believe in the math, not wolframalpha

[@vinayemani]

The greatest trick humanity ever learned from mathematics - If you don't know some value, call it x and proceed.

[@ull893]

Vinay Emani that's why we have an X even in sex !!! Lol.

[@me-legend8408]

XD

[@cemsentin]

You can be found possible roots of x^3+3x-14=0 without using ¨Rational Root Test¨. My solution of it is below: x^3+3x-14=0 x^3-8+3x-6=0 (x-2)*(x^2+2x+4)+3*(x-2)=0 (x-2)*(x^2+2x+4+3)=0 (x-2)*(x^2+2x+7)=0 Due to x>0, x=2. Other roots of x are, -1+Sqrt6*i and -1-Sqrt6*i, are complex.

[@warri7948]

cemsentin x^3+3x-14 isnt equal to x^3 - 3 + 3x - 6

[@goodplacetostart9099]

-3-6 are -9 not -14

[@davidbondy2250]

Forget the math, I want to learn that marker-switching trick.

[@lsbrother]

It's quite easy - he presses the 'pause' button each time

[@_-_-Sipita-_-_]

Why (7+√50)(7-√50) = -1???

[@keescanalfp5143]

@@_-_-Sipita-_-_, why .. 7×7 = 49, 7√50 - 7√50 = 0, √50 × (-√50) = -50. but why ..

[@quanpa]

@@keescanalfp5143 (a^2 -b^2) 7^2 - (√50)^2

[@keescanalfp5143]

@@quanpa, yess of course. did she zij/hij see, you think

[@georgeharamuniz1921]

At 13:36 you can see that wolfram alpha is assuming the principal root is being used. If you click the option to use the real-valued root instead you will get the answer of 2.

[@nathanisbored]

i thought the principal root was defined to be the one along the positive real axis, if it exists (and otherwise i assume it would default to the one in the first quadrant of the complex plane?)

[@PackSciences]

Thank you for explaining that, I didn't notice it George Haramuniz

[@toriknorth3324]

nathanisbored That's correct; the problem though is that the principle root and the real-valued root are only the same for positive numbers, but 7-sqrt(50) < 0. If we rewrote the problem as cbrt(7+sqrt(50))-cbrt(sqrt(50)-7) instead then Wolfram Alpha would give the answer as 2 using either option.

[@NuclearCraftMod]

nathanisbored: The principle root is the root whose magnitude is obtained by 'rooting' the absolute value of the number and whose complex argument is obtained by dividing the argument of the number by the root index. This means that, if the number is real and positive, then so is the principal root. However, if the number is real and negative, then the argument of the principal nth root will be π/n.

[@nathanisbored]

ok i understand now, the issue is that its taking the principal roots of each term separately, and THEN summing them together. For some reason I was thinking of the whole expression as a root itself

[@EdDavisTeaching]

If you look just under the input box after he hits equals there is a blue box that contains the text "Assuming the principal root | Use the real-valued root instead." Click on the option for the real-valued root and it gives you 2. I do not know why Wolfram Alpha defaults to the principle root. But it will calculate the real value correctly. The same is true by the way when you use Mathematica by the way. However, I am not sure how to get Mathematica to return the real-valued root.

[@gamerpedia1535]

Well here's my guess. When you get a cube root, there are three solutions. Say x^3 = n We can divide by n on both sides and bring it in to get (x/(cbrt(n))^3 = 1 So we'll just say x^3 = 1 When solving x^3 = 1, we can use Euler's identity x^3 = e^(i0) However we can add a 2πn to account for cyclicity x^3 = e^(i2πn) By exponentiating by 1/3, we get x = e^(iπn × 2/3) Which gives us 3 answers 1, -1/2 + i√(3)/2, -1/2 + i√(3)/2 So when we get the principal solution of x^3, to get the other results we need to multiply by the other two constants above. Now, when evaluating the cubic roots, I'm guessing that the principal root is not real, but rather includes an imaginary part. This would mean that by multiplying by one of the above constants would give you one of the alternate roots, that being the real root.

[@TheGregstar92]

When you get an integer as a root, that makes you reeeeaaaally happy.

[@zweiosterei]

The marker switch game is strong in this one.

[@nischay4719]

zweiosterei Was that a star wars reference? Lol

[@yogeshnagpal3671]

Yeah it was

[@M-F-H]

👍🏼 for detecting the WA "bug" (but don't forget that you discarded negative solutions of √ and complex solutions of ³√ which can be taken as multi-valued functions...). Also, you can do it for a ± √b in more generality and with less writing. You get in 2 lines x(x² - 3• ³√(a²-b)) = 2a, then with our a,b: x(x² + 3) = 14 = 2(4+3).

[@sameerplaynicals8790]

@@M-F-H i am a 12 yr old, is something wrong with me if i dont understand this?

[@M-F-H]

@@sameerplaynicals8790 No... I think my reply didn't concern this comment 😓. in the last part I say that its shorter to use a,b and plug in the numbers only in the end.

[@VandreiganMW]

On Wolfram Alpha, you're showing the principle root. If you click "Use the Real Root instead," you'll see your expected answer of 2. That being said, I agree that it's bad to lean on Wolfram Alpha and Mathematica too much. But they certainly are nice tools when used correctly!

[@andreaspurnomo1688]

photomath is better

[@nathannguyen7449]

andreas purnomo LOL! Thanks for the good laugh

[@andreaspurnomo1688]

@@nathannguyen7449 know right

[@andreaspurnomo1688]

😁

[@tonydai782]

@Jonathan Schwartz when you type in the expression and it is calculated, there will be a box under the search bar that asks if you want to use the real root instead of the principal root.

