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anonymous
 5 years ago
Hi, I need help solving an equation with a fraction: 0=(55x^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 pq formula... Any idea?
anonymous
 5 years ago
Hi, I need help solving an equation with a fraction: 0=(55x^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 pq formula... Any idea?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My first step was to isolate the 5 so that 0=(5(1x^2)/(x^2+1)).

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0Hi 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) = 55x^2 So the question is, how do we solve: \[55x^2=0\] Doing this isn't so bad, just factor out the 5 \[5(1x^2)=0\] and now solve \[1x^2 = 0\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0just 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
 5 years ago
Best ResponseYou've already chosen the best response.0but i wouldnt go so far as to say the bottom cannot equal to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, so use the L'Hospital rule?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, i was just talking in general

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in this case the denominator is strictly positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he said it must obey two rules. there are exceptions

mathteacher1729
 5 years ago
Best ResponseYou've already chosen the best response.0The reason you get 1 and +1 as your answer is because we can factor: \[1x^2=(x+1)(x1)\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0for example find 0 = (x^2 + x  6) /(x2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0err, solve that i mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, yes. You are right. I forgott that. Thanks a lot!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That'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
 5 years ago
Best ResponseYou've already chosen the best response.0But thanks a buch! You guys helped me a lot!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just 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) /(x2) you have 0/0 for x = 2, but we still have x=3 as a solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so my point is, look out for 0/0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, thanks. So it seems like the functions just have gaps at certain points...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes there could be a hole, or a jump discontinuity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an essential discontinuity or a removable (if i plug it) discontinuity
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