Algebraic Fraction Help

[iitsrii]

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

 

[Deleted member 4993]

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 $\displaystyle x \ne 0$

 

[JeffM]

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.

 

[Dr.Peterson]

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.

 

[Deleted member 4993]

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.

 

[JeffM]

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.