Newton's Method (problem solving)

[mateomy]

Homework Statement

Its not an actual problem, just a question. We're being lectured on Newton's Method, which I understand. But there's a section in the book (and related problems later in the chapter), that ask for you to find (2)^1\6 correct to eight decimal places. It goes on to say that it is equivalent to finding the positive root of the equation and breaks it down to x^6 - 2 = 0. I don't remember finding roots from my previous classes. I know it probably has something to do with variation in sign (Descartes?). Again, this isn't a problem, but can somebody just briefly explain to me the "this is equivalent to finding the positive root" part?

THANKS!

 

[Char. Limit]

Let's say that 2^(1/6) = x (which is true, because we define x that way. Then...

2^(1/6)=x
2=x^6
0=x^6-2

So you can see that the root of x^6-2 would be 2^(1/6). Then you use Newton's method to find the root.

 

[mateomy]

Got it. Thanks for taking the time to explain.

 

[Mark44]

The equation x⁶ - 2 = 0 (or equivalently, x⁶ = 2) has six roots. Four of them are complex and two of them are real, with one positive and one negative. Each of the roots of this equation is a sixth root of 2, or $$\sqrt[6]{2}$$