BELIEVE IN ALGEBRA, NOT CALCULATOR

[@aasyjepale5210]

i did some mental math, but hit a wall at trying to find the square root of 63,252,753,001

[@iamgroot3615]

that’s some impressive mental math assuming you’re telling the truth . Is there a trick or something

[@AngryAxew]

@@iamgroot3615 theres no trick hes probably lying

[@AngryAxew]

r/iamverysmart

[@marvinfung2050]

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

[@narayanankannan6787]

I mean it's OBVIOUSLY 251501.

[@yuliaokhremenko6613]

I just started learning English, but the explanations are clear and interesting even at my levels of English. Thanks a lot 😁👍

[@blackpenredpen]

Юлия Охременко I am Glad to hear!

[@enhace15anos.83]

x2

[@corona8073]

U r indian Chinese korean or ....???

[@donovanholm]

@@harelavv8806 the name may seem obviously Russian to some but not all

[@siddharthsoni2101]

@@blackpenredpen hii

[@Armbrust666]

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.

[@ClimateAdam]

Awesome! Good spot!

[@fanyfan7466]

Gábor Tóth holy shit you’re right! That’s crazy man

[@blackpenredpen]

Yup!!

[@kingbeauregard]

The most pathological case I can think of is -1 thru 2, and yes indeed I get 1, which is a perfect square.

[@pcklop]

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.

[@BryanLu0]

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

[@blackpenredpen]

Bryan Lu omg that cat!!!!

[@AmitBentabou]

Or even u=x^2+3x, then u^2+2u+1

[@matias12381]

digno de nyan cat, jajajajaj

[@mattat3847]

My life is a lie. I thought u subbing was only for integrals

[@RunstarHomer]

@@mattat3847 nah man, sub whenever it makes the problem simpler

[@sethdon1100]

Olympic math taught me that insanely hard problems often had elegant solutions, this is no exception.

[@blackpenredpen]

: ))))

[@hafizh8461]

@@leif1075???

[@hemandy94]

@@leif1075 people like these are called problem solvers...

[@drudi1]

@@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

[@jayasri6764]

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.

[@threadeater345]

1990: we'll have flying cars by 2019 2019: 2=1+1, wow I'm a genius

[@blackpenredpen]

LOL

[@ghotifish1838]

2+2 is 4, minus one that's three quick maths

[@Kyanzes]

Flying cars... you can't even have a sharpie that could change color. Say, red and black.

[@santinodemaria2818]

@@ghotifish1838 topical meme reference

[@unutentediyoutube3282]

Well it can also be 2=500-498

[@Hypoli]

My last words whispered in a final breath : "Don't forget the +1"

[@ሰላምሰይጣን]

😂

[@68plus1.]

LMFAOOO

[@naiknaik8812]

He never resumed the video

[@isaacqian-e3d]

He added the one wdym 🧐

[@leagueplays2100]

i put it in my calculator and got 251501, that was easy

[@bowtangey6830]

Boo!

[@Netherexio]

@The Balton American calculators are beasts

[@paradox9265]

@@Netherexio Agreed but Americans aren’t

[@Netherexio]

@@paradox9265 What do you mean?

[@СвятославИерусалимов]

It was easy, but not so beautiful like this)

[@blackpenredpen]

Did you know that 2 = 1 + 1?? I bet not! jk : )

[@williamadams137]

blackpenredpen No i don’t, i need a calculator to this

[@snejpu2508]

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. : )

[@theolbiterator5408]

No but I knew 2= 0.9+1.1.

[@chaitanyagadekar5025]

I Known 2+2 = 5

[@clubstepdj]

What i know is 5/2 = 2 with int data type

[@mr.n1933]

Dafuq did i jusf watch.i lost it when the 2=1+1

[@executorarktanis2323]

You are not nerdy enough

[@maheshagrawal7779]

no 2=1+1/2+1/4+1/8+1/16... you have got many misconceptions blackpenredpen!!!

[@iabervon]

When he writes 1, he's obviously just abbreviating 1/2+1/4+1/8+1/16+...

[@InDstructR]

@@iabervon and when he writes 1/2 he's abbreviating for 1/4+1/8+1/16+...

[@agces2001]

@@InDstructR And when he writes 1/4 he's abbreviating 1/8 + 1/16 + 1/32+...

[@InDstructR]

@ki kus won't stop me, And when he writes 1/8 he's abbreviating 1/16+1/32+1/64+1/128+...

[@shounakghosh8595]

Whoa that converged quickly

[@tanmay8017]

I remember solving this exact question in my JEE ( Mains ) exam.