[@Robi2009]

I always enjoy when some complicated sq roots, cube roots etc. end up equaling some integer value :)

[@XoIoRouge]

Same. When he magically solved for 2, I was like, "Wait, it can't be that simple. It's just an integer?!"

[@niveditabhaskar8193]

7+5√2= 2√2+1+3√2+6 =(√2)³+1³+3(√2)[1+√2] =(√2+1)³ Similarly 7-5√2=(1-√2)³ Therefore, (7+5√2)^1/3+ (7-5√2)1/3 = (1+√2) + (1-√2) =2

[@RodelIturalde]

I really dislike nested roots. Like this one and others that appear from Cardanos formula for dolving 3rd degree polynomials. Such a pain in the assumptions to solve.

[@reubenmanzo2054]

Realistically (technically, not the right term and you'll see why), there are 3 solutions. The cubic factorises to (x-2)(x^2+2x+7)=0. What this means is x^2+2x+7=0 is a valid solution to the problem, opening i*sqrt(6)-1 and -i*sqrt(6)-1 as solutions.

[@yf-n7710]

@@reubenmanzo2054 Not really. The other two solutions were artifacts, introduced when both sides were cubed in the first step. You can tell that 2 is the correct one because it has no imaginary part. I suppose there might be 3 solutions if you interpret a cube-root as meaning any of three values which, when cubed, provide the value on the inside. But generally, the cube root of a number refers to the real solution only unless stated otherwise. Just like how the square root refers to the positive solution only unless stated otherwise.

[@AnalKumar02]

Whoever came up with this problem is a genius

[@Yash.the.seeker]

nooo, this is not very hard

[@davidzheng8926]

@@Yash.the.seeker I think Anal was talking about the creation of the problem and not the answer.

[@Yash.the.seeker]

@@davidzheng8926 hhhhhuuuuuu

[@ygritte4829]

@@Yash.the.seeker lmao u low iq

[@Yash.the.seeker]

@@ygritte4829 😑

[@rustamkalimullin]

Russians do like this: Cuberoot(7-5sqrt(2))=cuberoot(1-sqrt(2))^3=1-sqrt(2). Cuberoot(7+5sqrt(2))=1+sqrt(2). Finally, (1+sqrt(2)) + (1-sqrt(2))=2.

[@MathNerd1729]

That's how I did it too! (I am not even Russian!)

[@ДаняКурсинский]

how do you know that 7-5√2=(1-√2)^3???

[@ДаняКурсинский]

i mean how can you even come up with smth like this

[@Slangs]

@@ДаняКурсинский I am really interested to know that as well

[@pedrodaccache4026]

@@ДаняКурсинский 7-5√2 = 1+6 -2√2 -3√2 = 1 -3√2 +6 -2√2 (thats actually a^3 - 3a^2b + 3ab^2 - b^3). Then, its all equal to (1-√2)^3

[@Jiggerjaw]

Your marker switching skills are deft af.

[@blackpenredpen]

Jiggerjaw thanks!!!!

[@nossonweissman]

😂

[@jcgodinez90]

😂😂😂

[@ull893]

His marker switching skills must be due to the chopstick skills he have !!!

[@46pi26]

Mohan7 That's actually a pretty legitimate assumption lol

[@cheeros]

Impressive black pen red pen powers at 2:00

[@blackpenredpen]

zerep sesiom lolllll thanks!!!!

[@blackpenredpen]

zerep sesiom Ahhhhhhh I see what u did in ur YT name now. Loll

[@Cockaine880308]

Truly impressive...

[@smokescreen9119]

500 iQ 😂😂

[@gabrielfois9781]

This was awesome, i couldn't understand how he was doing that, I thought that the board reflects bad the ligths jajajajsjs

[@windowsforvista]

I feel like I disappointed him every time I don't pause to try 😅

[@blackpenredpen]

windowsforvista it's okay. U can just enjoy the show

[@yrcmurthy8323]

Me2

[@harshranjan8526]

I saw the video not to watch the solution , but to compare it with mine(I had solved it long before I heared about this great and auspisious channel), my method was only slightly different😀

[@yrcmurthy8323]

@@harshranjan8526 Oh really

[@anuragguptamr.i.i.t.2329]

Try this then

[@georgemissailidis1504]

2:02 the way you swap between the black pen and red pen is so smooth, Jesus.

[@crimsonkaiser3162]

This equation literally blew my mind.

[@stevenvanhulle7242]

I think you should reconsider the use of the word "literally" here. I hope...

[@garmrdmr]

at the top it says: Assuming the principal root | Use the real‐valued root instead click the blue bit.. it shows 2! so dont complain about wolfram alpha result untilk you READ THE RESULT

[@oscard4801]

That means ... Let me think ... Emm ... That WolframAlpha is fine :) It's just that he does not read well :v I think he did not see and it was a simple mistake

[@JohnDoe-iu5xi]

@@oscard4801 why would it matter. the best part of the video is the fact that he solved the problem without wolfram alpha. who cares if he clicked the blue link. he solved it beautifully.

[@TheBaggyT]

Please could you explain what it means by the "principal root"? According to the equation (x^3 + 3x - 14 = 0), I get complex roots of -1+sqrt(6)i and -1-sqrt(6)i. Thanks.

[@U014B]

No, it doesn't show 2!, it shows 2. 2! is equal to 2×1, which is... uh, nevermind.

