Equation

123

New member
Joined
Nov 6, 2010
Messages
46
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.
:?:
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,145
123 said:
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

Senior Member
Joined
Aug 22, 2010
Messages
2,375
123 said:
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

New member
Joined
Nov 6, 2010
Messages
46
Subhotosh Khan said:
123 said:
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

Senior Member
Joined
Mar 8, 2009
Messages
1,577
\(\displaystyle \frac{x^2+1}{x}+\frac{x}{x^2+1} \ = \ \frac{29}{10}, \ I'm \ assuming \ that \ you \ want \ to \ solve \ for \ x.\)

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

\(\displaystyle Descartes' \ Rule \ of \ Signs:\)

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

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

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

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

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

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

lookagain

Senior Member
Joined
Aug 22, 2010
Messages
2,375
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

Senior Member
Joined
Feb 17, 2004
Messages
1,473
Hello 123: do you know "Descartes' Rule of Signs" or do you not?
Let us know so we don't "guess"; thank you.
 

123

New member
Joined
Nov 6, 2010
Messages
46
Denis said:
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.
 
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