How to Think Outside the Box? | Nice Algebra problem | (Math Olympiad Preparation)

[@htunwynn7523]

Very nice solving.

[@PreMath]

Thank you! Cheers!

[@XLatMaths]

I used complex numbers. Solving using QF, x = √3/2 ± i/2 which is e^±iπ/6 in its principal form, so x¹⁰⁰⁰ = e^500iπ/3. 500 is congruent to 2 (mod 6) so x¹⁰⁰⁰ is e^2iπ/3. X¹⁰⁰⁰ + 1/x¹⁰⁰⁰ is therefore 2cos(2π/3) which is -1.

[@almosawymehdi3416]

I did the same way, two minutes to solve this problem

[@talhahaque-sq2bc]

another answer of this question is -2

[@jamesharmon4994]

The way I would have attacked this is multiplying the given statement through with "x". This then results in x^2 + 1 = sqrt(3)x. Then, put everything on the left hand side gives x^2 - sqrt(3)x + 1 =0. Then use quadratic formula to find x. Then substitute that value into x^1000 and 1/x^1000.

[@geyorge6573]

bro have u tried solving the quadratic theres no real solutions

[@jamesharmon4994]

@@geyorge6573 so?? When you take them to even powers, it becomes real.

[@rajbinani]

I took same approach and did it in a minute without pen and paper. What’s wrong with this approach?? Works very well imo. Not sure why such a complicated approach in the video though

[@aminimam5118]

Just square both sides X+1/x = sqrt (3). You get x2+1/x2 =1. Square again, you get x4+1/x4=-1. Square again, you get X8+1/x8=-1. No matter how many times you square, the answer will be -1. Since 1000 is an even number, the answer is -1.

[@cosmolbfu67]

Yes but if equation is x^998 + 1/x^998 =? solution = 0

[@sunil.shegaonkar1]

X^4 + 1/ X^4 = -1, and further squaring this expression does not change. Because 1000 is divisible by 4, the 1000 powered expression is also minus 1.

[@alexgorman4864]

There is another way of doing it.. when you solve for x in the initial equation you can find that x= cost + isint , where t= 60 deg. then using De moiver's theorem x^1000 = cos(1000t) + isin(1000t) and 1/x= cost-isint so 1/(x^1000) = cos(1000t)-isin(1000t) . Hence sin terms cancelled and finally it becomes 2cos(1000t) =-1

[@mahinnazu5455]

Wow...thats great solution... Thank u my favourite Sir

[@bigm383]

Fabulous, thanks Professor!❤

[@paulortega5317]

As you increment the exponent, the values repeat after 12 terms. 1000 module 12 = 4. The 4th value in the series is -1.

[@Copernicusfreud]

I came up with the answer another way. If I did something wrong, please feel free to address it and provide a correct solution. Thanks. I worked with the first equation of (x + 1/x) = sq rt(3). I squared both sides and ended up with x^2 + 1/x^2 = 1. I took this equation and squared both sides and ended up with x^4 + 1/x^4 = -1. I took this equation and squared both sided and ended up with x^8 + 1/x^8 = -1. The pattern should repeat until reaching x^1000 + 1/x^1000, resulting in -1 as the answer. So, the answer is -1.

[@Duke_Of_Havoc]

Except this pattern won't reach 1000 since 1000 isn't a pure multiple of 2. The pattern you are following is 2,4,8,16,32,64 and so on. It won't reach 1000 as you can see.

[@ravikrpranavam]

Very well explained

[@manojkantsamal4945]

Respected sir, You are one of the best

[@advancedintention7169]

Thank you very much sir......

[@Roberto74B]

i did to consider 1/x as X/X^2 ... so X^3+X=X^2*sqrt_2(3) ..... next x^6+x^2+2x^4 = 3x^4 .... next x^6-x^4+x^2= 0 .... put Y=X^2 ... come out y^3-Y^2+1=0 ... one real solution and two immaginary solutions ... and at the end i did consider Y^500 + Y^-500 = ...... but the result don't is not the same i did to be a bit not convinced about to have, you, to arrive to define x^1000=-1/x^2 before you did flip the equation ... and next of the flip (nominator with the denominator) you have used the previous x^1000=-1/x^2 to continue on resolving ... maybe it's super correct ... but maybe no? ... you are a really super professor, my compliments, but in my mind there is something that retain it a remote not possible right logic ... have a nice day

[@hichamitani6433]

Amazing problem and solution Superb

[@henrbenbrown]

Fabulous!...thank you so much!!

[@dirklutz2818]

Tremendous!

