Simple Equation: Find x in 1+8x^{-1}+15x^{-2}

  • Thread starter thomasrules
  • Start date
In summary, to find x for the expression 1+8x^{-1}+15x^{-2}, you can simplify it to (x+3)(x+5)/x^2 where x can be any real number except 0. However, if they are asking for all real numbers of x, then x can be any number, even complex. The textbook's solution is (1+\frac{5}{x})(1+\frac{3}{x}).
  • #1
thomasrules
243
0
find x:

[tex]1+8x^{-1}+15x^{-2}[/tex]

I've got so far:

[tex]\frac{x^2+8x+15}{x^2}[/tex]
 
Last edited:
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  • #2
find x for
[tex]1+8x^{-1}+15x^{-2}[/tex]

and so far you got
[tex]\frac{x^2+8x+15}{x^2}[/tex]

x can be anything, all real number, or even complex number... apple, orange, banana... just put what ever you want
 
  • #3
sorry :) LOL ...ya a banana...ok just simplify
 
  • #4
If they're asking for all real numbers of x, then x can be anything but zero. Otherwise, like vincentchan said, x can be anything, even complex.
 
  • #5
sorry :) LOL ...ya a banana...ok just simplify.
 
  • #6
(x+3)(x+5)=x^2+8x+15
 
  • #7
vincent I know that but there is a x^2 underneath!

The answer in the book is

[tex](1+\frac{5}{x})(1+\frac{3}{x})[/tex]
 
  • #8
can you simplify from my solution to your textbook's one?
(x+3)(x+5)/x^2=(1+3/x)(1+5/x)
 
  • #9
;) got it thanks mr.
 

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