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 #6
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Even if you say set f to equal 1/0 it is useless because nothing needs to divide by zero. oh and doing polynomial equations with xf=x/0 you can get some pretty weird stuff. like how some times you can just cancle all but a few things out and get a single point. or f=0 when you started with some thing like, (fx-1)(x+f)=0

                                                                                                fx^2+xf^2-x-f=0

                                                                                                 (fx^2+xf^2-x-f)=((1/0)x^2+x(1/0)^2-x-(1/0))=0

                                                                                                 ((1/0)x^2+x(1/0)^2-x-(1/0))*0=0*0

                                                                                                 (1+f-1)=0

                                                                                                  (f+1-1)=f=0???????

but when you try to solve the factored version it comes out to be

fx-1=0 or x+f=0

x=1/f or x=-f

x=1/(1/0) or x=-(1/0)

sooo if you plug that in to the equations: (f*(1/(1/)-1)(1/(1/0)+f)=0

1/1/0 = 1/0 so

 

((1/0*1/0)-1)(1/0+1/0)=0

 

(f^2-1)(2f)=0

 

so f^2-1=0 or (2f)=0

and since 2f is just 1/0 + 1/0 we get undefined

but 1/0^2, if just 1/0 because 1*1/0*0

and then subtracting one doesnt help since we cant divide by zero.

 

btw i have found an equation that has a one point solution its just lost in my notes.

Mar 6, 2017
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Mar 6, 2017
 #1
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Mar 6, 2017

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