believe in the math, not wolframalpha
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- Опубліковано 16 сер 2017
- Believe in the math, not WolframAlpha. We will simplify this nested radical expression. This expression is usually from the cubic formula. Surprisingly, this expression gives us a whole number! Enjoy!
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The greatest trick humanity ever learned from mathematics - If you don't know some value, call it x and proceed.
Vinay Emani that's why we have an X even in sex !!! Lol.
XD
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.
cemsentin x^3+3x-14 isnt equal to x^3 - 3 + 3x - 6
-3-6 are -9 not -14
Forget the math, I want to learn that marker-switching trick.
It's quite easy - he presses the 'pause' button each time
Why (7+√50)(7-√50) = -1???
@@_-_-Sipita-_-_, why ..
7×7 = 49,
7√50 - 7√50 = 0,
√50 × (-√50) = -50.
but why ..
@@keescanalfp5143 (a^2 -b^2)
7^2 - (√50)^2
@@quanpa,
yess of course. did she zij/hij see, you think
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.
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.
When you get an integer as a root, that makes you reeeeaaaally happy.
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.
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?)
Thank you for explaining that, I didn't notice it George Haramuniz
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.
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.
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
The marker switch game is strong in this one.
zweiosterei Was that a star wars reference? Lol
Yeah it was
👍🏼 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).
@@M-F-H i am a 12 yr old, is something wrong with me if i dont understand this?
@@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.
Whoever came up with this problem is a genius
nooo,
this is not very hard
@@Yash.the.seeker I think Anal was talking about the creation of the problem and not the answer.
@@davidzheng8926 hhhhhuuuuuu
@@Yash.the.seeker lmao u low iq
@@ygritte4829 😑
2:02 the way you swap between the black pen and red pen is so smooth, Jesus.
Your marker switching skills are deft af.
Jiggerjaw thanks!!!!
😂
😂😂😂
His marker switching skills must be due to the chopstick skills he have !!!
Mohan7 That's actually a pretty legitimate assumption lol
Impressive black pen red pen powers at 2:00
zerep sesiom lolllll thanks!!!!
zerep sesiom
Ahhhhhhh I see what u did in ur YT name now. Loll
Truly impressive...
500 iQ 😂😂
This was awesome, i couldn't understand how he was doing that, I thought that the board reflects bad the ligths jajajajsjs
This equation literally blew my mind.
I think you should reconsider the use of the word "literally" here. I hope...
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.
That's how I did it too! (I am not even Russian!)
how do you know that 7-5√2=(1-√2)^3???
i mean how can you even come up with smth like this
@@user-mx9yg4oh9h I am really interested to know that as well
@@user-mx9yg4oh9h 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
I feel like I disappointed him every time I don't pause to try 😅
windowsforvista it's okay. U can just enjoy the show
Me2
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😀
@@harshranjan8526 Oh really
Try this then
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!
photomath is better
andreas purnomo LOL! Thanks for the good laugh
@@nathannguyen7449 know right
😁
@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.
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"
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.
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.
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..
@@twosomestars9254 learn maths pls
@@shreyasdas5130 I was not saying a fact to begin with. this is comment section, your opinion matters, not facts.
@@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.
I always enjoy when some complicated sq roots, cube roots etc. end up equaling some integer value :)
Same. When he magically solved for 2, I was like, "Wait, it can't be that simple. It's just an integer?!"
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
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.
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.
@@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.
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
Bloomex yes
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.
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
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.
+ 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.
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!
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.
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
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
@@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.
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.
No, it doesn't show 2!, it shows 2. 2! is equal to 2×1, which is... uh, nevermind.
@@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.
We need a video on how to switch markers like that, looks simple yet amazing.
It's on my channel trailer, which is here ua-cam.com/video/gA4Lcrko5jg/v-deo.html
I'm mkre curious about what sort of marker he's using that doesn't seem to dry out without a cap on!
Love watching your videos! Your enthusiasm makes every puzzle really fun to go through
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.
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
What really impresses me is how seamlessly you can switch between both pens. Amazing.
click on "Use the real‐valued root instead"
LT (r*T'(r))'=0 can i have this solution step by step from wolfram alpha ,sir
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.
An unusual way to solve a cubic, but if it works for you...
calculator: Am I a joke to you?
You can calculate this in calculator, it's for numbers only!!!
@Austin Martín Hernández yes the scientific one.
.
