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

[Raizy]

Homework Statement

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?

 

[HallsofIvy]

Homework Statement

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?

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 $\sqrt{x}- \sqrt{2}$.

 

[Hogger]

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

 

[kbaumen]

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

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.

 

[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?

 

[kbaumen]

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.

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.

 

[Hogger]

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.

This is what I meant.

 

[kbaumen]

This is what I meant.

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

 

[Raizy]

And how did "x- 2" in the denominator become "x+ 2"?

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...

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

Raizy did you just try simplifying the book's answer or is what you posted an answer you got by yourself?

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:

Homework Statement

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

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

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

 

[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$$\sqrt{x}$$ = x^3/2

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)$$\sqrt{x}$$