This problem becomes much simpler if you notice that both numerator and denominator are sums of terms in geometric progression: x^7+x^5+x^3 = x^3 * (x^4+x^2+1) = x^3 * (x^6-1)/(x^2-1) as x = ± 1 are not roots of the initial equation Similarly, you have x^6+x^5+x^4 = x^4 * (x^2+x+1) = x^4 * (x^3-1)/(x-1) Then factor out x^6-1 as a difference of two squares: (x^3-1)(x^3+1), then proceed to group terms with the same factor in common. Do not cancel out factors directly, since you can get some solutions to be missing at the end.
Do long division and get x -1 + (1/x) = 3 immediately. After solving the quadratic, rewrite this equation as (x^2 +x+1)(x^2-4x+1)=0 and find the two additional complex roots.
Actually, it only has four roots, as by its construction, the triple root of x = 0 must be excluded, otherwise the lefthand side of the equation would be undefined. Two of them are complex and two of them are real.
This problem becomes much simpler if you notice that both numerator and denominator are sums of terms in geometric progression:
x^7+x^5+x^3
= x^3 * (x^4+x^2+1)
= x^3 * (x^6-1)/(x^2-1)
as x = ± 1 are not roots of the initial equation
Similarly, you have
x^6+x^5+x^4
= x^4 * (x^2+x+1)
= x^4 * (x^3-1)/(x-1)
Then factor out x^6-1 as a difference of two squares: (x^3-1)(x^3+1), then proceed to group terms with the same factor in common.
Do not cancel out factors directly, since you can get some solutions to be missing at the end.
Do long division and get x -1 + (1/x) = 3 immediately. After solving the quadratic, rewrite this equation as (x^2 +x+1)(x^2-4x+1)=0 and find the two additional complex roots.
Nice but the pen kept scraping against the paper like nails on a chalk board, so was kind of hard to watch.
This equation has four roots. It would have seven roots, but by its construction, the triple root of x = 0 must be excluded.
When X⁷ + X⁵ + X³ / X⁶ + X⁵ + X⁴ = 81/27 , Then X1 ≈ 0.267949 and X2 ≈ 3.73205
you made my wife cry with this solution, and she wasn't happy with it
На икс в пятой надо было сокращать, тогда сразу бы получили кваратное уравнение относительно x+1/x
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Nice approach but some roots have been cancelled since you directly divide them
So there should be 7 roots instead of 2
Actually, it only has four roots, as by its construction, the triple root of x = 0 must be excluded, otherwise the lefthand side of the equation would be undefined. Two of them are complex and two of them are real.
@@davidbrisbane7206 Thanks for clarifying
divide numerator and denominator by x⁵
(x² + 1/x² + 1)/(x + 1/x + 1) = 3
x + 1x = u => x² + 1/x² = u² - 2
(u² - 1)/(u + 1) = 3
3(u + 1) = (u + 1)(u - 1)
(u + 1)(u - 4) = 0
u = -1 ∨ u = 4/3
u = -1
x + 1/x = -1
x² + x + 1 = 0 => complex roots
u = 4
x + 1/x = 4
x² - 4x + 1 = 0
x = (4 ± 2√3)/2
*x = 2 ± √3*
That's what I did, but u = -1 does not give any solutions (not even complex solutions) because that would result in dividing by zero
@@armacham yes. You are right.
x = 0 isn't a solution, t = x+1/x, (x^2+1+1/x^2)/(x+1+1/x) = 3, x^2+1/x^2+2 = (x+1/x)^2 = t^2, (t^2-1)/(t+1) = 3, t-1 = 3 and t+1 ≠ 0, t = 4, x+1/x = 4, x^2-4x+1 = 0, x = 2+-sqrt(3).
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curious where are you broadcasting from ?
I guess india or pakistan
Tanzania
i