I mean can you solve a question which has more than one equation and they are related like intersection of lines using calculator...???
Brains are smarter than calculators! (After all, brains invented the calculator.)
If you're dyslexic, you need a Brian. ;)
@@TheBaggyT lmao
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
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.
@@lionel4685 for practis
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
beautifully done. you're helping me see the bigger patterns to basic operations.
nobody:
blackpenredpen: believe in the math
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.
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.
medexamtoolsdotcom niceeee!! So classic to have an answer like 1.999999....
Well, you know, 1.9999.. is equal to 2, so it's ok
Floating point errors. What would computer science be without them!
Everything is better if x^(1/3) is changed to cbrt(x)
But if you enter this:
cbrt(7-sqrt(50)) + cbrt(7+sqrt(50))
The result is 2.
Fascinating
discrimination on powers and roots
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)
@@iliyasone _"you cannot raise negative numbers to non-integer degree"_ , incorrect
(-2)^3 = -8, therfore (-8)^(1/3) = -2
@@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
13:38 Isn't there some hyperlinked text at Wolfram Alpha saying *use the real-valued root instead* - or am I missing the point?
Yeah LOL he completely missed that
The question is why WA picks that one among 9 possible solutions.
Did you try clicking the link to use the real valued root, instead? I bet that's where your 2 went. Love the videos!
James Phillips love ur comment!! :)
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 :/
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.
@Doc Brown What's the other complex solution though?
@@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.
can we just appreciate how easy he makes the pen swapping seem
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!
Yours is the better method here.
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
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.
Can we use the Lagrange's resolvent to solve the cubic equation at the end? @blackpenredpen pls replyyyy
I got almost the same results, however I do have a doubt and it is that since cbrt(-1) has 2 imaginary roots too the final equations I got were
S^3 + 3S - 14 = 0 ;
S^3 - 3((1 + i sqrt(3))/2) - 14 = 0 ;
S^3 - 3((1 - i sqrt(3))/2) - 14 = 0
I will say however that I am not very experienced with complex numbers so I may be wrong
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.
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.
Yes, this is the method I used, since it's obvious from a glance that AB will be a nice integer.
@@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.
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.
Thanks for this and other videos I've been enjoying. Is there any reason to believe there is a rational (nevermind integer) root? Certainly no issue with checking...
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!
I like how complex mathematical formulas equal something so pure and simple
Turns out wolframalpha wasn't wrong so I keep trusting it more than anything.
Ciroluiro but.... when u actually see students who just put down whatever they see from WFA...
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
... 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.
"...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...
how's that a poor equation/assignment?
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.
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
Good point. How would you rectify this?
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
Thanks, UA-cam ads, for thinking that every math video on UA-cam is over a decade old and completely unhelpful...
Great video, by the way!
I like your channel. Really good stuff. You explain pretty well. Glad I found your channel. Thanks for all the videos.
Michel Ferreira thank you Michel!!!
This was just now recommended to me by UA-cam, but it’s always nice to have a video where I actually know how to do something lmao
I decided to pause when you asked, and make an attempt; it was _extremely_ satisfying to work through. Thanks.
u need to click on "the real-valued root" on top bro
in the second addend of the big sum the part under the cubic-root goes slightly under 0, because 7 minus 7.07 (root(50) is 7.07) equals a negative number; thats why it cannot solve it without complex numbers; but you can click the "use the real-valued root instead" to get the number 2
the calculator was actually in the RIGHT to not give 2 at first because it wasnt allowed to handle the second addend within its normal operating range
MDFlight you can not have the square root of a negative number, but you can have a cube root of one. For example the SQRT of -1 is i, but the cube root of -1 is -1.
(-1*-1*-1)=-1
(-2*-2*-2)=-8
Man, cubic roots are always defined, no need for complex numbers.
I think you miss important idea that the calculator has to handle all cases - in case of x^(1/3) the value of 1/3 does not belong to integers, so it has to change the operating range to handle complex numbers. Also note, that the answer provided by wolfram is totally accurate in this number set - same like sqrt(4) = 2, while both 2 and -2 squared equal 4, wolfram just took the '-2' option here which its totally allowed to do while dealing with complex numbers. Using cuberoot instead of x^(1/3) gives the expected answer right away.
www.wolframalpha.com/input/?i=cuberoot(7%2Bsqrt50)%2Bcuberoot(7-sqrt50)
noisy cat It distinguishes between "real valued root" and "principal root". Not sure what these are. I don't get why it doesn't simply give all the solutions at once.