[@Zulfar-bd9tc]

@@oscard4801 @Graham Richards Thing is though he's a professor. He's dealing with a lot of students in his class that just use Wolfram Alpha, and that's the purpose behind why he makes these kinds of videos. Because you're both right, Wolfram Alpha is fine, but a lot of the students in his class will most likely go to Wolfram Alpha, type in the equation, get the answer, not realizing wtf this is displaying and just write it down as an answer. I'm pretty sure he's aware of exactly what you're complaining about: a simple mistake of not clicking the Use the real-valued root. However, many of the students he teaches probably do exactly what this professor did and say ah so that's the answer, and instead of going wtf is this answer and not using it, they might round off the answer and write it down thinking there will be no issue at all writing down that as the answer rather than actually trying to solve the question, or, you know, clicking use the real-valued root instead.

[@88Nieznany88]

2 7+5sqrt2 is 1+3sqrt2+6+2sqrt2 which is (1+sqrt2)^3 which means a=1+sqrt2 We do same for b and b=1-sqrt2 a+b=2

[@blackpenredpen]

Bloomex yes

[@88Nieznany88]

blackpenredpen wolframalpha might be wrong on this, because i was taught that x^(1/3) does equal cube root of x only of x is positive.

[@theo1395]

Wolframalpha used the principal root which is not taught in my class. I googled it for a while. It seems not formally defined. It is similar to the concept "principal value" or "principal branch”. The principal root is the root which has the least principal argument defined in [0, 2pi). See the picture here if you are interested: en.wikipedia.org/wiki/Nth_root#/media/File:Visualisation_complex_number_roots.svg

[@Horinius]

WolframAlpha *arbitrarily* defines that principal value of a cubic root of a negative real number is a complex number, whereas, in all formal mathematical education and discussion, principal value for real number is still a real number, no matter if the starting value is positive or not. So, there is still a problem in WolframAlpha and that's at the level of computer/syntax.

[@theo1395]

+ Horinius: the principle cubic root of 8 is 2. However, the principle cubic root of -8 is 2e^(i pi/3)=1+cubic_root(3)i. the real cubic root of -8 is -2. *I'm curious how you make the word "arbitrarily" bold.

[@maxlarose75]

What really impresses me is how seamlessly you can switch between both pens. Amazing.

[@walter_kunz]

In WolframAlpha you have to use the Cbrt function which is the real-valued root: Cbrt[7+sqrt(50)]+Cbrt[7-sqrt(50)]. And the result is "2"

[@OneWeirdDude]

13:35 "Assuming the principal root | Use the real-valued root instead[.]"

[@CofeeAuLait]

We need a video on how to switch markers like that, looks simple yet amazing.

[@blackpenredpen]

It's on my channel trailer, which is here kzbin.info/www/bejne/nXKXfZaooNRooMk

[@edukid1984]

I'm mkre curious about what sort of marker he's using that doesn't seem to dry out without a cap on!

[@danielmacsai776]

Its much easier if you write it like this: (a+b)^3 = a^3 + b^3 + 3ab(a+b) And there you can substitute the x: = a^3 + b^3 + 3abx

[@isaacoshea]

in further maths we were practising expressing terms with a and b into terms with (a+b) and ab, so a^3 + b^3 = (a+b)^3 - 3ab(a+b), which rearranges into that. That probably made no sense, it's 4:00 am as of writing and I currently suffer from chronic fatigue, so yah.

[@snyggmikael]

@@lionel4685 for practis

[@bobbmsee]

click on "Use the real‐valued root instead"

[@palashdas823]

LT (r*T'(r))'=0 can i have this solution step by step from wolfram alpha ,sir

[@felonhighman5146]

I believe wolfram alpha just shows you one of the correct answer, one of the complex solutions, because the cubic root of something actually leads to three results, not just one. You just can't take it as a real number for granted.

[@Mernusify]

True. All you need to do to get the right answer is to multiply by one of the primitive roots of unity. For the case above you either multiply Wolfram Alpha's result by (-1+i*sqrt(3))/2 or (-1-i*sqrt(3))/2 and one of them will give you the right answer. Every number, no matter it be real or complex, has EXACTLY n n-th roots.

[@twosomestars9254]

Is complex answer really an answer? It's like the difference in real root and illusionary root. Here, the answer 2 is real, and that complex answer given by wolfa is illusionary..

[@shreyasdas5130]

@@twosomestars9254 learn maths pls

[@twosomestars9254]

@@shreyasdas5130 I was not saying a fact to begin with. this is comment section, your opinion matters, not facts.

[@twosomestars9254]

@@shreyasdas5130 my sentence has no correlation to learning match. It's like taking out my opinion whether it matters or not, even if it's wrong mathematically. Because I'm not having math knowledge to begin with.

[@emmeeemm]

I tried it before watching the video, and I got the same thing by essentially the same method. I approached the end with some slightly different phrasing, though. I did work it down to 14-3x=x^3. But I *just* moved the 3x to the other side, giving me 14=x^3+3x. I factored out the common x: 14=x(x^2+3). Knowing that 14 is composite (and seeing immediately that x=1 didn't work), I factored it into 2 and 7 (conveniently, its full prime factorization) and started testing. It turned out that 2 worked, as 2 * (2^2+3) = 2*(4+3) = 2*7=14.

[@RickDesper-v8z]

An unusual way to solve a cubic, but if it works for you...

[@franciscoabusleme9085]

Let the first term be "a" and the second one" b" and x=a+b. Notice that ab=-1 and a^3+b^3=14. Then a^2+b^2 =2+(a+b)^2 =2+x^2 And a^3+b^3=(a+b)(a^2-ab+b^2) =>14=x(3+x^2) And we then get the equation: x^3+3x-14=0

[@AllThisThen]

When I'm using a bunch of completely different irrational numbers based on numbers that have no common factors and wind up with a rational number answer I look for where I messed up.

