Can you solve this If x^2+x+1=0 then Find x^1999+x^2000=? | Olympiad Math Question!

Can you solve this If x^2+x+1=0 then Find x^1999+x^2000=? | Olympiad Math Question!
#challengingmathproblem #matholympiad

[tamle]

This is the wrong approach because x = 1 does not satisfy the expression x^2+x+1=0

[prabhudasmandal]

After finding x^3=1, one should not infer that x=1, because it doesn't satis fy the given expression. So, use x^3=1 to resolve the problem. Thanks.

[adityaprakash]

Multiplying it by x-1, we have generated an extra root. Infact, we should choose a different approach. The root of that expression is omega(w) and omega w^2. Now 1+w+w^2 = 0. Use it and get the answer eaisly.

[jfcrow]

Complex method. Roots e^(+/- 120 degree i ) =x get same answer.

[mputuchimezie]

Great 👌 solution
What motivated you to multiply both sides by (x-1)
?

[SladeMacGregor]

Thank you Route to Mathematics! Very nice! Good explaination.

[JSSTyger]

Yes. I solved x = [-1±sqrt(3)]2 or [-1/2]±i[sqrt(3)/2]. This gives you x = e^(iz) for some angle z (we find it to be 2π/3+2kπ or 4π/3+2nπ). We know that x^1999+x^2000 = x^1999(1+x) and from the original information, (x+1) = -(x²) and so x^1999(1+x) = -(x^2001). The final answer will be -[[e^(iz)]^2001] = -e^(2001iz). This value turns out to yield even integer multiples of π and so, in both cases it reduces to -e^(0) = -1.

[ricardodamasceno]

Simply gorgeous and creative resolution. Naturally, they're obtained two complex numbers like original equation's roots. Therefore, it seems to be tottaly justified the final value (-1). Not rarely, the more an algebraic expression looks like in simplest form (apparently), the more the resolution method must be sophisticated.

[johngreen]

Instead of multiplying both sides by x-1 just solve for x^2 and multiply both sides of the equation by x, this is legitimate since x does not equal zero
x^3=x*x^2=x(-x-1)=-x^2-x=1 so this gives us the desired equation x^3= 1.
It seems simpler to me.

[themathsgeek]

Great video, thanks! This is a classic problem regarding cube roots of unity and their properties

If x = 1 then x^2 + x + 1 = 3; Is it true ?
Thanks, Miss

[arnoldvillastique]

The value x =1 does not satisfy the given quadratic equation..

[HemRamachandran]

It is easy to do is you know what need to be done.

[rudraraj]

It is a mistake ... If x=1, then x^2+x+1=0 does not give a real root, where, 3 is not equal to zero... You are giving a paradox output...

[佐藤広-cp]

When I watch a similar video, I always wonder why anyone solves the problem, expressing x in polar form and using De Moivre's theorem. That is, x^2+x+1=0 ⇒ x=[-1±√{1^2-4*1*1}]/2={-1±√(1-4)}/2={-1±√(-3}/2=(-1±i√3)/2, and we use De Moivre's theorem on the unit circle in the complex plane. (-1±i√3)/2 is expressed in polar form, x=cos(±2π/3)+i*sin(±2π/3), so In other words, we can see that x makes one revolution by the cube. Even if the multiplier of x increases, you can understand that it only turns on the unit circle, then the calculation will be the same as the video.

[muthukumarasamymuthukumara]

X³=1 So cube route of X=1. How the answer is -1. I don't understand.

[pec]

You multplied by x-1 but x-1=0 it's à mistake...