An Interesting Rational Equation

[@tetramur8969]

The solution isn't finished. There are six complex roots of this equation, given by x = cos(2πn/7)+i*sin(2πn/7) for n = 1 to 6

[@raghvendrasingh1289]

👍

[@kinshuksinghania4289]

Absolutely spot on

[@moeberry8226]

You’re not finished either even with you finding 6 roots of unity since x=1 is discarded because it was brought but Syber multiplying by 1-x. You still have to check the denominator of 1+x^4 that it’s not zero. And if one of those sixth roots of unity makes it zero you will also have to discard it.

[@tetramur8969]

@@moeberry8226 Well, when we substitute x^4 = -1 into this sextic we get x^3 = 0 which gives us x = 0, but 0 is a root of neither 1+x^4 = 0 nor 1+x+x^2+x^3+x^4+x^5+x^6 = 0, so I can't exclude any of six roots I've found

[@moeberry8226]

@ I know that but you still have to show work because most people think this problem is a polynomial but it’s not.

[@pwmiles56]

The root x=1 is introduced by multiplying through by (x-1), so not necessarily a solution.

[@dan-florinchereches4892]

(1+x^4)(1+x^2)(1+x)=(1+x^4)(1+x+x^2+x^3)=1+x+x^2+x^3+x^4+x^5+x^6+x^7=1 x(1+x+x^2+x^3+x^5+x^6)=0 so x=0 or 1+...+x^6=0 by multiplying by x-1 x^7-1=0 x^7=1 so the other roots are all seventh roots of unity except 1 which we introduced so all are complex

[@iabervon]

I prefer to stay ahead of the game and check if there are solutions where the expression I'm multiplying or dividing by is zero. Before multiplying by x-1, check whether x=1 is a solution, and before dividing by x, check whether 0 is a solution. Then you can say it's true if x=0, false if x=1, otherwise, x⁷=1. Then when you find that 1 is a root, you just exclude it as not in the domain of values you're still searching.

[@scottleung9587]

I also got x=0 as the only solution, while the hexic equation I attained after cross-multiplying produces 6 complex solutions.

[@aarona3144]

Math being "edible" reminds me of an old game I would play on computers at school called "Number munchers". it was a lot of fun.

[@ChristopherEvenstar]

Yay! Thank you for showing us that tool. I did the, can x be negative? can x be positive? interrogation. I think your method might be more useful.

[@SyberMath]

You are very welcome! 😁

[@ismaelperbech]

We have the following equation: (1+x)(1+x^2)(1+x^4) = 1 Given that x doesn't solve x^4 = -1 which in this case it isn't Now, noticing that we are including the solution x = 1, we multiply both sides by (1-x). We get on the left hand side that: (1-x)(1+x)(1+x^2)(1+x^4) = (1-x^2)(1+x^2)(1+x^4) = (1-x^4)(1+x^4) = (1-x^8) So 1 - x^8 = 1 - x which is easy to solve: x^8 - x = 0 x(x^7 - 1) = 0 On the one hand x = 0 solves the equation. On the other hand, the seventh complex roots of unity also solves the equation but in this case we must exclude the 1 because it is not a solution. So we get, x = 0 and x = e^((2π + 2πk)/7) with k ranging from 1 to 6, giving us a total of 7 roots which corresponds to the degree of our initial polynomial.

[@ismaelperbech]

I missed an i in the exponent :)

[@mcwulf25]

Can't have x=1 because it arises from the (x-1) factor used to simplify the equation.

[@holyshit922]

Multiply by x-1 and you have x*(x^7-1) x^7=1 we can use 7th roots of unity

[@rakenzarnsworld2]

x = 0