Only Geniuses Can Solve This | A Challenging Olympiad Problem

[@Noname-rc8uc]

5:25 First mistake, it should be x = +- sqrt(-sqrt(11)), you forgot the negative solution 5:49 Second mistake, sqrt(-sqrt(11)) does not equal to ∜(-11), it equals to i∜(11). 6:00 Third mistake, ∜(-11) does not equal to -11^(1/4), (a^m)^n=a^(mn) law isn't applicable here as the minus sign isn't raised to the power of m. ∜(-11)=sqrt(i)i∜(11)=(1+i)i∜(11)/sqrt(2) 6:32 fourth and fifth mistake, x= +-∜(-11) is not the final form, and you're missing 2 roots out of x^4=-11 equation. Essentially, it's the same mistake you've pointed out at 0:52 Out of 8 roots to find, you found 2. You were on track to find 2 more, but you stopped too early, and through your mistakes you've made one of your roots double. This is terrible, it would look better if you solved it only for real roots.

[@xXSoonexXx]

First mistake is 2:35. (X⁴+11)*(X⁴-11) Is not zero. Actually it's never zero for all real numbers of X.

[@somedud1140]

@@xXSoonexXx Yes, for real numbers, but in this exercise we are including complex numbers. Which means that part wasn't a mistake. Fundamental theorem of algebra tells us that in complex number field, every n-th degree polynomial has n-roots.

[@cdiesch7000]

Even without Eulers equation, it is clear that the shown solutions are wrong. By continuous factorization, we arrive at 8 solutions: 4root(11); - 4root(11); 4root(11) i; - 4root(11) i; sqrt(2)/2 4root(11) (1+i); - sqrt(2)/2 4root(11) (1+i); sqrt(2)/2 4 root(11) (1-i); - sqrt(2)/2 4root(11) (1-i).

[@puffinbasher]

While a lot of people are having fun pointing out the mistakes - its worth pointing out that the solution here is really easy. Rearrange and make it obvious we are in the complex plain. (121 + 0i)^(1/8) From there its just a case or first using the properties of polar representations to determine the roots, so we calculate the absolute root at 11^(1/4) Given the target value is at a rotation of 0, the points are then trivially at the N*pi/4 directions. So, we have [unit]*11^(1/4) for four solutions and (+-1+-i)*2^(1/2)/2*11^(1/4) for the remaining four. Remember, in these kinds of equations, to sanity check end results!

[@william2496]

Most complicated way to get the wrong answer

[@dewah7775]

@@william2496 Just apply polar system and then the fundamental theorem of algebra lol, now you have the general solution.

[@nicholastrapellides-n4y]

Yes challenging. Vid should have the title 'geniuses don' t need to check their answers'. x=2 can't be a solution cos 2^8 is 256 and 256-121=135 not 0.

[@sttlok]

Apparently he is not a genius.

[@Gabriel-se6tj]

Um, he actually had x=2 and x = 2±2i for the function (x-4)^4 = x^4... So i guess he's right there bud, you should've checked with more attention to what he was doing, there were 2 distinct functions on this video.

[@xy-pk8gb]

8 ln x= ln 121 > ln x = ln 121/8 > ln x = .599 > x = 1.821 , -1.821, 1.821i, -1.82i. (1.82 is approximate 4 root 11 is more precise.)

[@salahlamsaoub7753]

I just had an aneurism seeind x^2 being negative AND THEN take the root of said negative number😭

[@xXSoonexXx]

(x^4+11)*(x^4-11) is not 0 lol... Try with any Real number.

[@cameronashford5744]

okay the video is horrendously bad, gets wrong solutions and takes a really stupid approach to get to the solution, but that one specific step (while not needed) is still technically correct. You said that the term is not equal to 0, but that is exactly the point of the equation: you are supposed to find the x for which it is. Now you said that it doesn't work with any real number, but obviously it won't work for most numbers, BUT it WILL for work for 8 specific ones (or 2 specific real numbers). If you set x to the fourth root of 11 and plug it in, the term will be 0 for example. The formula he used to expand the term is a^2 - b^2 = (a+b)(a-b). You can try that specific equation out with any values for an and b and it will work!