[@TheRedPython]

If you type ∛(7+√50)+∛(7-√50) into WolframAlpha it gives the answer 2 If you type (7+√50)^(1/3)+(7-√50)^(1^3) it doesn't . ∛(7+√50)+∛(7-√50) uses the real root, (7+√50)^(1/3)+(7-√50)^(1^3) uses the principle root

[@squdardt.9719]

nobody: blackpenredpen: believe in the math

[@l4kkyl333]

But if you enter this: cbrt(7-sqrt(50)) + cbrt(7+sqrt(50)) The result is 2.

[@cygorx]

Fascinating

[@keescanalfp5143]

discrimination on powers and roots

[@iliyasone]

No, this is correct, because the cub root is not quite the same as the 1/3 degree (you cannot raise negative numbers to non-integer degree)

[@rashidisw]

@@iliyasone _"you cannot raise negative numbers to non-integer degree"_ , incorrect (-2)^3 = -8, therfore (-8)^(1/3) = -2

[@iliyasone]

@@rashidisw there are a paradox, if we allow it. On the one hand, (-8)^(1/3) = 3 root of ((-8)^1) = -2 On the other hand, as 1/3 = 2/6 (-8)^(1/3)= (-8)^(2/6) = 6 root of ((-8)^2) = 2 => -2 = 2 This is why it isn't allowed

[@guest_informant]

13:38 Isn't there some hyperlinked text at Wolfram Alpha saying *use the real-valued root instead* - or am I missing the point?

[@ishaanivaturi2387]

Yeah LOL he completely missed that

[@txorimorea3869]

The question is why WA picks that one among 9 possible solutions.

[@medexamtoolscom]

I didn't even use wolfram alpha. I just used windows calculator and got 1.999999999... as the answer and I'm like ok, I guess it's 2 then.

[@blackpenredpen]

medexamtoolsdotcom niceeee!! So classic to have an answer like 1.999999....

[@VendettTM]

Well, you know, 1.9999.. is equal to 2, so it's ok

[@stapler942]

Floating point errors. What would computer science be without them!

[@peterhron]

WolframAlpha was correct, it is just a computer, though, and cannot know which solution you were looking for. That is why it asked you, whether you want to use the principal root, or the real valued cube root. You missed that, that's why it came with a complex solution, as people before me already pointed out.

[@themultilangualcoder8832]

This is the first time that I've solved alone one of math problems after watching four or five of your other ones! and I'm happy with that . thank you for your intresting work and content!

[@srinathtankasala]

The reason you get the incorrect answer is because you miss cube root of -1. at 7:30 you cannot take (-1)^1/3 outside the cube root. cube root of -1 has 3 roots so you should get three different cubic equations for X. That's why wolramalpha does not give 2 as the result

[@supermanifold]

Good point. How would you rectify this?

[@srinathtankasala]

TLDR: The issue occurs when wolframalpha tries to compute cube root of 7-sqrt(50). If you just type (7-sqrt(50))^1/3 into wolframalpha you will see that it gives a complex number. To rectify use -abs((7-sqrt(50))^1/3) instead. @@supermanifold Technically there are 9 possible solutions for "x", not 3. Namely: (7+sqrt(50))^1/3 = (1)^1/3 * abs((7+sqrt(50))^1/3) which has 3 possible values. (7-sqrt(50))^1/3 = (-1)^1/3 * abs((sqrt(50)-7)^1/3) which has 3 values so totally 3x3=9. To get the above keep (-1)^1/3 just as is. So his last equation becomes 14+3 * (-1)^1/3 * X = X^3. Take 14 to RHS and cube both sides to get a 9th order which has 2 as its real root and 4 pairs of complex roots

[@jamesphillips92jp]

Did you try clicking the link to use the real valued root, instead? I bet that's where your 2 went. Love the videos!

[@blackpenredpen]

James Phillips love ur comment!! :)

[@OHomemquecalculava]

Yeah, there's where it is. I was just waiting for him to notice the link in the center of the screen but he didn't, and it indeed gives 2 as an answer :/

[@NateROCKS112]

Wolfram-Alpha defaults to the principal root, which is the possible root with the smallest non-negative real component. That's why it displayed a complex number.

[@lyrimetacurl0]

@Doc Brown What's the other complex solution though?

[@Daniel-il4gr]

@@lyrimetacurl0 It would be the complex conjugate of the other complex root, 2.621...-i0.358... That is actual an important theorem in algebra, the complex conjugate root theorem. If a+ib is a root of the polynomial P, it's complex conjugate a-ib will be too.

[@Hexanitrobenzene]

I think it would be more efficient to modify binomial formula to A³ + 3AB(A+B) + B³ and then do the substitution. Also, as far as my experience goes, problems of this kind are constructed so that a perfect cube (square) is under a cube (square) root. In this case, the first try (1+√2)³=1+3√2 + 3*2 + 2√2 = 7+5√2 gives the answer right away.

[@Fuzeha]

That's true, but it takes a bit of a jump to get there intuitively unless you're well practised, so I think he taught it the right way.

[@RickDesper-v8z]

Yes, this is the method I used, since it's obvious from a glance that AB will be a nice integer.

[@RickDesper-v8z]

@@Fuzeha It's really the same formula. It just makes the math simpler if you first multiply AB and notice that the product simplifies in a very useful manner. That's true if you have AB(A+B) or A^2B + B^2A.

[@jimskea224]

In my complex analyss classes (and a few books on the subject I use ∛ to denote the principal cube root only and reserve the superscript notation z^(1/3) to denote the multivalued power function. So, by that notation, Wolfram Alpha's answer is correct.