[@vcvartak7111]

It's good you are also talking algebraic problems

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!

[@walidbinsiddik]

Yeah afcource i should know 6*167=1002

[@mvrpatnaik9085]

Nice Explanation

[@alster724]

Nice solution

[@md.shahadathossen5738]

That's cool !

[@IlyaNickkhah-bz1ii]

How x to the power of an even number (2k = 6) is equal to a negative number ( -1) ? X^6 is not equal to -1 or I have neglected sth and made a mistake so please correct me if I'm wrong

[@talhahaque-sq2bc]

Is the alternative answer (-2)

[@richardlollar7751]

X to the 6th equals a negative one? Are we including complex numbers?

[@RickDesper-v8z]

For x > 0, (x + 1/x) >= 2. So you have to include complex numbers.

[@JamesDavy2009]

In short, yes. Even so, with up to six complex roots that x could equal to, certain operations can yield real solutions such as i^2 = -1.

[@lavoiedereussite922]

Thank

[@NITINCHAUHA62]

Sir make a video on Mathematics facts for us ❤please🙏🙏 ..... In addition I like u ❤❤❤

[@kalpanadeka8226]

I solved it in my mind 😊

[@cameronweiss4044]

Same I learned this in 3rd grade

[@kalpanadeka8226]

@@cameronweiss4044 Noice

[@brendanmclearie4278]

Why can't I save it?

[@राजनगोंगल]

👍👍👍👍👍👍

[@j.c4007]

😮 how x⁶ can be equal to : -1. Even x is an odd number , the result is not an odd number ?

[@sohampinemath1086]

It has complex roots

[@lk-wr2yn]

Is that possible x^6= never negativ? So -1 as result is not possible?

[@JamesDavy2009]

x^6 = -1 means that x equates to a complex number.

[@lk-wr2yn]

@@JamesDavy2009 Thanks, I never saw this in school...

[@JamesDavy2009]

@@lk-wr2yn Complex numbers tend to be taught at the university level or upper-level high school if you're lucky.

[@howareyou4400]

@@JamesDavy2009 Wait, there are regions in the world that does NOT teach complex number in high school?

[@JamesDavy2009]

@@howareyou4400 Apparently. This was based on my experience. I did learn calculus and linear algebra in high school.

[@ultimatedoug2227]

Min value of x + 1/x is 2 so that original thing doesnt make sense

[@webscreener8797]

x^6= -1 impossible?!?

[@manojitmaity7893]

When x equals to i, x⁶ may be equal to -1.

[@LogintoMaths]

Its complex number

[@e65666]

I was doing the right thing the first time then decided to brute force it

[@pralhadraochavan5179]

Good morning sir

[@LogintoMaths]

I solved in 1 min

[@nantesloire]

x⁶ = -1 ???? and x +1/x =3½ never

[@28.quachphuc26]

Is there any other possible answer: I got 1

[@saumyaadhikari8078]

-1. Two minute job 😊😊

[@sattkrit_pathak]

x^6 = -1 ? they dont usually give imaginary numbers in there exams

[@rakeshprasad7313]

It's not a imaginary no 😮

[@RickDesper-v8z]

I think this presentation is intended to show that you can solve the problem only using basic algebra. But it takes a fairly long path to do so, and you have to pull ideas out of thin air.

[@Wahid7X]

I solve this by a pattern

[@utsabrajsarkar7776]

Too much struggle for a trivial problem. x+1/x=√3 => x²+1=√3x => x²-√3x+1=0. It's a polynomial with real coefficients and discriminant 3-4 x=z=cos t+I sin t for some t. Now, x+1/x=√3 yields cos t=√3/2 => t=±π/6 so, x^1000+1/x^1000=2 cos (1000π/6) the rest is easy.

[@matteogorghetto]

This is how i solved it. One thing though, The assumptions should have included that x can belong to the realm of complex numbers, usually, one assumes that x belongs to realm of REAL numbers, in which case there's no solution. Given details are not accurate on here.

[@ploi4590]

There is no such x!

[@itzrealzun]

No REAL x, but COMPLEX x exists

[@ploi4590]

If that is what you assume, then the answer can be found without using the algebra trick.@@itzrealzun

[@mohammadmatinafshordidibaz9966]

X^6=-1 it's not true

[@RickDesper-v8z]

It's not real. Which isn't the same thing. It's a non-real solution.

[@JamesDavy2009]

The fundamental aspect of a complex number lies in the imaginary unit i (j to engineers) such that i^2 = -1. The Mandelbrot Set uses complex numbers to trace out its shape.

[@lavoiedereussite922]

Thank