@RunstarHomer oof yea it's actually correct. sorry for the mistake!
"And now, here's the deal"… You know that when he pronounces that phrase things are 'bout to get complicated.
An alternative way to solve it is by letting x = 501.5 and change the expression into
sqrt((x - 3/2)(x - 1/2)(x + 1/2)(x + 3/2) + 1)
The inspiration for this is difference of squares, simplifying gives.
= sqrt( (x^2 - 9/4)(x^2 - 1/4) + 1)
= sqrt( x^4 - 10/4(x^2) + 9/16 + 1)
= sqrt(x^4 - 5/2(x^2) + 25/16)
= sqrt((x^2 - 5/4)^2) , *factors nicely, perfect square*
= x^2 - 5/4
= (501.5)^2 - 1.25
= 251501
Except the 501.5^2 part is not so pleasant by hand. Not the best alternative.
if only most teachers were like this guy, it actually makes me wanna learn math again and I'm 29 years old! not gonna lie that did look fun for some reason.
I know this kind of the prob, i use (n+1)(n+2)-1
I know this is not related to this video but I wanted to post this on a new video so you might see it :) your trick for integrals of thinking "wouldn't it be nice if..." has helped me so so much, so thank you :) love your videos!
Awww thank you!!!!!
The factorization was more if you defined a variable "a" that was equal to x^2+3x
Because since you multiply and you have left
(X^2+3X)(X^2+3X+2)+1
With the variable "a" you had left
(a)(a+2)+1
And that is equal to (a^2+2a+1)
And that is factorizable as (a+1)^2
Greetings from Mexico
I need to start a tradition of watching this video every year. It's just that good
You could also do like:
Consider 501 as “x” and 502 as “y”
You can rewrite the sentence like:
(x-1).(x+1).(y-1).(y+1) +1
That’s equal to:
(x^2 - 1^2).(y^2 - 1^2) +1
Or
(501^2 - 1).(502^2 - 1) +1
And there’s your answer xD!!
Shout-out to my colorblind fam who can never tell when he switches pens
I did it by recognizing that (x^2+3x)(x^2+3x+2) is easily simplified with a substitution of y=(x^2+3x+1). It simplifies to (y-1)(y+1) = y^2 - 1. Since we have a "+1" hanging out after the 4-term product, that gets rid of the "-1" in our simplified expression, yielding just y^2 under the radical sign. square root of y^2 = y. That means the solution is our substitution: y=(x^2+3x+1). Plugging in 500 for x gives us 251501.
you can also use substitution x^2+3x = t and you get t(t+2)+1=t^2+2t+1=(t+1)^2 and replace t: (x^2+3x+1)^2
Wow your method and my method are similar .....I had take the whole expression as y and then square it and then assume x to be 500 and multiplied and had taken x^2+3x to be z and at end I got y=z+1 that is y = x^2 + 3x + 1 and it's done
Hmmm ive got an easier way when you use 1 + 1 instead of 2
Here is how I do according to you:
(x^2+3x+2)(x^2+3x+1)
=(x^2+3x)^2 +2 then
√((x^2+3x)^2 +2) = x^2+3x+1
You can't solve squareroots of sums like that, eventhought your result is correct.
I find the solution in a different way:
k(k+1)(k+2)(k+3)=x^2-1
=(x-1)(x-2),
and the difference between these number is 2;
so multiplying
k(k+3)=k^2+3k
(k+1)(k+2)=k^2+3k+2
I have the numbers with desidered difference.
So x=k(k+3)+1, having the result with substitution 500->k.
3:43
If we assume x^2+3x to be t we get t + 1 whole squared which is a lot easier
Dude, I always had a good grasp on algebra as a kid and in highschool I always aced most algebra, but somehow my teachers (and I) missed this property of algebraic equations. So freaking cool. It has been nigh on 15 years since high school, but I am still learning new and cool algebra. Thanks so much blackpenredpen!
I'm doing this on the toilet, so I only hope I'm starting correctly, with (500)(502)=(501²-1) and (501)(503)=(502²-1)
But then again, we could cheat and go with (501)(502)=(501½²-¼) and (500)(503)=(501½²-9/4)?
Just say x^2+3x = y
y(y+2) + 1
= y^2+2y +1
=( y+1)^2
Comes out of root as
(Y+1)
= x^2+3x + 1
And put the values
The entry was epic and the whole video is interesting.
please give an example differentiation of complex functions
Him: Starts the video with 1+1=2
Me: *ok we are getting somewhere now
A generalisation of the algebraic expression - (X)(X+1)(X+2)(X+2)+1= (Y)^2
@4:15 wow I did not see it coming! Really good thank you!