[@unteren_text5425]

Bot

[@sttlok]

@@xXSoonexXx It does work for the real number 11^(1/4).

[@wes9627]

My IQ is roughly 100 and I'm here to prove you wrong. Using Euler's equation, which should be taught in high school, x^8=121*e^{i2jπ}, j any integer. The eight roots are x_j=∜(11)e^{ijπ/4}=∜(11)[cos(jπ/4)+i*sin(jπ/4)], j=0,1,...,7. Just plug in the numbers.

[@TymexComputing]

if you know the eulers formula you already are above 120 IQ ;) i only can using my 100IQ part of brain to prove that 2^8 - 121 not equal 0. because 2^8 -121 = 256 -121= 145 != 0 ;)

[@Foxtrot6624]

@@TymexComputingyou have a very 80iq way of understanding iq. It is entirely irrelevant. You do not need iq to do maths.

[@TymexComputing]

Of course you need IQ to recognize similarities and differences to apply the formulas you could remember, maths is not an mathematica (c) algorithm ;). More than 60% of population does not understand exponents, and more than 80% of population doesn't understand logarithms and these to numbers to 100iq human could pretend to be the same

[@Foxtrot6624]

@@TymexComputing More than x amount of people don’t understand logs because they either left education before needing to use them or haven’t used them in so long they’ve forgotten their formulae’s. It has next to nothing to do with iq

[@dewah7775]

@@wes9627 Yes, so you don't really need to be a genius. You just need to know a little mathematics.

[@TheStormbringer14]

bro just found a nonimaginary solution to an equation of the form x^2 = a, where a

[@dewah7775]

11^1/4, I stared at it for like 10 seconds and didn't even take my paper out.

[@csd1597]

What are the 7 other solutions?

[@OpposingFork]

What was your thought process then

[@dewah7775]

@OpposingFork x^8=121, so x^4=11, and 11 is obviously not a perfect sqaure, so we can say x=11^1/4.

[@wiilli4471]

@@dewah7775 Buddy this is exactly why you would get the problem wrong lmfao, there are 8 solutions.

[@davidkippy101]

But by that logic, 121^1/8 would be a valid solution

[@sharatchandrasekhar2711]

What an absolute bit of rubbish! The answer by inspection can be distilled down to the 4th root of +/-11

[@DhoklaAboveVadapav]

sqrt(sqrt(11))

[@RossJandorf]

X equals the fourth root of 11

[@brucekritt7036]

x = plus or minus the fourth root of 11

[@Myrslokstok]

Ah the sqrt(11) was smart!

[@shdz5984]

Isn't it just x = ⁸√121

[@Noname-rc8uc]

It's x= +- ⁸√121, but only if you're looking for real solutions

[@Turtol.1]

x= +-⁸√121 yes, but only on the real number line. There are answers for this equation unaccounted for in the real numbers that lay in the complex plane. The question in the video is slightly misleading. I believe that either the full question or the course would provide more context. In order for the student to understand they need to solve this question in the complex plane

[@puffinbasher]

@@Turtol.1 If your in the real space, typically the power function would need a +- type term, and represents a single number. In the complex space the +- term in front of a root does not do the same thing - for example, a sqrt(i) would be (1+i)/sqrt(2) and (-1-i)/sqrt(2), but +-sqrt(i) would actually have FOUR results, two of which when squared give negative i! For this reason, when in a complex or hypernion system, @shdz5954 is actually correct (weather he knew this or not)

[@dewah7775]

@@shdz5984 He tried to get all solutions, this is a single solution. He was wrong btw, and if he had just applied representation of numbers in polar coordinates and the fundamental theorem of algebra, then he could have gotten a general solution much quicker.