[@classicmelodyvetrivel710]

@Sanat R mains usually has easy questions

[@Avighna]

@Sanat R - Study Vlogs Sure, yeah, "easy question" 😬

[@Avighna]

@Sanat R - Study Vlogs Woah, really? What kinda questions do they ask? Could you send me a link?

[@steve2817]

1 + 1 = 3 And sinx/n=six=6.

[@rio_agustian_]

You stupid, 1 + 1 ≠ 3 3 = 2 + 1 π = 2 + 1 π - 1 = 2

[@CookieJar2025]

@@rio_agustian_ so π = 3 lol nice discovery

[@Kevin-14]

@@CookieJar2025 e = 3 = π

[@SameerKhan-nd5qb]

@@rio_agustian_ noob

[@SameerKhan-nd5qb]

@@Kevin-14 lool nooob

[@ganaraminukshuk0]

"If you're using a calculator, why are you watching this video?" Sanity check.

[@blackpenredpen]

hahahahaa

[@andrel8243]

I am a calculator, not a person

[@mysticdragonex815]

*laughs in Shakuntala Devi

[@rd8396]

Take x ^2 + 3x = a Then in step 2 a(a+2) + 1 a^2 + 2a + 1 = (a+1)^2

[@mat1305h]

Yes much easier, and you see it imediatly too.

[@Polarspy]

was about to say this, i think it's a lot more intuitive

[@milanmitreski7657]

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.

[@sanjaisrao484]

@@milanmitreski7657 Yes

[@akshetpatial5466]

You extra smart boy the time required here will be same

[@tiborgrun6963]

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

[@etemkaandelibas3649]

I didn't understand. Where did you use binomial expansion

[@jinja3113]

0 = 1-1 1 = 1*1 2= 1+1

[@yogeshpathak73]

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

[@Bayerwaldler]

@@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.

[@yogeshpathak73]

Oh ok... Didn't know that. Thanks.

[@DatSwif]

This is beautiful. I've been looking at it for five hours now

[@dr3w199]

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

[@juanbomfim22]

OMG ive almost done it completely. i just stopped at (n+1)(n-1) lol WD! i mean 'not that almost' lmao

[@joshuamason2227]

How do I play around with it

[@dr3w199]

@@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.

[@joshuamason2227]

@@dr3w199 okie

[@sunnykarwani3556]

Damn... It's a great method. Neat work. 💯

[@chaitanyagadekar5025]

0:09 That was PowerFul

[@mhm6421]

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

[@razvy3827]

that is what i wanted ti type nice 👍

[@Inujasa88]

0:10 is this a pewdiepie reference? 😂😂

[@albel2094]

He liked it!!!

[@randomdude9135]

@@albel2094 yupp

[@randomdude9135]

Yupp

[@workout9594]

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 :)

[@matthewmanzanares6798]

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

[@cheesecircle3033]

That's what I did as well

[@TheS1lentX]

Jeez thats smart *proceeds to use the calculator to prove that 251501 is the right answer*

[@thatssokwekwe]

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”

[@detachedmars158]

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).

[@kilindogma9711]

that's what i thought he was gonna do as well but what he did was cool as well.

[@ssdd9911]

why?

[@iabervon]

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.

[@isaacdeutsch2538]

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.

[@asa-ze1kn]

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..

[@trueriver1950]

Yes, that's using even more algebra than BPRP did.

[@zocker2586]

Well yes because using the variables is actually the logic behind the solution, it's just that it was invisible throughout the process :D

[@baranibarani4970]

Where r u from?

[@sanjanabiswas9774]

Agreed! Variables always help to proceed the solution.

[@enricomassignani]

I put x=500 but multiplied everything. In the end i got to sqrt((x+y)^2) with x=500 and y=501^2

[@blackflash9935]

5:48 “Back in my day kids would use *ALGEBRA* but now their brains are rotting from these darn *CALCULATORS* ”

[@almightyhydra]

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.

[@cypherx7247]

I also did it in this way...but that way was also fine...its all about which method comes in your head first

[@lasergamer2869]

Dang that’s genius

[@star_ms]

Seemingly elementary problems can have wonderfully elegant solutions! All we need is to substitute a number with x, and the magic begins.

[@lucasxue2211]

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

[@snatchngrab8262]

The world needs more teachers like you. I'm more impressed by your teaching skills than any math. Much respect.

[@agabe_8989]

0:01 that's my life philosophy now

[@bucinoulje7505]

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!

[@bucinoulje7505]

for those wondering the question was 202120212019(202120212021)(202120212023) all over 100010001 x (202120212021 squared +4)

[@jekoddragon6227]

now do it with CALCULUS

[@davidappell3105]

Why do you think this is funny?