I understood everything until the bit at 4:35 - 5:04
What do you mean by "factoring out"?
When a number repeats itself in an addition you can factor it out, basically do the inverse of distributive property. So we have x²+3x+1 repeating in both therms. You can factor it out and you will be left with x²+3x+1 ( x²+3x +1), equivalent to x²+3x+1( x²+3x) + x²+3x+1 (1)
@@AE-rg5rc but thats just squaring it and you don't have two of the sake expression..you don't jave twobx squared plus 3x plus 2 you only,have one
@@leif1075 please Learn some math, this is for grade 6 atleast in asian countries.
@@ViratKohli-jj3wj I know some math, thanks very much..I had a valid question
this guy is a genius!
My way was : set term = y. Square both sides to get x(x+1)(x+2)(x+3) + 1 = y^2. Do minus - 1 -> x(x+1)(x+2)(x+3) = y^2 - 1 which is (y-1)*(y+1): aim: Get left side to the same form-> multiply x with (x+3) and (x+1) with (x+2) to get (x^2 + 3x) * (x^2 + 3x + 2) which is equal to (x^2+3x+1 - 1) * (x^2+3x+2-1 + 1) having the third binome and then calculating x^2 + 3x + 1 = y by putting in 500. Pretty much the same but another way!
Another nice solution is to assume x=501.5
And rewrite the equation which would give
x⁴-(5/2) x²+(9/16) +1 which is basically (x²-5/4) ²
The square and square root will cancel and give x²-5/4
Taking lcm would give us
((2x)² - 5) /4
(2x)²=1003² which can be computed very easily as 1003=1000+3
And then we just have to subtract 5 and divide by 4
There is one more solution
1.Equate expression to x
2. 500*501*502*503 = x^2 - 1
3. (500*503)(501*502) = (x-1)(x+1)
These multiplication can be done easily
4. Equating x
We will get the required result
I love his smile when he showed how easy it is FOR A GENIUS LIKE HIM.
I ended up expanding out the (x^2+3x)*(x^2+3x+2)+1 directly into (x^2+3x)^2+2*(x^2+3x)+1,
which is a perfect square as (x^2+3x+1)^2. If you make the variable substitution
y=x^2+3x, it becomes even more obvious as y^2+2y+1=(y+1)^2.
Ok, eu não sei se alguém já postou uma resposta similar (são 1,2 mil comentários), mas, vamos lá:
Eu tentei fazer o cálculo de cabeça (não, não sou um gênio, mas o tempo do vídeo sugeriu que havia algum macete) então, depois de algumas tentativas, percebi um padrão:
1. Tome a raiz quadrada do produto de números (eles nem precisam ser sucessivos!) e observe o seguinte detalhe:
1.1. Para números sucessivos, a raiz quadrada exata deles será o produto deles adicionado à 1^n (ex. Sqrt[(50*51*52*53)+(1^4)]
1. 2. Multiplique os termos centrais e subtraia, do resultado, o número adicionado elevado à potência entre eles: (51*52) - (1^2) = 2652 - 1 = 2651.
OBS. 1: Essa regra vale para qualquer sequência de números e tamanho deles, com um detalhe: para você encontrar a raiz exata terá que "fechar" cada sequência, mas isso é bem simples, quando você percebe o padrão; por exemplo, para encontrarmos a raiz quadrada exata do produto de 20*23*26*29, você já deve ter percebido que a minha sequência é uma P.A de razão 3, logo, o número a ser somado é 3^4 (a razão aparece para cada termo); assim, sqrt[(20*23*26*29)+(3^4)] = 23*26 - (3^2) = 589.
Obs. 2.: Nas sequências com o número 5, eu percebi que aconteceu uma mágica: a última parte do resultado correspondia ao 2o número dos termos, e a primeira parte correspondia à metade do primeiro termo+1; logo sqrt[(500*xyz*502*503)+n] = (500/2)+n (1a parte do número) xyz (segunda parte do número) = (500/2)+1 501 = 251 501.
Essa "mágica acontece em toda repetição com a sequência de 5 (50, 500, 5000,....) Vá em frente, experimente!
Enfim, não sei se mais alguém chegou à essa conclusão, eu só li as 100 primeiras mensagens e não haviam citado esse modelo (pode ter parecido longo mas, no papel, vc verá que é bem rápido, inclusive pode ser feito mentalmente).
Um salve do Brasil!
