1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Simplifying Radicals

  1. Mar 9, 2006 #1
    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 [tex]\sqrt{49y^{12}[/tex]. I cant decide if I have to root it out like this [tex]\sqrt{7y^{6}[/tex] times [tex]\sqrt{7y^{6}[/tex] 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. jcsd
  3. Mar 9, 2006 #2
    [tex]\sqrt{49y^{12}} = 7y^{6}[/tex]
  4. Mar 9, 2006 #3
    Yessh, I'm not thinking clearly. Thanks for knocking me some sense ksinclair13.
  5. Mar 9, 2006 #4
    You're welcome :).

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

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

    [tex]\sqrt{49y^{12}} = \sqrt{(7)^{2}(y^{6})^{2}} = 7y^{6}[/tex]
  6. Mar 9, 2006 #5
    Forgive me for being ignorent but I don't understand that method at all.
  7. Mar 9, 2006 #6
    So, ksinclair13, if I'm understanding what you said [tex]\sqrt{a^{4}[/tex] would be [tex]a^{2}[/tex] and not [tex]a\sqrt{a^{2}[/tex]
  8. Mar 9, 2006 #7
    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:

    [tex]\sqrt{49y^{12}} = \sqrt{49} * \sqrt{y^{12}} = 7*y^{6} = 7y^{6}[/tex]

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

    [tex]\sqrt{25} = \sqrt{5^{2}} = 5[/tex] because the [itex]\sqrt[/itex] and [itex]^{2}[/itex] cancel.
  9. Mar 9, 2006 #8
    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?
  10. Mar 9, 2006 #9

    [tex]\sqrt{x^{a}} = x^{\frac{a}{b}}[/tex]
    [tex]\sqrt[2]{x^{4}} = x^{\frac{4}{2}} = x^{2}[/tex]
    [tex]\sqrt[2]{y^{12}} = y^{\frac{12}{2}} = y^{6}[/tex]
    Last edited: Mar 9, 2006
  11. Mar 9, 2006 #10
    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.
  12. Mar 9, 2006 #11
    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.
  13. Mar 9, 2006 #12
    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.
  14. Mar 10, 2006 #13


    User Avatar
    Homework Helper

    We do not give out COMPLETE SOLUTIONS here, at PhysicsForums!!! :grumpy: :grumpy: :grumpy: :grumpy:
    You may want to skim through some rules here!!!
  15. Mar 10, 2006 #14
    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
  16. Mar 10, 2006 #15
    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.

  17. Mar 10, 2006 #16
    Agreed. You learn by seeing and doing, not always by hearing what people tell you to do.
  18. Mar 10, 2006 #17
    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.
  19. Mar 10, 2006 #18
    You were telling me which one was correct, I knew how to do it I just didn't know which one was correct.
  20. Mar 10, 2006 #19
    Here is another way to look at this type of problem.

    Say : [tex]\sqrt{49y^{12}[/tex]

    this equals
    [tex]\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}[/tex]

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

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

    therefore, [tex]{7y^6[/tex]
    Last edited: Mar 10, 2006
  21. Mar 10, 2006 #20


    User Avatar
    Homework Helper

    Are you sure that [tex]\sqrt{49 y ^ {12}} = 7 y ^ {12}[/tex]?
  22. Mar 12, 2006 #21
    That's not what he had...he had 7y6.:confused: Anways, his method is very helpful, that's what I was taught. Here is an example:

    Sqrt. 24x2y5

    First, factor the 24x2y5 into its prime factors.

    Sqrt. 2 * 2 * 2 * 3 * x * x * y * y * y * y * y

    Now we can simplify the square root using the factorization of 24x2y5 and pull out the roots. Every two factors under the square root are equivalent to one factor outside the square root.

    2 * x * y * y Sqrt.2 * 3 * y

    The x's and y's outside the radical can be multiplied to make xy2.

    2xy2 Sqrt. 6y

    So the square root of 24x2y5 simplifies to 2xy2 times the square root of 6y.


    Does that help any?:smile:
  23. Mar 12, 2006 #22
    None whatsoever, but I've memorized most of the common radicals and I should be able to build up from there.
  24. Mar 12, 2006 #23
    I hope so because I know how hard math can be if you start missing things here and there...it makes future math courses quite hard. Let me see if I can explain this one more time, yeah?:smile:

    Sqrt. 25

    Have you ever done what I believe is called a "factor tree"? You write 25 down on a piece of paper and then start breaking it down. In this number's case, you're only going to be able to write down two 5's. Just in case you need to refresh your memory (I don't know what it is that's confusing you so I figured I'd just go down the line.:rofl: ), http://nlvm.usu.edu/en/nav/frames_asid_202_g_3_t_1.html" [Broken] is a good site you can use to practice.

    The square root of 25 is 5 because 5 squared (Five times itself or 52) is, well, 25.:rofl:

    Sqrt. x6

    Ignoring the Square Root part for now, that basically means:

    x * x * x * x * x * x

    When finding the square root of a number, you're looking for a number that, multiplied by itself, gives you the number inside the sign. Just in case I've confused you...

    Sqrt. 100

    What number times itself equals 100? In other words, what number squared (ex. 32 = 3 * 3 = 9 so 3 squared is 9) equals 100? Answer= 10. See a pattern? When squaring something, you're always looking for a number you can multiply by itself. 2 * 2, 3 * 3, 4 * 4, 5 * 5, etc. You're always multiplying two numbers together.

    Thus, you circle pairs of the same term. You have six x's...

    x * x * x * x * x * x

    You have three pairs of x's. That means you can simplify Sqrt. x6 to simply x3.:smile:

    Let's try mixing the two types of problems now.

    Sqrt. 25x6

    First, break it down.

    Sqrt. 5 * 5 * x * x * x * x * x * x

    Circle pairs of similar terms.

    5 * 5 * x * x * x * x * x * x



    If that doesn't help you, try looking at this site http://regentsprep.org/Regents/math/radicals/Lsimplify.htm".:smile:
    Last edited by a moderator: May 2, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook