# Having trouble simplifying conjugate

1. Jun 4, 2004

I am having trouble simplifying

(4[(SQRT(x+2)) – (SQRT2))]/x

I multiply the conjugate and I come up with 4/((SQRT(x+2)) + SQRT2)

I'm not sure if I'm on the right track

2. Jun 4, 2004

### Icarus

Quite frankly, by the concept of "simplified" that I am familiar with, the only simplification of the original that is possible is simply to multiply the 4 through:

[4*sqrt(x+2) - 4*sqrt(2)]/x

"Simplified" usually means rational denominators - your multiplying by the conjugate took you in the opposite direction.

What is this for? If the problem is just to simplify the original expression, then what I gave is as far as you can go. But if you need to do something else afterward, then this "simplified" form may not be the easiest to work with. It depends on the application.

3. Jun 4, 2004

Actually its a limit problem I'm working on I'm having trouble with algebra.

Find the limit by analytic methods:

lim x->0 (4[(SQRT(x+2)) – (SQRT2))]/x

The first part of the problem asked me to estimate the limit by using a table and I came up with 1.414.

I am trying to multiple the conjugate...

4. Jun 4, 2004

### master_coda

Actually you've already done all the hard work. Notice that the form you now have, 4/((SQRT(x+2)) + SQRT2), no longer gives you 0/0 if you evaluate it at x=0. So you can just plug in x=0 and get the result of the limit. You'll get 4/(2*SQRT(2))=2/SQRT(2)=SQRT(2).

Last edited: Jun 4, 2004
5. Jun 4, 2004

### Math Is Hard

Staff Emeritus
Did you multiply (sqrt (x+2) + sqrt (2) ) to your numerator and your denominator?
You should end up with 4(x) / [x((SQRT[x+2]) + (SQRT[2]))]
cancel an x on top and bottom. now you can plug in 0s anywhere x is (only one place)and you get
4/ (SQRT[0+2] + SQRT[2])
now what have you got?

6. Jun 4, 2004

### Math Is Hard

Staff Emeritus
whoops! mc beat me to it!

7. Jun 4, 2004

### Icarus

Master coda is right. Your derivation was correct, and in this form the limit is trivial.

(An excellent example of where "simplified" is not the form you want.)

8. Jun 4, 2004

### TALewis

ladyrae, why didn't you post this in your earlier limit thread, which you already created to get help with this problem? It's much easier to keep track of your questions if you keep them in one place.

9. Jun 4, 2004