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

You would need to use (x-2)\sqrt{x} = (x-2)x1/2 = x3/2 - 2x1/2I hope you can follow what I am saying. It is a little difficult to explain without seeing you face to face.In summary, Raizy asked for help simplifying a fraction and provided an attempt at a solution. They multiplied by the denominator's conjugate and FOILed to get their final answer. However, they incorrectly simplified the first and third terms in the numerator, resulting in an answer
  • #1
Raizy
107
0

Homework Statement



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

The book's answer: [tex]\frac{x\sqrt{x}-x\sqrt{2}+2\sqrt{x}-2\sqrt{2}}{x-2}} [/tex]

My answer: [tex]\frac{x+2\sqrt{x}-x-2\sqrt{2}}{x+2}[/tex]

I combined like terms in the numerator... at least I think I did it correctly?
 
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  • #2
Raizy said:

Homework Statement



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

The book's answer: [tex]\frac{x\sqrt{x}-x\sqrt{2}+2\sqrt{x}-2\sqrt{2}}{x-2}} [/tex]

My answer: [tex]\frac{x+2\sqrt{x}-x-2\sqrt{2}}{x+2}[/tex]

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 [itex]x\sqrt{x}- x\sqrt{2}= x(\sqrt{x}- \sqrt{2})[/itex] and [itex]2\sqrt{x}- 2\sqrt{2}= 2(\sqrt{x}- \sqrt{2})= 2(\sqrt{x}-\sqrt{2})[/itex] and then factor out [itex]\sqrt{x}- \sqrt{2}[/itex].
 
  • #3
a quick check is let x = something. I suggest 1 or 0.
 
  • #4
Hogger said:
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.
 
  • #5
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
Hogger said:
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.
 
  • #7
kbaumen said:
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.
 
  • #8
Hogger said:
This is what I meant.

Yeah, now I see that. My bad. I should have read more carefully.
 
  • #9
HallsofIvy said:
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... :grumpy:

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

Hogger said:
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: [tex]\frac{x+2}{\sqrt{x}+\sqrt{2}}[/tex]

The book's answer:[tex]\frac{x\sqrt{x}-x\sqrt{2}+2\sqrt{x}-2\sqrt{2}}{x-2}} [/tex]

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):

=[tex]\frac{(x+2)(\sqrt{x}-\sqrt{2})}}{(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})}[/tex]

Step 2 - FOIL:

=[tex]\frac{x\sqrt{x}-x\sqrt{2}+2\sqrt{x}-2\sqrt{2}}{x-2}}[/tex]

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 [tex]x\sqrt{x} \ \mbox{and} \ 2\sqrt{x}[/tex] would simplify to [tex]x+2\sqrt{x}[/tex]

and that

The second and fourth term in the numerator, which are [tex]-x\sqrt{2} \mbox{&} -2\sqrt{2} \ \mbox{would simplify to:} \ -x-2\sqrt{2}[/tex]
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: [tex]\frac{x+2\sqrt{x}-x-2\sqrt{2}}{x-2}[/tex]

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 :wink:
 
Last edited:
  • #10
Hi
The rule you have broken is that x[tex]\sqrt{x}[/tex] or x[tex]\sqrt{2}[/tex] 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[tex]\sqrt{x}[/tex] = 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[tex]\sqrt{x}[/tex] + 2[tex]\sqrt{x}[/tex] = (x+2)[tex]\sqrt{x}[/tex]
 

1. What does it mean to divide radical expressions?

Dividing radical expressions means to divide two expressions that contain a radical symbol, such as square roots or cube roots. This is a common operation in algebra and is used to simplify and solve equations.

2. How do you divide radical expressions?

To divide radical expressions, you need to first simplify each expression. Then, you can divide the two simplified expressions using the rules of exponents. If the radicals have the same index, you can divide the numbers inside the radical and keep the same index. If the radicals have different indices, you can change them to the same index before dividing.

3. Are the rules for dividing radical expressions the same as the rules for dividing regular numbers?

No, the rules for dividing radical expressions are not the same as the rules for dividing regular numbers. When dividing radical expressions, you must also take into account the rules of exponents, such as the product rule and the quotient rule.

4. Can you provide an example of dividing radical expressions?

Sure, for example, if we have the expression (sqrt(8))/(sqrt(2)), we can first simplify each radical to get (2sqrt(2))/(sqrt(2)). Then, using the quotient rule of exponents, we can divide the numbers inside the radical to get 2. Therefore, the simplified expression is 2sqrt(2).

5. How do I know if two divided radical expressions are equivalent?

You can check if two divided radical expressions are equivalent by simplifying them and comparing the simplified forms. If the simplified forms are the same, then the divided radical expressions are equivalent. You can also solve for the variable in the original expressions and see if they both yield the same value, which would also indicate that they are equivalent.

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