Interessante esta generalização do resultado para sucessões aritméticas em geral.
Basicamente, está a dizer que:
x(x+1)(x+2)(x+3)+1=(x²+3x+1)²
generaliza para:
x(x+a)(x+2a)(x+3a)+a⁴=(x²+3ax+a²)²
Repare que nesta identidade o membro direito é simétrico em x e a.
Trocando x e a na identidade, igualando os membros esquerdos, e fazendo um pequeno rearranjo, obtemos uma identidade nova:
a(a+b)(a+2b)(a+3b)-b(b+a)(b+2a)(b+3a)=a⁴-b⁴
Ou seja, a diferença entre o produto dos primeiros quatro termos de duas sucessões aritméticas, onde o primeiro termo de uma é a diferença comum da outra, é igual à diferença entre as quartas potências dos termos iniciais.
Fiz um comentário separado (em inglês) sobre este resultado.
@@MichaelRothwell1, obrigado. Realmente, essa igualdade até é abordada no ensino fundamental, e peço desculpas se não a associei diretamente :-). Em minha defesa: a) Terminei o ensino médio há muitos anos; b) só as fórmulas consideradas "corriqueiras" eram demonstradas, as demais, apenas memorizadas; c) há muitos anos que deixei as ciências exatas, seguindo o caminho das ciências sociais aplicadas. Mas anotarei sua demonstração e estudarei em momento oportuno. Grato!
@@VyctorFranca Não tem nada que pedir desculpa. Para mim, é toda matéria nova. Nunca vi estas fórmulas. Eu é que lhe agradeço para chamar a minha atenção a estas ideias.
I think you forgot about the absolute value; Square root of a 2nd power produces absolute value result because both positive and negative values are true.
Dunkoro it is obvious that the number inside the square root is positive，so ignore the absolute value symbol
me : multiplies it out, then square root it
#bprp #YIAY
Or I think on the second step, you could do a plus 1 - minus 1 trick. This is what I’ve learned from others. So take the left out integer, which is one, and regard it as c, since you’re doing an expression containing a, b, c. And within similarity, if there’s an expression including only a and b, it should be in standard, so + c. Then the other would be opposite.
So c = 1, (x^2 + 3x + c)(x^2 + 3x - c) = (x^2 + 3x + 1)(x^2 + 3x + 2 - 1) = (x^2 + 3x + 1)(x^2 + 3x + 1) = (x^2 + 3x + 1)^2.
Just a trick heard from others.
the easier way would be to solve it symmetrically:
lets say x equals 501,5 in this case.. then the product would be (x-1,5)(x+1,5)(x-0,5)(x+0,5)+1=
(x^2-2,25)(x^2-0,25)+1=
x^4-2,5*x^2+0,5625+1
which is obviously
(x^2-1,25)^2
Everyone know 1+1=2
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But only he knows 2=1+1
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But only I know that 2 also = 2*1
Why I shouldn't belive my calculator?
It gave me the answer as 251501
Edit: I never had that much likes in my life
but... was it fun?
@@aifesolenopsisgomez605 Fun is not something one considers when wrestling with numbers, but this... Does put a smile on my face.
Because in exams, you get your marks based on working and not on the answer
I think he meant that algebra is more useful than a calculator. The title doesn't read "you shouldn't believe your calculator", it says you shouldn't believe IN your calculator, and rather resort to algebra instead.
LOL, Wasn't expecting that much comments, sure algebra is better than directly using calculator and is more enjoyable to do so , but it took me time to realize what he meant.
Besides, I am a 7th grader, so calculator is not allowed in school for us. We can use from 9th grade.
You could set y = x^2 + 3x and then it would be y*(y+2) + 1 which equals (y+1)^2 and then the sqrt and the square cancel. I think its easier than coming up with a way to factor the original expression
It's only easier if you see it.
I think that the BPRP solution is an easier spot, but I agree yours is better if you are lucky enough to spot it
Solved it!
For a clean solution to exist I assumed that
x·(x + 1)·(x + 2)·(x + 3) + 1 is a perfect square for any integer x.
Playing around with x = 1 & x = 2 it is quickly apparent that
x·(x + 3) + 1 is a contender for the solution.
It is easy to prove that this is the solution by expanding out
[x·(x + 3) + 1]² and showing that it is equivalent to x·(x + 1)·(x + 2)·(x + 3) + 1
I doubt I would have spotted the algebraic manipulation that BPRP used without knowing the solution first. I also learn something new i.e. the product of four consecutive integers plus 1 is always a perfect square.
Thank you for the video - I enjoyed this one.