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ok, its me again. sorry to bother you, but this one really caught me up.
http://www2.johnabbott.qc.ca/~daniel.gatien/AS1/practice1.pdf
on question 2, i am quite stuck...it says below that the answer is -1/4, but i dont know how to get to that answer. if i just plug in 2 by substitution, then i get a 0/0 which is of course indeterminate.
i think i need to factorise it, but that is confusing me more.
so for (x-2)/(4-x^2), like it says in the problem, what exactly do i need to do? square the numerator and denominator? multiply both sides by the numerator or denominator? i have tried all three of those methods, but the only answers i ended up with were 1/2, 1/4 and -1/16. here is what i think i need to do, but please tell me where i go wrong:
starting from step 2:
[(x-2)(x-2)]/[(4-x^2)(4-x^2)]
multiplying out brackets gives:
(x^2-2x-2x-4)/(16-4x^2-4x^2-x^4)
then plugging in the value for x, which is 2 as stated in the question:
-8/-32
which simplifies to:
1/4
but i figure i messed up somewhere. in order to get -1/4, not 1/4, i can only have one of the numerator or denominator as a negative, not both.
so, how do i solve it?
thanks guys.
Well, this is how I solve the problem
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As coffeeking said, the easiest way to solve this one is to factor the denominator and cancel out the numerator.
As for your method, it is not legal to square the expression.
Wrap it in bacon
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