[@ClueyyHD]

can we just appreciate how easy he makes the pen swapping seem

[@leonardobarrera2816]

I don’t have an easy homework as that, in my brain I was thinking that it only can have a complex solution

[@Near_Void]

I like how complex mathematical formulas equal something so pure and simple

[@not_vinkami]

Everything is better if x^(1/3) is changed to cbrt(x)

[@marcinozga5098]

13:49 Click "Use the real-valued root instead"

[@dhov760]

This video really has given me some confidence in my mathematics adventures, I’m a Bio/Chem Major and have always loved math but haven’t touched it in awhile and I feel as if I could’ve completed this problem which was seemingly hard to begin with. Thanks bud.

[@niveditabhaskar8193]

7+5√2= 2√2+1+3√2+6 =(√2)³+1³+3(√2)[1+√2] =(√2+1)³ Similarly 7-5√2=(1-√2)³ Therefore, (7+5√2)^1/3+ (7-5√2)1/3 = (1+√2) + (1-√2) =2

[@michaelbauers8800]

Lovely algebra. This sort of thing doesn't seem to come up much, but it's still great to know how to work it out, and your explanation was crystal clear. Obviously you have to take care, with x, when there's multiple roots, but only one is a valid solution.

[@galladeguy123]

Something interesting I noticed: What you solved for in this video is a value of 7 for x in this equation: y=cbrt(x+sqrt(x^2+1))+cbrt(x-sqrt(x^2+1)) If you solve for x, you get x=(y^3+3y)/2 We can then prove that if y is an integer, x will also be an integer. To do this, we just need to prove that y^3+3y is even for all integer values of y, since an even divided by 2 is always an integer. This will be true when y^3 and 3y are either both odd or both even, since they must add up to an even number. y^3=y*y*y, and since an odd times an odd is odd and an even times an even is even, y^3 will be odd if y is odd and even if y is even. This is also true for 3y, since an even times an odd is even and an odd times an odd is odd. Thus, we can say that y^3+3y is even, and that by extension x is an integer for all integer values of y. Therefore, there are infinitely many integer solutions to the equation at the top of the post.

[@elenenify]

no. wrong. there are not.

[@galladeguy123]

Why not? Unless I'm missing something, I don't think there was a flaw anywhere in my reasoning.

[@elenenify]

@@galladeguy123 Because in the original equation y has only one value, it is not a function. I mean if you treat it as a function of x then of course you are right.

[@galladeguy123]

When I said original equation, I meant the one I had posted, not the one in the video. I should have made that more clear.

[@velsni]

More in general, you can get infinite solutions for sqr3(m+sqrt(n))+sqr3(m-sqrt(n))=2 by taking m=7+3k, n = k^3+12k^2+45k+50, for k =0,1,2 ... etc

[@folklinoff]

Notice that 7+sqrt(50) can be also written as 7+5*sqrt(2) which is a formula of a cube 1^3+3*(1^2)*sqrt(2)+3*1*(sqrt(2)^2)+sqrt(2)^3 and that is simply (1+sqrt(2))^3. Also 7-sqrt(50) is (1-sqrt(2))^3. Further-easier. (1+sqrt (2))^3^(1/3)+1(-sqrt(2)^3^(1/3)=1+sqrt(2)+1-sqrt(2)=2. Easy! But I can notice that your solution more universal if take another numbers. Great job!

[@forcelifeforce]

Yours is the better method here.

[@twy_]

Since you use ^(1/3) instead to obtain the cubic root, so the principal root is returned. Try out (-1)^(1/3) and you will obtain 0.5 + 0.866... i (1/2+(√3̅/2) i) instead of -1. In polar representation of complex plane, -1 is represented as (1,180°(π)), so wolfram alpha should return (1,60°(π/3)) by default instead of (1,180°(π)) or (1,300°(5π/3))), the 180°(π) returns the result of 60°(π/3), different from that a positive real number which is 0°(0), returning 0°(0). In order to obtain the cubic root, choose real‐valued root or CubeRoot(-1) or cbrt(-1) instead

[@xilaifan6905]

Good answer! Solved my problem. Thx!

[@notovny]

I decided to pause when you asked, and make an attempt; it was _extremely_ satisfying to work through. Thanks.

[@matthewtanous7905]

WolframAlpha returns the principal root by default, which is the root with the largest real component NOT the purely-real root. There’s a link to get the pure real root near the top of the screen there.

[@dalek1099]

The definition of cube root is different in Complex Analysis than the definition in Real Analysis. In Real Analysis, the answer will be 2 but in Complex Analysis the answer will be what Wolfram Alpha gives. Wolfram Alpha always uses Complex Analysis as a default. Maybe you should have mentioned whether you were working in Real Analysis or Complex Analysis or your definition of the cube roots. In Real Analysis, you can only have one cube root but in Complex Analysis the most sensible cube root is complex.

[@drewmandan]

And there are infinitely many octonian roots (I think).

[@deepakgowda89]

cuberoot (7 +√50) + cuberoot (7 -√50), use this expression you'll get 2 in Wolframalpha. It is always good to solve math problems rather than depending on software tools.

[@niveditabhaskar8193]

7+5√2= 2√2+1+3√2+6 =(√2)³+1³+3(√2)[1+√2] =(√2+1)³ Similarly 7-5√2=(1-√2)³ Therefore, (7+5√2)^1/3+ (7-5√2)1/3 = (1+√2) + (1-√2) =2

[@boostaddict_]

I have a math exam on Thursday and something like this will probably be on it lol. Good video.

[@sagnikpanja7191]

Wolfram Alpha is using the principle branch cut but if you use the command cuberoot or sure(x,n) instead of raising the power to 1/3 you will get your required rational value as x³=c has 3 solutions.

