Please, check my little algebra work

[metalilty]

Simplify the following, giving the result without fractional indices:

 

[(x^2 -1)^2 * √(x+1)]/ (x-1)^3/2

 

Attempt at solving:

 

There are no common bases to add the indices and no common indices to multiply out the bases so I tried this and got it wrong, please show me where though:

 

[(x-1) (x+1)]^2 * (x+1)^1/2] / (x-1)^3/2

 

=[(x-1)^2 * (x+1)^2 * (x+1)^1/2] / (x-1)^3/2

 

multiplied the indices by 2 and got rid of the fractions

 

=[(x-1)^4 * (x+1)^4 * (x+1)] / (x-1)^3

 

= (x-1) *(x+1)^5

 

By the book it should be : (x+1)^2* √(x^2-1)

 

Thank You.

[Realitycheck]

nm, trying to do it in my head

Edited September 12, 2011 by Realitycheck

[Schrödinger's hat]

=[(x-1)^2 * (x+1)^2 * (x+1)^1/2] / (x-1)^3/2

 

multiplied the indices by 2 and got rid of the fractions

 

=[(x-1)^4 * (x+1)^4 * (x+1)] / (x-1)^3

 

= (x-1) *(x+1)^5

 

By the book it should be : (x+1)^2* √(x^2-1)

 

Thank You.

 

Nearly there. Remember that you can't just square everything and get the same expression, so your last step should be:

((x-1) *(x+1)^5)^(1/2)

or √(x-1)*√((x+1)^5)

Can you see how to get from there to the last step? (might be easier to go from the one above

[math]\sqrt{(x-1)(x+1)^{5}}[/math]

Can you somehow get a [math]x^2-1[/math] term in there?

Also you can use latex on this forum by putting it in between

[math]

and

[/math]

tags.

[cfa99]

The workings should be correct

 

Charles

www.cfalabs.com

[CJY]

My solution, hope this helps

 

[(x^2 -1)^2 * √(x+1)]/ (x-1)^3/2

 

(x-1)^2 * (x+1)^2 * (x+1)^1/2 * 1/(x-1)^3/2

 

(x+1)^5/2 * (x-1)^1/2 ~~~> use index laws

 

I think the rest is straightforward. You can do it yourself.

Edited October 5, 2011 by CJY