Solving 1+8x^{-1}+15x^{-2}=0

[thomasrules]

find x:

$$1+8x^{-1}+15x^{-2}$$

I've got so far:

$$\frac{x^2+8x+15}{x^2}$$

 

[vincentchan]

find x for
$$1+8x^{-1}+15x^{-2}$$

and so far you got
$$\frac{x^2+8x+15}{x^2}$$

x can be anything, all real number, or even complex number... apple, orange, banana... just put what ever you want

 

[thomasrules]

sorry :) LOL ...ya a banana...ok just simplify

 

[Jameson]

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.

 

[thomasrules]

sorry :) LOL ...ya a banana...ok just simplify.

 

[vincentchan]

(x+3)(x+5)=x^2+8x+15

 

[thomasrules]

vincent I know that but there is a x^2 underneath!

The answer in the book is

$$(1+\frac{5}{x})(1+\frac{3}{x})$$

 

[vincentchan]

can you simplify from my solution to your textbook's one?
(x+3)(x+5)/x^2=(1+3/x)(1+5/x)

 

[thomasrules]

;) got it thanks mr.