# To divide radical expressions; quick question, are these equivalent?

1. Mar 19, 2009

### Raizy

1. The problem statement, all variables and given/known data

There is no problem, ^title^, just a quick question :)

The book's answer: $$\frac{x\sqrt{x}-x\sqrt{2}+2\sqrt{x}-2\sqrt{2}}{x-2}}$$

My answer: $$\frac{x+2\sqrt{x}-x-2\sqrt{2}}{x+2}$$

I combined like terms in the numerator.... at least I think I did it correctly?

2. Mar 19, 2009

### HallsofIvy

Staff Emeritus
What terms did you combine? There are no "like terms" in the numerator! And how did "x- 2" in the denominator become "x+ 2"?

You can write $x\sqrt{x}- x\sqrt{2}= x(\sqrt{x}- \sqrt{2})$ and $2\sqrt{x}- 2\sqrt{2}= 2(\sqrt{x}- \sqrt{2})= 2(\sqrt{x}-\sqrt{2})$ and then factor out $\sqrt{x}- \sqrt{2}$.

3. Mar 21, 2009

### Hogger

a quick check is let x = something. I suggest 1 or 0.

4. Mar 21, 2009

### kbaumen

Not quite a good idea. Better do the factorization that Hallsofivy recommended. Putting in some arbitrary numbers can give contradictory results. For example 2x - 1 = 3 if x = 2 but 2x - 1 =/= 3 if x = 0. Even if this approach sometimes gives you the answer, there are many occasions when it won't so better not get used to it. Just take Hallsofivy's advice.

5. Mar 21, 2009

### Hogger

I meant to compare if one function is equal to another. If you think 2x2 = 2 + x + x, you can check it by letting x = 3. After rereading the op I think I misunderstood what happened. Raizy did you just try simplifying the book's answer or is what you posted an answer you got by yourself?

6. Mar 21, 2009

### kbaumen

That's not the way to compare if functions are equal. Except if you find that their domains are equal and than compare all values of x from their domain, which would take an infinite amount of time.

Now than I think about it again, that's the case if you need to prove that the functions are equal. But this time if we find that with at least one value of x they aren't, we've proven empirically that they're not equal, which is what the OP was asking.

7. Mar 21, 2009

### Hogger

This is what I meant.

8. Mar 21, 2009

### kbaumen

Yeah, now I see that. My bad. I should have read more carefully.

9. Mar 21, 2009

### Raizy

ARGHH!! I think I screwed up copying something, and I always throw out my work sheets since I can't bare to even try and "study" off these pieces of paper with a bunch of x's and ='s and -'s and +'s and... :grumpy:

(edit)... ahH! found it. Now i need to re-do the question...

I'll be back :D ... I actually got the right answer but I tried to take one step further because if you would read down below, I thought the numerator would simplify even further.

Okay here it is again:

1. The problem statement, all variables and given/known data

Simplify: $$\frac{x+2}{\sqrt{x}+\sqrt{2}}$$

The book's answer:$$\frac{x\sqrt{x}-x\sqrt{2}+2\sqrt{x}-2\sqrt{2}}{x-2}}$$

3. The attempt at a solution

Step 1 - Multiply by the denominator's conjugate (Latex won't align, I'm not sure what I'm doing wrong):

=$$\frac{(x+2)(\sqrt{x}-\sqrt{2})}}{(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})}$$

Step 2 - FOIL:

=$$\frac{x\sqrt{x}-x\sqrt{2}+2\sqrt{x}-2\sqrt{2}}{x-2}}$$

Now here's what I thought about Step 2 (which turns out to be the correct answer based on the book's answer):

The first and third term in the numerator, which are $$x\sqrt{x} \ \mbox{and} \ 2\sqrt{x}$$ would simplify to $$x+2\sqrt{x}$$

and that

The second and fourth term in the numerator, which are $$-x\sqrt{2} \mbox{&} -2\sqrt{2} \ \mbox{would simplify to:} \ -x-2\sqrt{2}$$
Which finally, I end up with (I have copied the down the wrong denominator in the original post, the denominator should have a negative sign) as follows:

Final answer: $$\frac{x+2\sqrt{x}-x-2\sqrt{2}}{x-2}$$

Since my answer is apparently not equivalent, what mathematical rule have I broken?

Did I ever mention latex is a pita to use? I better get used to it

Last edited: Mar 21, 2009
10. Mar 22, 2009

### MrB3nn

Hi
The rule you have broken is that x$$\sqrt{x}$$ or x$$\sqrt{2}$$ are not equal to x. They can't be simplified. They are what they are. The best you can do is write it in a different form i.e. x$$\sqrt{x}$$ = x3/2

I think you have tried to take the common multiple out of the two terms as a factor but you must always use a bracket when you do this:
x$$\sqrt{x}$$ + 2$$\sqrt{x}$$ = (x+2)$$\sqrt{x}$$