[@alnitaka]

Wolfram Alpha says "using the principal root". Instead, click on "using the real-valued root", and you will get 2. Also, if instead of raising the two terms to the 1/3 power, take cuberoot() of them - Wolfram Alpha does have this function, and using that gives 2. Also I note something interesting. As soon as blackpenredpen typed in (7+sqrt(50))^(1/3), the red number below showed 2.414213... That's sqrt(2)+1( !) That may explain this phenomenon of having a complicated expression like (7+sqrt(50))^(1/3)+(7-sqrt(50))^(1/3) turn into just a 2. It's because 7+sqrt(50); i.e., 7+5sqrt(2), is the cube of sqrt(2)+1.

[@xnick_uy]

Wolframalpha is just "too wise". You have to input the formula as cbrt(7+sqrt(50))+cbrt(7-sqrt(50)) to get the real solution. Note that the issue here is that the expression x^(1/n) is not a function of x, and wolframalpha decides to report only one of the possible solutions of the equation y = x^(1/n) (the principal value).

[@liquid_cheese9930]

blackpenredpen: Writhes a bunch of cool math and receives a beautiful answer Wolframalpha: That's dope, but how about clicking "the real-valued root instead"

[@charlescox290]

That's amazing. A small critique, when you referenced Pascal's triangle, you didn't mention that it is be specifically the third row. Making that association might have made it click for people that didn't quite grab the concept yet.

[@michaelempeigne3519]

why not call 7 + sqrt ( 50 ) = a^3 and 7 - sqrt ( 50 ) = b^3 then it becomes cbrt ( a^3 ) + cbrt ( b^3 ) = a + b Now call a + b = x ( a + b )^3 = x^3 a^3 + b^3 + 3ab ( a + b ) = x^3 14 + 3ab *x = x^3 to find ab we do : a^3 * b^3 = ( ab )^3 == > 49 - 50 = (ab)^3 == > ab = - 1 so as a result x^3 + 3x - 14 = 0 and x = 2 comes out.

[@pransuvadsmiya4242]

It can even be solved by keeping (7+√50)^1÷3 =X+√Y then cube both the sides then make the rational part of lhs=rhs and same with irrational one.

[@denemedeneme5729]

let's assume that a€Z and b is √something: (a+b)^3 = 7+√(50). that means a^3+3ab^2=7 and b^3 +3ba^2 = 5√2 a=1 and b=2√2 and the answer of the question is 2.

[@KcMsPik7278]

What??!!!

[@michelferreira333]

I like your channel. Really good stuff. You explain pretty well. Glad I found your channel. Thanks for all the videos.

[@blackpenredpen]

Michel Ferreira thank you Michel!!!

[@bougmayamba4988]

Exactly right. We all know that sqr50=5sqr2 which means we can have 3sqr2+2sqr2. 2sqr2 = (sqr2)^3. In 3sqr2 we have 3 ×sqr2 ×1. And we know that (a+b)^3 = a^3 +b^3 + 3 a^2 ×b + 3 b^2 ×a. In our case we have a^3 which is (sqr2)^3. 3sqr2 ×1 should be 3b^2 ×a becoz if a=sqr2 , 3 a^2 ×b should have been whole number not radical. So b=1 and in 7 we have 1 + 6 which is b^3 + 3a^2 ×b = 1+ 3×(sqr2)^2 × 1. Finally we got ((1+sqr2)^3)^1/3 + ((1-sqr2)^3)^1/3 = 1+1+sqr2 -sqr2=2.

[@bougmayamba4988]

Sqr means square root of.

[@Johnxxxxxxx]

This was just now recommended to me by KZbin, but it’s always nice to have a video where I actually know how to do something lmao

[@JohnDoe-iu5xi]

beautifully done. you're helping me see the bigger patterns to basic operations.

[@mkrish1977]

7:06 cube root of -1 is not just -1. How do we handle imaginary roots?

[@ZUB3RBR]

(-1)*(-1)*(-1) = -1

[@clementr]

@@ZUB3RBR e^(iπ/3) and e^(-iπ/3) are also roots of - 1: e^(iπ/3)^3=e^(iπ)=-1 e^(-iπ/3)^3=e^(-iπ)=-1

[@keyowah1]

u need to click on "the real-valued root" on top bro

[@AkshayMuraliNerd098]

I like the way you switch between the pens :)

[@blackpenredpen]

thanks!!!

[@abhayagrawal8461]

Ur videos may let me clear my Olympiad...thanks

[@yubodeng2103]

I think wolfram alpha is correct because when we raise sth. to the third power, we create extraneous root. In the expression,7 is less than sqrt(50), so the cube root of that should be a+bi.

[@Dorddis]

X=2 is the first zero... Other two zeroes are still possible..... Answers may vary

[@blackpenredpen]

i love "Answers may vary"

[@bs140598]

Glad u pointed it out first he clearly dont know the algebra

[@mohitjain3496]

There are actually upto 9 possible answer to original expression. Cuberoot or -1 itself has 3 values and thus 3 equations where each has up to 3 zeroes.

[@michelkhoury1470]

Yes

[@Tomyb15]

Turns out wolframalpha wasn't wrong so I keep trusting it more than anything.

[@blackpenredpen]

Ciroluiro but.... when u actually see students who just put down whatever they see from WFA...

[@georgehnatiuk5806]

Blackpenredpen, Note: A = cube root [ 7 + sqrt(50) ] = 1 + 5*sqrt(2) B = cube root [ 7 - sqrt(50) ] = 1 - 5*sqrt(2) hence A + B = 2

[@seanehle8323]

... then you realize that you gave a poor assignment, sigh, and make a better curriculum for next semester. This is the cycle of teaching. Knowing which equation to solve and how to apply the solution is what to teach. Being able to solve equations is the smallest part of math. Understanding the conceptual relationships and interwoven logic which compels the utility of an equation is far more important.

[@ib9rt]

"...when u actually see students who just put down whatever they see from WFA..." Then it's good to ask students trick questions like this, as you will catch the ones who cheat...

