1. Mar 9, 2006

### wScott

Started this today and I'm okay with simplifying radicals that don't include variables, but I'm having trouble solving ones with variables, like $$\sqrt{49y^{12}$$. I cant decide if I have to root it out like this $$\sqrt{7y^{6}$$ times $$\sqrt{7y^{6}$$ or just root it out the way it is since it is in itself a perfect square. Please help, I didn't get very clear instructions in class.

2. Mar 9, 2006

### ksinclair13

$$\sqrt{49y^{12}} = 7y^{6}$$

3. Mar 9, 2006

### wScott

Yessh, I'm not thinking clearly. Thanks for knocking me some sense ksinclair13.

4. Mar 9, 2006

### ksinclair13

You're welcome :).

You were correct in saying that $\sqrt{49y^{12}} = \sqrt{7y^{6}}*\sqrt{7y^{6}}$. That's a good way to show your work, but your final answer will be $7y^{6}$.

My old algebra teacher made us show our work like this:

$$\sqrt{49y^{12}} = \sqrt{(7)^{2}(y^{6})^{2}} = 7y^{6}$$

5. Mar 9, 2006

### wScott

Forgive me for being ignorent but I don't understand that method at all.

6. Mar 9, 2006

### wScott

So, ksinclair13, if I'm understanding what you said $$\sqrt{a^{4}$$ would be $$a^{2}$$ and not $$a\sqrt{a^{2}$$

7. Mar 9, 2006

### ksinclair13

Showing that step is basically pointless. It's easier to just do it in your head. He just wanted us to show at least some work so that we wouldn't simply copy all the answers out of the back of the book.

Here's a better way to think about these problems:

$$\sqrt{49y^{12}} = \sqrt{49} * \sqrt{y^{12}} = 7*y^{6} = 7y^{6}$$

Here's a more simple example of what my teacher was trying to get across:

$$\sqrt{25} = \sqrt{5^{2}} = 5$$ because the $\sqrt$ and $^{2}$ cancel.

8. Mar 9, 2006

### wScott

Well I think we're learning how to multiply radicals tomorrow but what you say makes perfect sanse to me. Now wht about that question I just asked?

9. Mar 9, 2006

### ksinclair13

Correct.

$$\sqrt{x^{a}} = x^{\frac{a}{b}}$$
$$\sqrt[2]{x^{4}} = x^{\frac{4}{2}} = x^{2}$$
$$\sqrt[2]{y^{12}} = y^{\frac{12}{2}} = y^{6}$$

Last edited: Mar 9, 2006
10. Mar 9, 2006

### wScott

Thanks a lot for your help ksinclair13. I will no doubt be on here sometiem soon with another question, I look forward to your help.

11. Mar 9, 2006

### ksinclair13

No problem :). However, you shouldn't necessarily look forward to my help, for I haven't learned advanced mathematics yet. You should look forward to the very qualified members of this forum helping you.

12. Mar 9, 2006

### wScott

True, but I always value you opinion of the underdogs.. no offence intended.

Well I've got to go do my science homework and read on New France for my Canadian History project. See you later.

13. Mar 10, 2006

### VietDao29

Aaarrgghhhh!!!!!!!!
We do not give out COMPLETE SOLUTIONS here, at PhysicsForums!!! :grumpy: :grumpy: :grumpy: :grumpy:
You may want to skim through some rules here!!!

14. Mar 10, 2006

### wScott

It may have been a complete solution Viet, and he probobly should of given me the directions on how to do it rather than giving me the answer right up front. But I just needed to know how to do it, since I already gave him two of my workings for a question "similar", not exactly the same, to that of a question on my homework for that night. And thanks to his confermation I got 72 questions out of 76 right on my homework check.

Edit:Sorry, I just looked over the rules again, and it said to try my very best. I looked over all my notes from that days class and i couldn't figure out how to do it correctly so I posted a question similar to that it. I'm sorry if we broke the rules but I was just getting confermation on how to do a question, it was like a "lesson" for me.

Last edited: Mar 10, 2006
15. Mar 10, 2006

### topsquark

The Socratic teaching method is one of the best, but even it sometimes needs to allow that occasionally doing an example is more educational than guiding one through it.

-Dan

16. Mar 10, 2006

### wScott

Agreed. You learn by seeing and doing, not always by hearing what people tell you to do.

17. Mar 10, 2006

### ksinclair13

It all worked out in the end. I should've just said everything that I said later in my first post instead of just posting the answer. It's just that he basically already had the correct answer, I was just writing it more clear for him.

18. Mar 10, 2006

### wScott

You were telling me which one was correct, I knew how to do it I just didn't know which one was correct.

19. Mar 10, 2006

### konartist

Here is another way to look at this type of problem.

Say : $$\sqrt{49y^{12}$$

this equals
$$\sqrt{7 \times 7 \times y \times y \times y \times y \times y \times y \times y \times y\times y\times y\times y \times y}$$

Since 2 is implied in front of the $$\sqrt$$ how many groups of 2's do you see in this radicand?

you see one group of TWO 7's
and
SIX groups of TWO y's

therefore, $${7y^6$$

Last edited: Mar 10, 2006
20. Mar 10, 2006

### VietDao29

Are you sure that $$\sqrt{49 y ^ {12}} = 7 y ^ {12}$$?
:)