[@gabrielpinhal8325]

@@davidappell3105 because suffering is funny

[@sanchit6107]

@@davidappell3105 Its FUNI

[@pfever]

Sorry...Time over! give me your exam!

[@martinepstein9826]

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

[@SanjayKumar-wh6cd]

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

[@imadkhan1825]

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 😎😎😎👍👍

[@SawkTheFighter13]

Why are your videos so entertaining? I'm so glad I came across this channel.

[@ozonejgs2887]

I am so impressed with myself, I actually used the same method you did before watching the video =D

[@blackpenredpen]

Jan Wrobel nice!!!!

[@yannisdekonoha]

When he drops the "Check this out", you know crazy stuff will happen on the board

[@LudwigvanBeethoven2]

Every body knows 1+1=2 but i know 1+1 =/= 3

[@blackpenredpen]

♫♪Ludwig van Beethoven♪♫ Hahahhaha

[@jgsh8062]

I’ve got you all beat with 1+1 > 0

[@JDguy11222]

@@jgsh8062 nah mine's better 1+1≠1+1

[@laudine878]

Doing a PhD in Literary Studies, but stuff like this is why I absolutely love maths ♥

[@MrShad]

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!

[@-Mohammed_S]

Your explanation is awesome . I like your teaching very much. Thanks

[@DavidS-qn3jm]

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).

[@tanmaygupta7778]

3:43 If we assume x^2+3x to be t we get t + 1 whole squared which is a lot easier

[@ammardaffa]

I know this kind of the prob, i use (n+1)(n+2)-1

[@ultrio325]

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

[@eannacoleman957]

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.

[@tempomi760]

Blew my mind! Earned yourself a new subscriber! Keep up the good work!👍

[@jdsingh3607]

Now this video makes me like algebra

[@FermionClasses]

kzbin.info/www/bejne/jomVdXxvepatjJo

[@cococh3607]

I got the point. Convert the number into variables. For example 500=x or 2=1+1 = a (convert what you want) Thank you for the useful tips... I realized that the algebra is so amazing at the complicated situation.

[@msmmath87]

Really good solution! GOOD Teacher👍

[@ugursoydan8187]

a very good perspective and a very good solution. thank you!!!

[@LiegeNorth]

this guy is a genius!

[@Hecti161]

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

[@hellopeter121]

What Everybody knows : 1+1=2 What BPRP knows : 2=1+1 . . . . . What I know : 1+1=2 and 2=1+1 😇😇😇😇😇😇😇😇😇😇😇

[@adrianau5231]

Actually why not: √(500)(501)(502)(503)+1 = x 500.501.502.503 +1 = x² 500.501.502.503 = x²-1 (500.503) (501.502) =(x-1)(x+1) 251500 × 251502 = (x-1) (x+1) x = ±251501 but we have to reject -251501 because it is negative in a square root

[@paawanjethva]

Everybody knows e^{iτ}=1 . . . . But I know 1=e^{iτ}

[@blackpenredpen]

Nice!!!

[@paawanjethva]

@@日本人じゃありません That's e^{iπ}. τ=2π

[@fgvcosmic6752]

My mans using tau! Up top!

[@peterg644]

@@日本人じゃありません he's using tau not pi

[@ridwansetiadi8393]

If somebody can't understand 3:59, you can try another method; you can also substitute the line 3 of 'Obs' into another variable, for example: y If: y = x²+3x Then the equation becomes: = (y+2)(y)+1 = y²+2y+1 Factorize that into this: = (y+1)² So, sqrt[(y+1)²] = y+1 Since y = x²+3x The equation becomes: = x²+3x+1 Since x = 500 The result is: = (500)²+3(500)+1 = 251501

[@DarkRedZane]

I have never been so hyped at 2 = 1+1 before.

[@gedlangosz1127]

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.

[@kinyutaka]

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)?

[@backyard282]

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

[@joaoaugusto9221]

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!!

[@ZiebartPatel]

A generalisation of the algebraic expression - (X)(X+1)(X+2)(X+2)+1= (Y)^2

[@whyit487]

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...

[@blackpenredpen]

Why It? Yea me too. I try to mix things up a bit.

[@deathzonesimang8043]

If you guys want to know the “secret”, it is if the sum of the digits in any number add up to a multiple of 3, it is divisible by 3. E.g 567: 5+6+7=18. 18 is a multiple of 3 thus 567 is divisible by 3.

[@deathzonesimang8043]

@Cerebro Spinal I was referring to the part at the end of the video

[@dev.imperatus6105]

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

[@XWurstbrotX]

You can't solve squareroots of sums like that, eventhought your result is correct.

[@jcb3393]

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.

[@-a5624]

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!

[@blackpenredpen]

Awww thank you!!!!!