Brandon Heintz thanks broh, i was struggling with that part
"Use the real‐valued root instead"
I learnt those in grade 10 (not the actual question but the use of formula, surds and the remainder theorem) and the way you taught was really interesting
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.
Camouflaged Will u never know... lollll
Camouflaged Will OMG, a whovian! YANA, I understood your reference.
He's too intelligent to be an Ood.
Well definitely hears the song of mathematics.
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.
no. wrong. there are not.
Why not? Unless I'm missing something, I don't think there was a flaw anywhere in my reasoning.
@@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.
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.
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
such a question came in my test yesterday, i had been searching the internet upside down and now i got this in my recommendation. Thank you so much!
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"
I have a math exam on Thursday and something like this will probably be on it lol. Good video.
Nice advertising for wolframalpha though lol
See Alan Falleur answer. input "cuberoot(7+sqrt(50)) + cuberoot(7-sqrt(50))" instead.
there is an easier method; assume k=expression. then substitute t as 7+root(50) then, 7-root(50)=-1/t therefore, k=(t^1/3)-{1/(t^1/3)} now cube both sides using (a-b)^3=a^3-b^3-3ab(a-b) a^3+b^3 becomes 14 ab becomes 1 and (a-b) becomes k again. solve the cubic to get the answer.
Really enjoyed this video. Thank you!!
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.
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
13:49 Click "Use the real-valued root instead"
13:35 "Assuming the principal root | Use the real-valued root instead[.]"
Where can I get more exercises like this?
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).
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
Good answer! Solved my problem. Thx!
How did you rule out the other two solutions of the 3rd degree equation?
I love this channel. I seen random videos on yutube and often I watched your video. Today I'll subscribe in happiness...
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.
And there are infinitely many octonian roots (I think).
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.
Sqr means square root of.
I have seen a very similar problem like this which also resulted in an integer. It was cbrt(9 + 4*sqrt(5)) + cbrt(9 - 4*sqrt(5)), and the answer ended up being 3. Perhaps there is some connection between the values inside of the cube root that will make the sum an integer?
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.
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.
you should've told that
7 + (50)^(1/2) = (1 + (2)^(1/2))^3
7 - (50)^(1/2) = (1 - (2)^(1/2))^3
to not make confuses before you go into algebra trick or after.
@blackpenredpen since 7-√50 is negative how can it be under a cubic root?
Only roots with an even index are restricted to positive radicands. Odd indexed roots (like cubic root) can accept negative radicands
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.
One of my favourite UA-cam channels!
Partha Dey thank you!!!!!
X=2 is the first zero...
Other two zeroes are still possible.....
Answers may vary
i love "Answers may vary"
Glad u pointed it out first he clearly dont know the algebra
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.
Yes
Is this something similar to cardano's method for solving cubic equations but used in reverse?
How can I know how many solutions they are with X³ ? Thanks a lot (I'm French)
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
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.
What??!!!
You can solve it in an even better way Just take a look
³√a +³√b =X
³√a +³√b +(-x)=0
Now use the law that if x+y+z=0 then, x³+y³+z³=3xyz hope it helps @blackpenredpen
hey root of negative 1 is complex no. why u directly take outside, that part mess me up so could you again explain me in detail
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.
It is lovely to be able to express your opinions, isn't it? I like his solution a lot more.
Your one is shorter cuz you're comparing but his is the one easier to understand
I like the way you switch between the pens :)
thanks!!!
How can you put out the negative sign outside from square root
6:57
Do we have to take the negative out?
Because I worked with the negative inside the cube root, but unfortunately, it led to the equation: x^3-3x-14=0, which yields no integer solution.
Because, if you don't take the negative out, the number inside the cube root would be (-7 - √50) and (-7 + √50) which is different from the initial x value. So we can't change it to x. Cmiiw
I enjoyed this so much! I highly recommend taking a look at photomath's way of solving the problem. In a way, I see it at a true Order of Operations problem as it is solved without the use of variables but still used formulas. It's just an interesting take.
I'll link it to you if you are interested.
I am in grade 7 and a similar problem is found in Maths Olympiad homework thanks ❤
it works when you put in (7+sqrt(50))^(1/3)+(7-sqrt(50))^(1/3)=x
it asks if you want to use the principal root or the real-valued root, if you select real valued root it gives you x =2
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.