anonymous
  • anonymous
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?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
My first step was to isolate the 5 so that 0=(5(1-x^2)/(x^2+1)).
mathteacher1729
  • mathteacher1729
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. :)
anonymous
  • anonymous
hey

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anonymous
  • anonymous
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...
anonymous
  • anonymous
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
anonymous
  • anonymous
but i wouldnt go so far as to say the bottom cannot equal to zero
anonymous
  • anonymous
Hmm, so use the L'Hospital rule?
anonymous
  • anonymous
no, i was just talking in general
anonymous
  • anonymous
Oh, ok.
anonymous
  • anonymous
in this case the denominator is strictly positive
anonymous
  • anonymous
he said it must obey two rules. there are exceptions
mathteacher1729
  • mathteacher1729
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.
anonymous
  • anonymous
for example find 0 = (x^2 + x - 6) /(x-2)
anonymous
  • anonymous
err, solve that i mean
anonymous
  • anonymous
Oh, yes. You are right. I forgott that. Thanks a lot!
anonymous
  • anonymous
x=-3
anonymous
  • anonymous
yes
anonymous
  • anonymous
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...
anonymous
  • anonymous
But thanks a buch! You guys helped me a lot!
anonymous
  • anonymous
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
anonymous
  • anonymous
so my point is, look out for 0/0
anonymous
  • anonymous
Ok, thanks. So it seems like the functions just have gaps at certain points...
anonymous
  • anonymous
yes there could be a hole, or a jump discontinuity
anonymous
  • anonymous
an essential discontinuity or a removable (if i plug it) discontinuity

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