## 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?

1. anonymous

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

2. 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. :)

3. anonymous

hey

4. 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...

5. 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

6. anonymous

but i wouldnt go so far as to say the bottom cannot equal to zero

7. anonymous

Hmm, so use the L'Hospital rule?

8. anonymous

no, i was just talking in general

9. anonymous

Oh, ok.

10. anonymous

in this case the denominator is strictly positive

11. anonymous

he said it must obey two rules. there are exceptions

12. 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.

13. anonymous

for example find 0 = (x^2 + x - 6) /(x-2)

14. anonymous

err, solve that i mean

15. anonymous

Oh, yes. You are right. I forgott that. Thanks a lot!

16. anonymous

x=-3

17. anonymous

yes

18. 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...

19. anonymous

But thanks a buch! You guys helped me a lot!

20. 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

21. anonymous

so my point is, look out for 0/0

22. anonymous

Ok, thanks. So it seems like the functions just have gaps at certain points...

23. anonymous

yes there could be a hole, or a jump discontinuity

24. anonymous

an essential discontinuity or a removable (if i plug it) discontinuity