Question

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?

Answer 1

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

Answer 2

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. :)

Answer 3

hey

Answer 4

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...

Answer 5

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

Answer 6

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

Answer 7

Hmm, so use the L'Hospital rule?

Answer 8

no, i was just talking in general

Answer 9

Oh, ok.

Answer 10

in this case the denominator is strictly positive

Answer 11

he said it must obey two rules. there are exceptions

Answer 12

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.

Answer 13

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

Answer 14

err, solve that i mean

Answer 15

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

Answer 16

x=-3

Answer 17

yes

Answer 18

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...

Answer 19

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

Answer 20

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

Answer 21

so my point is, look out for 0/0

Answer 22

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

Answer 23

yes there could be a hole, or a jump discontinuity

Answer 24

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