Easy Question

  • #1
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:

Answers and Replies

  • #2
609
0
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
243
0
sorry :) LOL ....ya a banana.....ok just simplify
 
  • #4
793
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
243
0
sorry :) LOL ....ya a banana.....ok just simplify.
 
  • #6
609
0
(x+3)(x+5)=x^2+8x+15
 
  • #7
243
0
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
609
0
can you simplify from my solution to your text book's one?
(x+3)(x+5)/x^2=(1+3/x)(1+5/x)
 
  • #9
243
0
;) got it thanks mr.
 

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