Solving a Maple Mystery: Why Does x=root(x^2+1)/x=.5root(2+root(20))?

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Homework Help Overview

The discussion revolves around understanding why Maple software outputs the equation x = √(x² + 1) / x as x = 0.5√(2 + √20). Participants are exploring the algebraic manipulation and reasoning behind this result.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting various algebraic methods, including squaring both sides and rearranging terms. Some are questioning the interpretation of the problem and the steps taken to derive the final expression.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing partial workings and others expressing uncertainty about specific steps. A substitution method has been mentioned, indicating a potential direction for further exploration.

Contextual Notes

One participant questions whether a value has been stored in the variable x, which may affect the interpretation of the problem. There is also a mention of using the ABC formula for solving a quadratic equation derived from the original expression.

fredrick08
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Homework Statement



can someone please tell me why maple says that x=root(x^2+1)/x=.5root(2+root(20))??

ive tried trig and rearranging, but i have idea why this is so..
 
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Have you got a value stored into the variable x?

EDIT: Oh i think i may have misinterpreted the problem.

Heres some working:
http://img4.imageshack.us/img4/8552/algebra.gif

Probably is a quicker way, but these are the steps i took.
 
Last edited by a moderator:
fredrick08 said:

Homework Statement



can someone please tell me why maple says that x=root(x^2+1)/x=.5root(2+root(20))??

ive tried trig and rearranging, but i have idea why this is so..

Square both sides, and you get a quadratic in x2. Solve.

[tex]\begin{align*}<br /> x &= \frac{\sqrt{x^2+1}}{x} \\<br /> \intertext{Hence}<br /> x^2 &= \sqrt{x^2 + 1} \\<br /> x^4 &= x^2 + 1 \\<br /> x^2 &= \frac{1 \pm \sqrt{5}}{2} \\<br /> \intertext{(Must be positive, so dismiss the negative part)}<br /> & = \frac{2 + \sqrt{20}}{4} \\<br /> \intertext{Hence}<br /> x & = \pm 0.5 \sqrt{2 + \sqrt{20}} \\<br /> &= \pm \sqrt{0.5 + \sqrt{1.25}}<br /> \end{align*}[/tex]​
 
thankyou very much, I am still unsure how you get from x^4 to x^2=2+root(20)/4.. but i know its right so thx
 
He uses a substitution p=x^2 and then solves p with the ABC formula.
 
ok yes thankyou
 

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