Equation

[123]

Hello! I need help with this one:
$\displaystyle \mathit{\frac{x^{2}+1}{x}+\frac{x}{x^{2}+1}=\frac{29}{10}}$
I try with something like that to do: $\displaystyle \mathit{\frac{(x^2+1)(x^2+1)+x^2}{x(x^2+1)}}=\frac{29}{10}$
and ...
But it's not work for me. :?:

 

[Deleted member 4993]

Hello! I need help with this one:
$\displaystyle \mathit{\frac{x^{2}+1}{x}+\frac{x}{x^{2}+1}=\frac{29}{10}}$
I try with something like that to do: $\displaystyle \mathit{\frac{(x^2+1)(x^2+1)+x^2}{x(x^2+1)}}=\frac{29}{10}$
and ...
But it's not work for me. :?:

What do you need to do? solve for "x"?

What is exactly meant by "not working for me"?

Does it mean you are not getting the correct answer?

Or you cannot proceed beyond that step?

 

[lookagain]

Hello! I need help with this one:
$\displaystyle \mathit{\frac{x^{2}+1}{x}+\frac{x}{x^{2}+1}=\frac{29}{10}}$
I try with something like that to do: $\displaystyle \mathit{\frac{(x^2+1)(x^2+1)+x^2}{x(x^2+1)}}=\frac{29}{10}$
and ...
But it's not work for me. :?:

User 123, I wouldn't try to work with the resulting fourth degree equation that would come from your correct
later (last) equation, so I did the following.


A possible suggestion:

$\displaystyle Let \ z \ = \ \frac{x^2 + 1}{x}. \ \ Then \ \ \frac {x}{x^2 + 1} \ = \ \frac {1}{z}.$

$\displaystyle Solve\ \ z \ + \ \frac{1}{z}\ = \ \frac{29}{10}\ \ by \ converting\ it \ into \ a \ quadratic \ equation.$
$\displaystyle \ Then \ substitute \ those\ z-values \ back \ into \ the\ equation\ \ z \ = \ \frac{x^2 + 1}{x},$

$\displaystyle and \ solve \ for \ the \ appropriate \ x-values.$

 

[123]

What is exactly meant by "not working for me"?

Does it mean you are not getting the correct answer?

Or you cannot proceed beyond that step?

Sorry, my english isn't good.

lookagain thanks. I solve it. Answer I get: 2 and 0,5. It's correct now, I think.

 

[BigGlenntheHeavy]

$\displaystyle \frac{x^2+1}{x}+\frac{x}{x^2+1} \ = \ \frac{29}{10}, \ I'm \ assuming \ that \ you \ want \ to \ solve \ for \ x.$

$\displaystyle \frac{(x^2+1)^2+x^2}{x(x^2+1)} \ = \ \frac{29}{10}, \ \frac{x^4+3x^2+1}{x^3+x} \ = \ \frac{29}{10}, \ 10x^4+30x^2+10 \ = \ 29x^3+29x.$

$\displaystyle Descartes' \ Rule \ of \ Signs:$

$\displaystyle Let \ f(x) \ = \ 10x^4-29x^3+30x^2-29x+10, \ 4 \ changes.$

$\displaystyle f(-x) \ = \ 10x^4+29x^3+30x^2+29x+10, \ 0 \ changes.$

$\displaystyle Hence, \ three \ possibilities: \ (4+,0-,0I),(2+,0-,2I), \ and \ (0+,0-,4I).$

$\displaystyle Ergo, \ f(x) \ = \ (x-2)(2x-1)(5x^2-2x+5), \ second \ of \ the \ three \ possibilities.$

$\displaystyle Note: \ x \ = \ 2 \ and \ x \ = \ \frac{1}{2} \ in \ real \ number \ land, \ see \ graph.$

[attachment=0:1k2oruej]ccc.jpg[/attachment:1k2oruej]

 

[lookagain]

User 123, you wouldn't have learned Descartes' Rule of signs if you're in
beginning algebra, as I have seen this covered in college algebra.

 

[Denis]

Hello 123: do you know "Descartes' Rule of Signs" or do you not?
Let us know so we don't "guess"; thank you.

 

[123]

Hello 123: do you know "Descartes' Rule of Signs" or do you not?
Let us know so we don't "guess"; thank you.

I don't know this rule. So, it's little 'messy' to understand BigGlenntheHeavy solution for this task.




*Sorry for wrong topic.