[@blackpenredpen]

how's that a poor equation/assignment?

[@caznax8270]

wait, *ARE YOU SECRETLY AN OOD?* for those who don't watch doctor who, a ood is an alien that has a orb in its hand.

[@blackpenredpen]

Camouflaged Will u never know... lollll

[@EmilioRomero]

Camouflaged Will OMG, a whovian! YANA, I understood your reference.

[@mikastrae]

He's too intelligent to be an Ood.

[@MrJdcirbo]

Well definitely hears the song of mathematics.

[@nilanjanshyamchowdhury6926]

Love how he switches between the red and black markers

Hey, I don't really know what is going wrong but if you just press "the real-value root", it will give you the correct answer, you also can do that if you input "cube root of (7+root 50) +cube root of (7-root 50)"

[@QYong-rq6iw]

Nice advertising for wolframalpha though lol

[@zebrg]

See Alan Falleur answer. input "cuberoot(7+sqrt(50)) + cuberoot(7-sqrt(50))" instead.

[@spiderjerusalem4009]

If you still don't believe it's 2 Here 7+sqrt(50) = (1 + sqrt(2))³ 7 - sqrt(50) = (1-sqrt(2))³ so, it would be = (1+sqrt(2)) + (1-sqrt(2)) = 1 + 1 + sqrt(2) - sqrt(2) = 2

[@alexiijserbin1791]

Thanks for the task, I didn't solve it completely, but had an idea to make like you did, that the new expression after ^3 starts too look quite same as the initial one. Really enjoyed the solving part, idk why, but for me it it's beautiful! Thanks!

[@toopytoopy8547]

I found a shortcut with almost no calculations: let's note a=(7+sqrt(50))^(1/3) and b=(7-sqrt(50))^(1/3). We want to find their sum s=a+b. Notice that a^3+b^3=14 (evident) Also we can easily calculate the product ab=[(7-sqrt(50))(7+sqrt(50)]^(1/3)=[49-50]^(1/3) =-1. Let us now consider the polynom P(X)=(X-a) (X-b). By developing it, P(X)=X^2- sX- 1. In fact, i replaced the product ab by its value which is -1. 0=P(a)+P(b)=(a^2+b^2)-s(a+b)-2 Which means that a^2+b^2=s^2+2 Also, 0=a P(a)+b P(b)= (a^3+b^3)-s(a^2+b^2)-(a+b) But remind that a^3+b^3=14 Then, 0=14-s(a^2+b^2)-s=14-s(s^2+2)-s Then s^3+3s -14=0 * At this point, he made a kind of 'trick' ( to not say big error) : in fact, we can verify that s:=2 does satisfy the equality * but eventually there 3 possible solutions for it and anyone of then could be the searched sum of a and b. His argument that the 2 other solutions are imaginary has to be proved . To do so, we manage to factorize s^3+3s-14 by (s-2) using the identification method and we get that s^3+s-14=(s-2)(s^2+2s+7) Thus the 2 remaining solutions are the zeros of s^2+2s+7 which is evidently having imaginary zeros. Consequently, we have a+b = 2 :) Notice that this method can be generalized : suppose that a=(y+sqrt(y^2+1))^(1/3)and b=(y-sqrt(y^2+1))^(1/3) such that y is an entire number (in the video he took y=7) then using my method you will found that their sum s=a+b is a solution of s^3+s-2y = 0. For example, if y=1 --> s= 1 if y=7 --> s= 2 (case of the video) if y=15 --> s= 3 if y=34 --> s=4 if y=75 --> s= 5 etc..... If you have any remark/question about my method please react to my comment :)

[@rikschaaf]

The problem seems to be with the (7-sqrt(50))^(1/3) part, where wolfram alpha takes the wrong complex solution. (7+sqrt(50))^(1/3) ~= 2.4142 and |(7-sqrt(50))^(1/3)| = 0.4142, but according to WA, instead of (7-sqrt(50))^(1/3) being equal to -0.4142, it's equal to 0.2071 + 0.3587 i (eg a 120 degrees clockwise rotation away from -0.4142 in the complex plane). EDIT: in 14:36, you can see that it's using the principal root, rather than the real-valued root.

[@brucehakami4489]

Thank you for your videos and your enthusiastic presentation, which I find very interesting. Usually you proceed at a blistering high speed, which is exciting and wonderful. This one looses momentum when you cube x numerically. If you do it symbolically, it would be faster and more transparent: (A+B)^3 = A^3 + B^3 + 3.A.B.(A+B) before you substitute numbers.

[@gedlangosz1127]

Mr Blackpenredpen - a lovely problem but I’m afraid your solution to it is not. I paused the video and very quickly came up with the following. The key is that we are given there is an elegant solution so we know the expression can be simplified. ∛(7 + √50) + ∛(7 - √50) Start by realising that ∛(7 + √50) + ∛(7 - √50) = ∛(7 + 5√2) + ∛(7 - 5√2) Now the only way this will simplify is if the expression inside each of the radicals is a perfect cube. Let’s try and look for this. Can (7 + 5√2) be expressed as (α + k√2)³ ? You give us the expansion (a + b)³ = a³ + 3a²b + 3ab² + b³ Now in (α + β √2 )³ , we must have α = 1. If α ≥ 2, then we could not have 7 in the radical because 2³=8 Now we need β so that (1 + β√2)³ ≡ (7 + 5√2) By comparing this with (a + b)³ = a³ + 3a²b + 3ab² + b³ We can easily find that β = 1 So (7 + 5√2) = (1 + √2)³ similarly (7 - 5√2) = (1 - √2)³ So ∛(7 + √50) + ∛(7 - √50) = ∛(7 + 5√2) + ∛(7 - 5√2) = ∛(1 + √2)³ + ∛(1 - √2)³ = 1 + √2 + 1 - √2 = 2 No need to cover the white board in lots of horrible expressions and no need to try and guess the solutions to a cubic.

