Finding rational roots to an equation (univariate) [closed]

Question

So I'm using Mathematica 7 for Students and I'm using the Root function to find the roots of an equation over the rational numbers:

$$225x^6 -343 = 0$$

I created this equation from trying to prove that the number $\frac{\sqrt7}{\sqrt[3]15}$ is irrational or rational, and using the rational roots theorem, I came up with 24 possible roots that I don't want to test out manually. Using Root gives me the following output for some reason:

Root[225*x^6 - 343 == 0, x]
Root::nup: -343 + 225x^5 is not a univariate polynomial >>

How is this polynomial not univariate? I'm pretty sure it is, since it only has $x$ in it. Also, another gem occurs when I try the same thing with $x^2 = 1$:

Root[x^2 - 1 == 0, x]
Root::nup: x^2 - 1 == 0 is not a univariate polynomial >>

What is going on?

Answer 1

Use Solve, not Root.

Solve[225*x^6 - 343 == 0, x]

(* ==>
{{x -> -(-(1/15))^(1/3) Sqrt[7]}, {x -> (-(1/15))^(1/3) Sqrt[
    7]}, {x -> -(Sqrt[7]/15^(1/3))}, {x -> Sqrt[7]/15^(
   1/3)}, {x -> -(((-1)^(2/3) Sqrt[7])/15^(1/3))}, {x -> ((-1)^(2/3)
     Sqrt[7])/15^(1/3)}}
 *)

Root is for representing roots of polynomials, not for computing them. Also, the syntax you used is simply incorrect (please check the docs for Root). The correct syntax to represent the 1st root of this equation would be:

 Root[225*#^6 - 343 &, 1]

You'll notice that this doesn't give a form in radicals automatically, it just returns unevaluated. Use ToRadicals to do the conversion.