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.
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.
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
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.
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
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.
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.
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
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
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.
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.
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.
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.
Very nice solving.
Thank you! Cheers!
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.
I did the same way, two minutes to solve this problem
another answer of this question is -2
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.
bro have u tried solving the quadratic theres no real solutions
@@geyorge6573 so?? When you take them to even powers, it becomes real.
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
Very well explained
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.
Wow...thats great solution...
Thank u my favourite Sir
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.
Yes but if equation is x^998 + 1/x^998 =?
solution = 0
Fabulous, thanks Professor!❤
Respected sir, You are one of the best
Amazing problem and solution
Superb
Thanks for video.Good luck sir!!!!!!!!!
Thank you very much sir......
Fabulous!...thank you so much!!
Nice Explanation
Tremendous!
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
Nice solution
It's good you are also talking algebraic problems
As you increment the exponent, the values repeat after 12 terms. 1000 module 12 = 4. The 4th value in the series is -1.
That's cool !
Thank
Yeah afcource i should know 6*167=1002
Is the alternative answer (-2)
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.
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.
I solved it in my mind 😊
Same I learned this in 3rd grade
@@cameronweiss4044 Noice
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
👍👍👍👍👍👍
Sir make a video on Mathematics facts for us ❤please🙏🙏 ..... In addition I like u ❤❤❤
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
X to the 6th equals a negative one? Are we including complex numbers?
For x > 0, (x + 1/x) >= 2. So you have to include complex numbers.
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.
Why can't I save it?
Good morning sir
I was doing the right thing the first time then decided to brute force it
x^6= -1 impossible?!?
When x equals to i, x⁶ may be equal to -1.
Its complex number
Is that possible x^6= never negativ? So -1 as result is not possible?
x^6 = -1 means that x equates to a complex number.
@@JamesDavy2009 Thanks, I never saw this in school...
@@lk-wr2yn Complex numbers tend to be taught at the university level or upper-level high school if you're lucky.
@@JamesDavy2009 Wait, there are regions in the world that does NOT teach complex number in high school?
@@howareyou4400 Apparently. This was based on my experience. I did learn calculus and linear algebra in high school.
😮 how x⁶ can be equal to : -1.
Even x is an odd number , the result is not an odd number ?
It has complex roots
Is there any other possible answer: I got 1
x⁶ = -1 ???? and x +1/x =3½ never
-1. Two minute job 😊😊
x^6 = -1 ? they dont usually give imaginary numbers in there exams
It's not a imaginary no 😮
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.
I solved in 1 min
I solve this by a pattern
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.
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.
There is no such x!
No REAL x, but COMPLEX x exists
If that is what you assume, then the answer can be found without using the algebra trick.@@itzrealzun
X^6=-1 it's not true
It's not real. Which isn't the same thing. It's a non-real solution.
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.
Thank