[@mesballo2224]

Your way to solve this is pretty AWESOME!! First I multiplied all together, i got x^4 + 6x^3 + 11x^2 + 6x + 1 then i calculate this polynomial for x=1 x=2 x=3 ....all the time i got a square!! I was really surprise!! I didn't expect x(x+1)(x+2)(x+3)+1 to be a perfect square for all x at all!!! This is incredible!! Thanks for sharing your knowledge you are very inspiring to me

[@sadeekmuhammadryan4894]

There are things to learn from each of your videos 😁❤️

[@deadvirgin428]

Well yes, that's the point.

[@gaetanocantisani3762]

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.

[@aikenkazim5318]

please give an example differentiation of complex functions

[@ericli2723]

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

[@cfgauss71]

Except the 501.5^2 part is not so pleasant by hand. Not the best alternative.

[@Kino-Imsureq]

BPRP know 2 = 1+1 I know 2 = 2 what happened to the comment button its gray

[@jinja3113]

I know 2 = two

[@reza1219]

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.

[@mango417]

"And now, here's the deal"… You know that when he pronounces that phrase things are 'bout to get complicated.

[@denisikonomi2019]

If you guys want a tip to make the solution simpler Set 501 as x instead of 500 That way (x-1)(x)(x+1)(x+2)+1 is under the square root Which many can see the only tedious multiplying we’d have to do is (x^2 - 1)(x^2 + 2) You’ll then find the equation to be x^2 + x - 1 Which gives the same result Just helps save space and makes the problem less tedious

[@noverdy]

Everybody know e^2.pi.i = 1 . . But I know 1 = e^2.pi.i

[@mundane3809]

Wrong it's - (e ^ pi × i)

[@nikolas9105]

@@mundane3809 Nice try but thats -1 ignoring your name

[@mundane3809]

@@nikolas9105 no e ^ ( pi × i ) = -1 So if you make -1 negative, it become positive.

[@RunstarHomer]

@@mundane3809 you are correct but the original comment was also correct. e^2πi = 1.

[@mundane3809]

@@RunstarHomer oof yea it's actually correct. sorry for the mistake!

[@SpongeDude8]

More impressed with how someone came up with the question

[@moosemoosington1441]

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!

[@谢天陈]

what about using x. First allow the equation equal to x, then square both sides and then both sides minus 1. We get (x+1)(x-1)=(500)(501)(502)(503). Then I found out that (500)(503)is two less than (502)(501) and the answer would be 500*503+1=251501. The last step can be done be simple calculation, no calculator needed.

[@Mothuzad]

Shout-out to my colorblind fam who can never tell when he switches pens

[@MrJdcirbo]

I multiplied everything out and got stuck... This is a brilliant solution. One day I will achieve this type of mathematical intuition. Lead the way, blackpenredpen!!!

[@DriffPL]

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.

[@bttfish]

Dunkoro it is obvious that the number inside the square root is positive，so ignore the absolute value symbol

[@tomerwolberg37]

You could have made it simpler using the (a-b)(a+b)=a^2-b^2 formula. (x^2+3x)(x^2+3x+2)+1 = (x^2+3x+1-1)(x^2+3x+1+1)+1 = (x^2+3x+1)^2-1^2+1 = (x^2+3x+1)^2

[@ishmeetsingh1146]

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

[@vudomath]

At 3:45, you could also treat the first factor as (x^2+3x+1-1) so together with the second factor you have (x^2+3x+1-1)(x^2+3x+1+1) = difference of squares ((x^2+3x+1)^2 - 1. Plus the extra 1 on the outside you get the perfect square.

[@黎真-e3f]

The conclusion of this question is : [ First number + second number ^2 ]

[@martinteichert9209]

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

[@enzoTHEferrari]

I understood everything until the bit at 4:35 - 5:04 What do you mean by "factoring out"?

[@AE-rg5rc]

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)

[@leif1075]

@@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

[@ViratKohli-jj3wj]

@@leif1075 please Learn some math, this is for grade 6 atleast in asian countries.

[@leif1075]

@@ViratKohli-jj3wj I know some math, thanks very much..I had a valid question

[@user-yg97f5hfvh]

You may substitute x^2+3x = t and make t(t+2)+1 = t^2+2x+1 so square root of t^2+2x+1 is t+1=x^2+3x+1(same)

[@GodzillaFreak]

Just do the multiplication by hand.

[@avisinha9331]

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

[@alexsandroagustini714]

Him: Starts the video with 1+1=2 Me: *ok we are getting somewhere now

[@curtisw0234]

Why not substitute y=x^2 + 3x then you can expand that to get y^2 + 2y + 1 = (y+1)^2. Perfect for cancelling out the square root