[@Bollibompa]

It is lovely to be able to express your opinions, isn't it? I like his solution a lot more.

[@aritradasgupta4461]

Your one is shorter cuz you're comparing but his is the one easier to understand

[@orangenostril]

I made it to 8:06 on my own (by distributing everything lol) but I was like "oh noo, I'm just back where I started!" and gave up

[@tural3918]

Oh no! My brain! It's broken!

[@jamesmnguyen]

I sort of have the cubic formula memorized and when I saw the thumbnail, I immediately recognized that it was formated like Cardano's formula (for a depressed cubic of x^3+px+q). The 7 is equal to -q/2, which results in q=-14. The sqrt(50) then allows you to calculate p from (q^2)/4+(p^3)/27=50 which results to cbrt(27(50-(q^2)/4)) = cbrt(27)cbrt(50-196/4) = 3cbrt(50-49) = 3. You then get the cubic equation x^3+3x-14=0 which 9:50 explains how to solve that.

[@SonArdhynata]

I suggest you to try (-1)^(1/3) in wolfram alpha and compare it with cbrt(-1), since there is the step here that showing the cube root of -1. I guess we are missing something in our understanding here.

[@udic01]

the 1st expression is equal to (1+sqrt(2)) the 2nd is equal to (1-sqrt(2))

[@raunakagarwal7485]

But dude as it is a cubic equation so it may have 3 factors. So the result of this may be also equal to {-2 ± 2i√6}/2

[@TJStellmach]

Not with the principal cube roots, which is what that notation usually means.

[@pierredonias8940]

"Use the real‐valued root instead"

[@err0rz633]

You can bypass long division since we know that a=1 in the quadratic and the sum of the x^2 coefficients = 0; -2+b=0 ==> b=2. You then have x^2+2x+c, and we know that c = -14/-2=7. So you have x^2+2x+7 with two complex roots.

[@ΑΝΤΩΝΗΣΠΑΠΑΔΟΠΟΥΛΟΣ-ρ4τ]

04:34 if you have voice recognition for google assistant open ("hey Google"/"okay google" activation) his words will activate it

[@theuseraccountname]

You can solve it by using the native cube root function in Wolfram Alpha: cuberoot(7 + sqrt(50)) + cuberoot(7 - sqrt(50))

[@lumiinouss4744]

I’m doing A level maths and I really struggle with intergration, in your 100 integrals world record i learnt so many more techniques I appreciate it a lot and now I find myself in your other videos

[@rezamiau]

it's just Cardano method's answer to that cubic equation, which we knew the real answer of it (2).

[@TJStellmach]

Yes, noticing that the original expression has the form of Cardano's formula, you can quickly back-solve for the cubic polynomial and skip the first 11 minutes.

[@pookie8856]

Thankyou for leaving me so I could do the question myself in privacy at the beginning of the video

[@blackpenredpen]

You are the first person who thanked me for that! Thank you too!

[@pookie8856]

Haha you're welcome. You left the screen and I was like where did he go?

[@xiaoyang1519]

The Wolfram Alpha defines n-th root in the complex sense. It is a multivalue function. One can also show that (7+50^(1/2))^(1/3)=sqrt(2)+1 and (7+50^(1/2))^(1/3)= -sqrt(2)+1 in the real sense. But there are two other solutions for each of them by multiplying the 3rd primitive root of unity. Somehow, the WA takes the values in the first quadrant. We can see it is (3*sqrt(2)+)/2 + i*(sqrt(6)-sqrt(3))/2 which is the same as the numerical values given by thh WA.

[@mikapeltokorpi7671]

Wolfram Alpha answer is complex answer for vector length of 7.

[@bs140598]

I love ur comment was searching for some real mathematicians

[@larrygao5854]

On CASIO non-scientific calculator, it also shows the answer of 2. But i was wondering how does a calculator work out this answer.... I think should be something about binary digit right?

[@coolgarrett17]

larry gao it approximates sqrt(50) using taylor series then adds 7 to that. Then it approximates the cube root of that sum using taylor series. It then does the same thing for the second term (but subtracting this time) and then it adds those together to get 2

[@larrygao5854]

@@coolgarrett17 so basically it's approximation all the way to estimate the answer is 2 right? (Using taylor series)

[@lythd]

@@larrygao5854 well he didnt respond over 6 months but if youre still wondering yes that is exactly what happens.

[@kbqz]

@@lythd hello, i have a question about my calculator. In my Casio calculator there is a function called RSolve with change a sequence from her '' recurring form'' (Un+1 =... Un, U0 or U1 =... ) to the explicit form (Un =...n) (I don't know the exact words to describe the type of the sequence since English in not my mother tongue sorry...) do you know how the calculator do this? It is so interesting...

[@lythd]

kingbadlex Gaming I mean I’m only 14 but I believe what happens it will take Un+1 - Un and multiply each of those terms by x and add U0. So if it’s U0 = 4 and Un+1 = Un + 2 it will get 2x+4. Hope that made sense.

[@alexkidy]

Here in Brazil we call this MATH MAGICS !!!!

[@staircase72]

Wolfram also is showing at the top of the page exactly why it found the wrong solution. It says "Assuming the principal root | Use the real‐valued root instead". It is choosing the definition for cube root of a negative number that gives a nice complex continuation rather than the one that gives a real solution for real inputs.

[@fabrizzioorderique2343]

Love watching your videos! Your enthusiasm makes every puzzle really fun to go through