Algebraic Fraction Help

iitsrii

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Oct 19, 2018
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Simplify
x+1
---------
1-x^-2






Would I rationalise the denominator and multiply the top by 1+x^2
or multiply the top by x^2 and bottom I'm not sure
 
Simplify
x+1
---------
1-x^-2






Would I rationalise the denominator and multiply the top by 1+x^2
or multiply the top by x^2 and bottom I'm not sure

\(\displaystyle \displaystyle{\dfrac{x \ + \ 1}{1 \ - \ \frac{1}{x^2}}}\)

I would first multiply numerator and denominator by x2 , assuming \(\displaystyle x \ne 0\)
 
Simplify
x+1
---------
1-x^-2






Would I rationalise the denominator and multiply the top by 1+x^2
or multiply the top by x^2 and bottom I'm not sure
\(\displaystyle x \ne 0\) is implied by the original expression, so it can be multiplied by \(\displaystyle \dfrac{x^2}{x^2}\).

Moreover \(\displaystyle x\ne \pm 1.\)

\(\displaystyle \dfrac{x + 1}{1 - x^{-2}} = \dfrac{x + 1}{1 - \dfrac{1}{x^2}} = \dfrac{x + 1}{1 - \dfrac{1}{x^2}} * \dfrac{x^2}{x^2} =\\

\dfrac{x^2(x + 1)}{x^2-1} = \dfrac{x^2(x+1)}{(x + 1)(x-1)} = \dfrac{x^2}{x-1} \text { if } x \ne 0 \text { and } x \ne \pm 1.\)

You may of course multiply by \(\displaystyle \dfrac{1 +x^2}{1+x^2}\), but what good does that do?

\(\displaystyle \dfrac{x+1}{1-x{-2}} = \dfrac{x +1}{1-x^{-2}} * \dfrac{1 +x^2}{1+x^2} = \dfrac{x^3+x^2+x+1}{1 +x^2-x^{-2} - x^0} =\\

\dfrac{x^3 +x^2+x+1}{x^2-x^{-2}} \text { if } x \ne 0 \text { and } x\ne\pm1.\)

Now it is true that can be simplified further.

\(\displaystyle x \ne 1 \implies \dfrac{x^3 +x^2+x+1}{x^2 -x^{-2}} = \dfrac{\dfrac{x^4-1}{x-1}}{\dfrac{x^4-1}{x^2}} = \dfrac{x^4-1}{x-1}*\dfrac{x^2}{x^4-1} =\\

\dfrac{x^2}{x-1} \text { if } x \ne 0 \text { and } x \ne \pm 1.\)

But that is a very roundabout approach.
 
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Why not simply factor the denominator?

\(\displaystyle \dfrac{x + 1}{1 - x^2} = -\ \dfrac{x + 1}{x^2 - 1} = -\ \dfrac{x + 1)}{(x + 1)(x - 1)} = - \ \dfrac{1}{x - 1}.\)

No need to worry about x = 1 because that was precluded by the original problem.

Now if, as SK seems to think, the problem really is to simplify

\(\displaystyle \dfrac{x + 1}{1 - \frac{1}{x^2}}\),

you cannot multiply by \(\displaystyle \dfrac{x^2}{x^2}\) unless the problem says \(\displaystyle x \ne 0.\)

But you may multiply by \(\displaystyle \dfrac{1 + x^2}{1 + x^2}.\)

But I don't see the point of doing so.

\(\displaystyle \dfrac{x + 1}{1 - \frac{1}{x^2}} = \dfrac{x + 1}{\dfrac{x^2 - 1}{x^2}} =\\

\dfrac{x + 1}{1} * \dfrac{x^2}{x^2 - 1} = \dfrac{x + 1}{1} * \dfrac{x^2}{(x + 1)(x - 1)} = \dfrac{x^2}{x - 1}.\)

Are you assuming there was a typo in the problem, and it really says x2 rather than x-2 in the denominator?

As given, the problem does imply that x is nonzero, as zero is not in the domain. After simplifying, of course, it is necessary to say that explicitly.
 
Are you assuming there was a typo in the problem, and it really says x2 rather than x-2 in the denominator?

As given, the problem does imply that x is nonzero, as zero is not in the domain. After simplifying, of course, it is necessary to say that explicitly.
I had skipped mentioning \(\displaystyle \ x \ne \pm 1\) - hoping that OP will catch that. At x=-1, the given function would have a removable discontinuity.
 
Last edited by a moderator:
Are you assuming there was a typo in the problem, and it really says x2 rather than x-2 in the denominator?

As given, the problem does imply that x is nonzero, as zero is not in the domain. After simplifying, of course, it is necessary to say that explicitly.
No. I just misread it. Thanks.

I'll fix my post.
 
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