the algebraic expressions x-2 divided by x² -9is undefined when x is ..?
It's undefined when the denominator is 0. This occurs when \[x^2-9=0 \rightarrow x=\pm 3\]
if the denominator is zero than wouldnt the answer be zero?
No, the denominator is zero for x-values such that \[x^2-9=0\] By what you're saying, if the answer is zero, x=0. But then the denominator would be \[0^2-9=-9\]which is fine. You can divide your expression by -9.
\[\frac{x-2}{x^2-9}=\frac{something}{0}\]is undefined.
You have to find the x-values that make the denominator 0.
Is that clearer?
not really
Is it the 'undefined' thing, i.e. *why* you have to find x where x^2-9 = 0?
yea.. i dont understand that..
Okay. In division, you can't ever divide by 0. This is because, if we did it, we wouldn't actually end up with something that made sense. I'll give you a simple example.
ok
You're happy with\[\frac{6}{2}=3\] yeah?
yes
So if I then write,\[6=\left[ ? \right]\times 3\]what number would you put in there?
2
Right. Now,
if we imagine \[\frac{6}{0}=\left[ ? \right]\](i.e. *some* number)
we could write
\[6=0 \times \left[ ? \right]\]
What number would you put in now?
thats impossible
EXACTLY
We attempted to divide 6 by 0 and ended up with an IMPOSSIBILITY.
That's why division by zero is undefined.
Is that better?
This will happen no matter what you try to divide by zero.
yeahh
So that's why you have to find all the x-values that will end up making x^2-9 = 0 ... because everything falls apart there.
If x is 3 or -3, x^2-9 = 3^2-9 = 9 - 9 = 0 or x^2-9 = (-3)^2-9 = 9-9=0
\[\frac{x-2}{x^2-9}=\frac{3 -2}{3^2-9}=\frac{1}{0}\]when x=3
\[\frac{x-2}{x^2-9}=\frac{-3-2}{(-3)^2-9}=\frac{-5}{0}\]when x=-3.
oooooooooooooh alrighttt tanks
In both cases, you'll end up with the same rubbish you get with \[\frac{6}{0}\]
I think it's clicked now :)
Feel free to become a fan :P
:D
lol
Good luck with your algebra. Try and link it to numbers if you have trouble.
okeydokey
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