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anonymous

  • 5 years ago

Hi, I need help solving an equation with a fraction: 0=(5-5x^2)/(x^2+1). I know how to get the zeros by locig reasoning but I don't get it in a form which allows me to use other tools like the p-q formula... Any idea?

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  1. anonymous
    • 5 years ago
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    My first step was to isolate the 5 so that 0=(5(1-x^2)/(x^2+1)).

  2. mathteacher1729
    • 5 years ago
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    Hi Panavia, An equation of the form (top ) / (bottom) = zero Must obey TWO rules: 1) (bottom) CANNOT equal zero! 2) (top) MUST equal zero. So in this case: 1) (bottom) = x^2 +1 , which is never zero, so we're good for rule one. 2) (top) = 5-5x^2 So the question is, how do we solve: \[5-5x^2=0\] Doing this isn't so bad, just factor out the 5 \[5(1-x^2)=0\] and now solve \[1-x^2 = 0\] This is pretty easy, just add x^2 to both sides \[1=x^2\] and the answer should be right in front of you. :)

  3. anonymous
    • 5 years ago
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    hey

  4. anonymous
    • 5 years ago
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    Yes, that's what I ment with logical reasoning. But when I plot the graph I get two zeros. One at +1 and one at -1. The -1 is the one which puts me off. Like you said, when I put the -1 into the bottom portion, I get a 0 in the denominator...

  5. anonymous
    • 5 years ago
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    just a comment, you could have p(x)/q(x) = 0/0 , in which case you can look to factor out something or analyze it further , look at the limit as x approaches your zeroes

  6. anonymous
    • 5 years ago
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    but i wouldnt go so far as to say the bottom cannot equal to zero

  7. anonymous
    • 5 years ago
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    Hmm, so use the L'Hospital rule?

  8. anonymous
    • 5 years ago
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    no, i was just talking in general

  9. anonymous
    • 5 years ago
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    Oh, ok.

  10. anonymous
    • 5 years ago
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    in this case the denominator is strictly positive

  11. anonymous
    • 5 years ago
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    he said it must obey two rules. there are exceptions

  12. mathteacher1729
    • 5 years ago
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    The reason you get -1 and +1 as your answer is because we can factor: \[1-x^2=(x+1)(x-1)\] Or when we solve \[1=x^2 \Rightarrow x=\pm1\] Substituting -1 into the denominator does not yield zero: \[(-1)^2+1 =1+1 =2\] Remember when you square negative one, or multiply -1 times itself, you get a positive.

  13. anonymous
    • 5 years ago
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    for example find 0 = (x^2 + x - 6) /(x-2)

  14. anonymous
    • 5 years ago
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    err, solve that i mean

  15. anonymous
    • 5 years ago
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    Oh, yes. You are right. I forgott that. Thanks a lot!

  16. anonymous
    • 5 years ago
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    x=-3

  17. anonymous
    • 5 years ago
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    yes

  18. anonymous
    • 5 years ago
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    That's pretty easy. In this case you can really solve the equation. But in my case it seems like there is no such solution and you can only figure out the zeros by logic reasoning...

  19. anonymous
    • 5 years ago
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    But thanks a buch! You guys helped me a lot!

  20. anonymous
    • 5 years ago
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    just because you have zero in the denominator does not mean you must stop there and say no solution, like 1/x = 0 (no solution). for my example 0 = (x^2 + x - 6) /(x-2) you have 0/0 for x = 2, but we still have x=-3 as a solution

  21. anonymous
    • 5 years ago
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    so my point is, look out for 0/0

  22. anonymous
    • 5 years ago
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    Ok, thanks. So it seems like the functions just have gaps at certain points...

  23. anonymous
    • 5 years ago
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    yes there could be a hole, or a jump discontinuity

  24. anonymous
    • 5 years ago
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    an essential discontinuity or a removable (if i plug it) discontinuity

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