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    t(t-3)=t^2-4

    if I were to solve this normally I would miss a solution (t=infinity) , this because t^2's cancel out but why is this the case? Can someone explain?

    thanks
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    t = infinity isn't a solution in the reals.

    So long as you don't divide by zero, you should be fine.
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    ok maybe not in this case , what if it t was tan\theta , then there would be a soultion right?
    why would this soultion go missing when cancelling both the tan^2's

    tan\theta(tan\theta-3)=tan^2\theta-4
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    Think of it as an equation in cot theta.
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    (Original post by DFranklin)
    Think of it as an equation in cot theta.
    yup I know how to find the solution , I just dont get why a solution would go missing when cancelling both the tan^2's ..
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    Well, I would say it doesn't. I wouldn't say "tan theta = infinity" is a solution to what you've posted. (It's possible other people might see things differently, but what I am certain about is that very very few people on here can sensibly discuss "t = \infty" solutions; it causes so many mistakes that I would avoid it if at all possible).

    But if you had an actual geometrical problem, then you'd have to ask yourself "OK, what happens when tan theta = infinity". Whether or not that's a valid solution depends on the problem.
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    fair enough ..thanks anyway to simon as well
 
 
 
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