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

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To solve for \( x + \frac{1}{x} \) given \( x^2 + \frac{1}{x^2} = 23 \), the discussion emphasizes squaring the expression \( x + \frac{1}{x} \). It is established that \( \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \). By substituting the known value, the equation simplifies to \( \left(x + \frac{1}{x}\right)^2 = 25 \). Therefore, \( x + \frac{1}{x} \) equals either 5 or -5.
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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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