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

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

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.

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

Jiggerjaw thanks!!!!

😂

😂😂😂

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

Mohan7 That's actually a pretty legitimate assumption lol

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 😑

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

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.

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

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

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.

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

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.

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.

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.

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

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.

It's on my channel trailer, which is here kzfaq.info/get/bejne/nadkf8aj0NWZmps.html

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

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

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

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

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!

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.

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

Yours is the better method here.

Yeah LOL he completely missed that

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

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.

Michel Ferreira thank you Michel!!!

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?

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

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.

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

Partha Dey thank 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

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

Sqr means square root of.

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

Good answer! Solved my problem. Thx!

thanks!!!

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

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

That would be true if the -1 was under a square root. But here, the -1 is under a cube root, so (-1)^(1/3) is equal to -1, since (-1) x (-1) x (-1) = -1

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

What??!!!

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