Question

@mathmale Just Number 25(: I've done part but don't know if it's right - I'll put it up.

Answer 1

OK. Waiting with bated breath....

Answer 2

N: Have you done this type of problem before? What do you find challenging about this particular problem?

Answer 3

As you probably know already, the third root of a variable x can be expressed as x^(1/3). Is that familiar?

Answer 4

Well, I looked in my book to see what they got and they came out to have 150 as part of the problem and I have no idea how to get that. Plus, I don't really know where to go.

Answer 5

OK. I've typed #25 into the Equation Editor, using the exponent 1/3 because the Editor doesn't seem able to write "third root of ... " symbolically.

Problem #25 can be expressed as\[(\frac{ 6x ^{2}y ^{4}}{ 10x ^{7}y })^{(1/3)}\]

Answer 6

Okay, I see.

Answer 7

6 is not a cube, and neither is x^2. However, y^3 is a cube and is a factor of y^4. Does it make sense to you to write: (y^4)^(1/3) = [(y^3)^(1/3)]*[y^(1/3)?

Answer 8

If so, then (y^4)^(1/3) reduces to y*y^(1/3). Comfortable with that?

Answer 9

Yes, Would 6x^2 still be under the root?

Answer 10

Yes, Nicole, that's right. Not just 6x^2, but 6*x^2*y. Make sense?

In the denominator: 10 is not a cube and does not have a factor which is a cube.
x^7 can be re-written as x(x^6).

Answer 11

I'm going to put this into the Equation Editor for clarity. Unfortunately, that's a pretty slow process. You can respond in any way you like while I'm tied up with that.

Answer 12

Thank you! I'm a bit confused but I think I will get it if I can see it(:

Answer 13

The original cube root expression reduces to
\[(\frac{ 6x ^{2}y }{ 10xy })^{(1/3)}*\frac{ y }{ x ^{2} }.\]

Does this make sense to you? If not please describe which part or parts you find confusing.

Answer 14

Oh! I see!

Answer 15

Great.
Now the last thing I typed in Equation Editor could be the answer to this problem, but your teacher would subtract points because you'd still have a radical in the denominator. We thus need to go one more step further, and that's to "rationalize the denominator." What does that phrase mean to you?

Answer 16

Um, get ride of all x and y?

Answer 17

This action means to modify the answer given above so that there is no longer any radical in the denominator.
Look: Our denom. is now 10xy. Were we to multiply that by (10xy)^2, we'd get (10xy)^3, which is a perfect cube. OK?

Answer 18

And therefore, the cube root of (10xy)^3 is simply 10xy, with no radical, right?

Answer 19

Okay, I think I get it.

Answer 20

But how do you get 150?

Answer 21

Honestly, I don't know where 150 would show up in this particular problem.
I'm going to do it again as a double check, but I believe we have the correct answer, which, by the way, MUST include x and y.

Answer 22

The book tells me that the answer is: y cube root 150x over 10x^2

Answer 23

Without commenting (yet) on that, I'd like to introduce a slightly different approach which may actually simplify the work of reducing the radical.

Answer 24

Okay

Answer 25

Nicole, please go backf or another look at the problem you've presented.
Look at what's under the radical sign.
What's under that is called the "radicand," by the way.
The radicand can be reduced before we do anything else, and doing tht saves work!

Answer 26

Reduced how?

Answer 27

Here's what to do: Reduce 6/10 to 3/5.
Reduce y^4 / y to y^3.
Reduce x^2 / x^7 to 1 over x^5.

Answer 28

Separate x^5 into [x^3] * [x^2].

Answer 29

so it would be like this? 3xy^3/5x^5 ?

Answer 30

We end up with\[(\frac{ 3y ^{3} }{ 5(x ^{3})(x ^{2})})^{(1/3)}\]

Answer 31

Yes, that's right. In my expression I've factored that x^5.

Answer 32

Okay, I see. I don't know how they got 150 but I'm just going to put that. Maybe they made a mistake..

Answer 33

Almost there. Nicole, what is the cube root of y^3?
What is the cube root of x^3?

Answer 34

x and y?

Answer 35

Right! So, right off, we can factor (y/x) out from under the radical.
What's left, then, is (y/x)*cube root of (3 over 5x^2).
Agreed?

Answer 36

Agreed(: I assume you have to go now?

Answer 37

No, I'll stay with you until we've finished.

Answer 38

There's more after the the cube root of 3y/5x(x^2)?

Answer 39

You'll have (y/x)(cube root of (3y over 5x^2). Please verify this for yourself; I may have made some simple errors immediately before this stage in the problem solution.

Answer 40

Oh, I forgot the y/x on the outside.

Answer 41

\[\frac{ y }{ x }(\frac{ 3 }{ 5x ^{2} })^{(1/3)}\]
is where we should be now.

Answer 42

Now, one last step: we have to "rationalize the denominator."

Answer 43

Yes, that is what I have.

Answer 44

Cool. Look at the denominator under the radical sign: 5x^2. By what quantity must we multiply 5x^2 to obtain a perfect CUBE?

Answer 45

2 or 4? 4 cause it makes 2?

Answer 46

5 is not a perfect cube. But if we multiply 5 by 25, we get 125, which IS a perfect cube.
Likewise, x^2 is not a perfect cube. But if we multiply it by x, we get x^3, which is a perfect cube. That's what we have to do here.

Answer 47

So now we're at the point where we have (y/x)(3 over 5x^2)^(1/3).

Answer 48

I'd like for you to please multiply both that 3 and that 5x^2 by 25x. See why?

Answer 49

Oh because the cube root of 125 is 3.

Answer 50

Actually, it's 5. But you know that already.

Answer 51

(y/x) 75/5x?

Answer 52

So now we end up with (y/x)*(75x over 125x^3). Right?

Answer 53

Yes, I cubed did before - sorry ha

Answer 54

I'm sorry, I left out the exponent, (1/3).

Answer 55

So, Nicole, what is the cube root of 1 over 125x^3?

Answer 56

5x? or 1/5x?

Answer 57

Sorry, I'm going to make you choose one!

Answer 58

5x

Answer 59

Right. :)

Answer 60

Yay! Thank you(:

Answer 61

So now your expression reduces to (y/x)*(1 over 5x)*cube root of (75x). Agreed or disagreed?

Answer 62

Agreed

Answer 63

And the product (y/x)*(1 over 5x) simplifies to what?

Answer 64

y/5x^2

Answer 65

Yes, exactly.

So our final answer is (y over 5x^2)*(cube root of 75x).
Whatcha think?

Answer 66

I think I got it(: Thank you!

Answer 67

We make a GREAT team. I've enjoyed this thoroughly. Looking forward to "seeing" you again here on OpenStudy.

Answer 68

Have a good night!

Answer 69

Please review the various steps of the process we went through and try applying it to one more problem, on your own. Good night to you, too! Bye, Nicole!