A Nice Polynomial Equation | Math Olympiads

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[Limbo]

I solved it by dividing everything with x^3 and substituting u=x+1/x

[pwmiles]

Multiply by (x^2 - 1)
x^8 - 1 = 0
x^8 = 1
x = e^(2 n pi i / 8) = e^(n pi i / 4)
n = 0, 1, 2, 3, 4, 5, 6, 7 are the distinct roots. However n=0 and n=4 give x=+/-1 which are not roots of the original equation -- they are roots of x^2 - 1 = 0 and are excluded.

[nasrullahhusnan]

x⁶+x⁴+x²+1=0
By letting y=x² it becomes 1+y+y²+y³=0, which is sum of 4 elements of geometric series: 1, y, y², y³. The sum is
1+y+y²+y³=(1-y⁴)/(y-1)
As the sum is 0 then y⁴-1=0
(y²+1)(y+1)(y-1)=0
giving y=±1 or y=±i
For y=±1 --> x²=±1 --> x≈±1 or x=±i
For y=±i --> x²=±i
x=±sqrt(i) or x=±sqrt(-i)

[RexxSchneider]

Let y = x^2. So y^3 + y^2 + y + 1 = 0. Using RRT, y = -1 is a solution. So divide through by (y+1) and we get (y+1)*(y^2+1) = 0. That gives y = -1 or y^2 = -1.
So the three roots of the cubic are y = -1 and y = ±i. Now x = √y and that gives us x = ±i or x = (±1 ±i) /√2.

[TheMoogleKing]

It's a geometric series with ratio x^2, first term 1 and 4 terms, giving (1-x^8)/(1-x^2) = 0, so it's the 8th roots of unity but not 1 or - 1.

[ManjulaMathew-wbzn]

Multiply the original equation by x^2-1 as x=+/-1 doesn’t satisfy the equation.
It simplifies to x^6-1=0
Find the sixth roots of 1 excluding 1 and-1.
x=e^nPIi/3 where n=1, 2, 4 and 5.

[misterdubity]

I did method 2 in my head and converted the eqn to x^8 - 1 = 0, so the solution is the 8 roots of unity except then delete 1 and -1 since those were introduced by multiplying by (t-1) or (x^2-1).

[yoshinaokobayashi]

x^2=-1 is one answer then given equation is (x^2 +1 )(x^4 +1)=0 then x^4=-1, answers are x= exp( +- pi()/2), exp( +- pi()/4), exp( +- 3*pi()/4).

[_ilsegugio_]

multiply everything by (x^2-1) and exclude solutions for x^2-1=0

[GoUtes]

I haven't taken a math class in a while but when I saw this I was like "This looks impossible unless we start talking about imaginary numbers". The only reason I came to that realization is because I noticed the exponents are all even numbers, meaning that plugging in a real number doesn't satisfy the equation.

[Qermaq]

Fun! If you do a lecture video on C, be sure to include when we want to consider the non-principal root and when we do. I think most people are confused on that. I know I often am.

[groscolisdery]

if x is solution, then -x is also solution
let y=x^2
a quick guess is x=I

y^3+y^2+y+1=0
or
y^2*(y+1)+(y+1)=0

(y^2+1)*(y+1)=0

y ={-1, i, -i}

therefore

x={ i, e^πi/4, e^-πi/4, -i, -e^πi/4, -e^-πi/4 }

[damyankorena]

Multiply by x^2-1
Then just don't consider the numbers 1 and -1

[Nameless-qehu]

I solved this one by factoring by difference of squares several times (using double negatives and knowing off the top of my head what i^1/2 is)

[bobbyheffley]

All roots of this equation are complex numbers.

[scottleung]

My answers are a little different from yours. I got +-i, +-sqrt(i), and +-i*sqrt(i). Are those still correct?

[Achill]

substitute u=x^2
u=-1 is a solution
divide (u^3+u^2+u+1)/(u+1)
= u^2+1
which has roots i and -i for u
x = i, -i, (1+i)/2, (1-i)/2,
(-1+i)/2, (-1-i)/2

[mcwulf]

I took alternate terms rather than pairs to get the same factorisations. Then I saw they were complex roots and lost interest 😂😂😂

[giuseppemalaguti]

x1=i, x2=-i, x3=sqrt(i), x4=-sqrt(i), x5=sqrt(-i), x6=-sqrt(-i)

[hassanlabraiki]

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