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

[mathlearn]

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

Any Ideas on how to begin? (Happy)

 

[Opalg]

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 $x+\frac{1}{x}$ ?

 

[mathlearn]

What happens when you square $x+\frac{1}{x}$ ?

Hey (Wave) Opalg,

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

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

 

[HallsofIvy]

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?

 

[mathlearn]

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)

 

[topsquark]

Hey (Wave) Opalg,

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

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

What happens when you square $x+\frac{1}{x}$ ?

Have you tried Opalg's suggestion? Multiply it out and see what you get.

-Dan

 

[mathlearn]

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