# Simplification not so simple for me

1. Dec 16, 2008

### Titans86

"simplification" not so simple for me...

(1+9x/4)^1/2 = 1/2(4+9X)^1/2 ?

I'm working on calculating line length using integration but my textbook keeps doing tricks like these... I can't seem to figure it out, can someone please explain this to me...

2. Dec 16, 2008

### jgens

Re: "simplification" not so simple for me...

Sure, try reexpressing the 1 so the both 1 and 9x/4 have a common denominator, then reexpress the squareroot and simplify.

3. Dec 16, 2008

### Defennder

Re: "simplification" not so simple for me...

$$\sqrt{1+\frac{9}{4}x} = \sqrt{\frac{1}{4}(4+9x)} = \sqrt{\frac{1}{4}} \sqrt{4+9x}= \frac{1}{2}\sqrt{4+9x}$$

4. Dec 16, 2008

### Titans86

Re: "simplification" not so simple for me...

Bah, So elementary.

Thanks guys!

5. Dec 16, 2008

### Dick

Re: "simplification" not so simple for me...

(1+9x/4)=(1/4)(4+9x), ok so far? Take the square root of both sides. (1/4)^(1/2)=1/2, so (1+9x/4)^(1/2)=(1/2)*sqrt(4+9x). They just factored a 1/4 out.

6. Dec 16, 2008

### Titans86

Re: "simplification" not so simple for me...

follow up question if I may... (I tried using this simplification to my demise)

I have $$\sqrt{1+\frac{1}{x^{2}}}$$

I tried he trick and got $$\left(\frac{1}{x}\right)\sqrt{x^{2}+1}$$

Yet it doesn't match... where did I go wrong?

Also, I need to be worrying about the +- when I do these things also no?

7. Dec 16, 2008

### Mentallic

Re: "simplification" not so simple for me...

Well there are multiple ways of expressing these, so you'll need to show us what the textbook has provided. Your simplification is correct though.
You don't need to worry about the $$\pm$$, since that is only when you have an equation in the form x^2=y and then you make x the subject.

8. Dec 16, 2008

### Dick

Re: "simplification" not so simple for me...

It should really be written (1/|x|)*sqrt(x^2+1)=sqrt(1+1/x^2). Is that what you mean by going wrong?

9. Dec 16, 2008

### Titans86

Re: "simplification" not so simple for me...

I plotted it through my Ti and two different curves resulted...

edit: ah yes, the absolute would resolve it...

(my algebra is so poor...)

10. Dec 17, 2008

### mutton

Re: "simplification" not so simple for me...

But that is what is happening here. $$\sqrt{x^2} = |x|$$.