Find the Solution to $x^2+\frac{1}{x^2}=23$: x+$\frac{1}{x}$

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Discussion Overview

The discussion revolves around the problem of finding the value of \(x + \frac{1}{x}\) given that \(x^2 + \frac{1}{x^2} = 23\). Participants explore the relationship between these expressions and consider how to manipulate them mathematically.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant asks how to begin solving the problem of finding \(x + \frac{1}{x}\) from the given equation.
  • Another participant suggests squaring \(x + \frac{1}{x}\) as a potential approach.
  • There is a discussion about the expression \(\Bigl(x + \frac{1}{x}\Bigr)^2\) and its expansion, with one participant noting it equals \(x^2 + 2 + \frac{1}{x^2}\).
  • Participants mention that \(x^2 + \frac{1}{x^2}\) can be used to find \(x + \frac{1}{x}\) by rearranging the squared expression.
  • One participant expresses difficulty in deriving an answer despite the suggestions provided.

Areas of Agreement / Disagreement

Participants generally agree on the method of squaring \(x + \frac{1}{x}\) and its relationship to \(x^2 + \frac{1}{x^2}\). However, there is some confusion regarding the terminology used, such as referring to the squared expression as a "sum of two squares." The discussion does not reach a consensus on the final answer.

Contextual Notes

There are unresolved aspects regarding the manipulation of the expressions and the assumptions made in the calculations. Some participants express uncertainty about the steps leading to the final result.

mathlearn
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If $x^2+\frac{1}{x^2}=23$. Find $x+\frac{1}{x}$

Any Ideas on how to begin? (Happy)
 
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mathlearn said:
If $x^2+\frac{1}{x^2}=23$. Find $x+\frac{1}{x}$

Any Ideas on how to begin? (Happy)
What happens when you square $x+\frac{1}{x}$ ?
 
Opalg said:
What happens when you square $x+\frac{1}{x}$ ?

Hey (Wave) Opalg,

Thank's for the catch , It's a sum of two squares

$\Bigl(x+\frac{1}{x}\Bigr)^2.=\Bigl(x+\frac{1}{x}\Bigr) \Bigl(x+\frac{1}{x}\Bigr).$
 
mathlearn said:
Hey (Wave) Opalg,

Thank's for the catch , It's a sum of two squares

$\Bigl(x+\frac{1}{x}\Bigr)^2.=\Bigl(x+\frac{1}{x}\Bigr) \Bigl(x+\frac{1}{x}\Bigr).$
Yes, of course, \left(x+ \frac{1}{x}\right)^2= \left(x+ \frac{1}{x}\right)\left(x+ \frac{1}{x}\right) but that is not the "sum of two squares"!

What did you get as an answer for this problem?
 
HallsofIvy said:
Yes, of course, \left(x+ \frac{1}{x}\right)^2= \left(x+ \frac{1}{x}\right)\left(x+ \frac{1}{x}\right) but that is not the "sum of two squares"!

What did you get as an answer for this problem?

Hey (Wave) HallsofIvy,

Still I am unable to derive an answer..(Happy)(Smile)
 
Last edited:
mathlearn said:
Hey (Wave) Opalg,

Thank's for the catch , It's a sum of two squares

$\Bigl(x+\frac{1}{x}\Bigr)^2.=\Bigl(x+\frac{1}{x}\Bigr) \Bigl(x+\frac{1}{x}\Bigr).$

Opalg said:
What happens when you square $x+\frac{1}{x}$ ?
Have you tried Opalg's suggestion? Multiply it out and see what you get.

-Dan
 
Opalg said:
What happens when you square $x+\frac{1}{x}$ ?

I think this is what you meant originally,

$\Bigl(x+\frac{1}{x}\Bigr)^2=x^2+2+\frac{1}{x^2}$

Now rearranging,

$\Bigl(x+\frac{1}{x}\Bigr)^2=x^2+\frac{1}{x^2}+2$

It is given that $x^2+\frac{1}{x^2}=23$

$\Bigl(x+\frac{1}{x}\Bigr)^2=23+2$

$\Bigl(x+\frac{1}{x}\Bigr)^2=25$

Many Thanks :rolleyes:
 

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