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Algebra II Simplifying Radicals Using Product and Quotient Properties

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Simplify. [itex]\sqrt[3]{\frac{5}{4}} [/itex]

    The answer according to the textbook is: [itex]\frac{\sqrt [3]{10}}{2} [/itex]

    2. Relevant equations
    --


    3. The attempt at a solution

    Separated numerator and deonominator into individual cube roots and multiplied both by [itex]\sqrt[3] {4} [/itex].
    [tex]\frac{\sqrt [3] {5}}{\sqrt [3] {4}}\cdot \frac {\sqrt [3] {4}}{\sqrt [3] {4}} [/tex]
    Multiplied.
    [tex] \frac {\sqrt [3] 20}{4} [/tex]

    And that is as far as I've gotten. Did I do the process wrong? Because I do see that '20' and '4' could be simplified to '10' and '2' respectively, but not from the way I did it. I shall endeavor to not bend the rules of algebra to get the right answer, so what way should I have done it instead?

    Also, if I could request the "tiniest" hints possible and then if I don't get it, increasingly larger hints; I would greatly appreciate it. (I'm trying my best to challenge myself to grasp the subject as much as I can on my own. And I do have people readily around me that are very good at math, but the problem is they just give the answer and process away without letting me earn it. That's why I'm here at PF.) So, please and thanks in advance!
     
  2. jcsd
  3. May 23, 2012 #2
    ...but [itex]\sqrt[3]{4}^2 \neq 4[/itex]
    [tex]\sqrt[3]{4}=\sqrt[3]{2^2} = \sqrt[3]{2}^2 [/tex]
     
  4. May 23, 2012 #3

    Mark44

    Staff: Mentor

    Instead of multiplying by the cube root of 4 over itself, multiply by the cube root of 16 over itself.
     
  5. May 23, 2012 #4
    Or try changing the original fraction inside the cube root to an equivalent fraction, so that when you take the cube root of the numerator and denominator separately, you get [itex]\sqrt[3]{10}[/itex] in the numerator and [itex]2[/itex] in the denominator.

    Hint : [itex]\sqrt[3]{2^{3}} [/itex]
     
  6. May 23, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Or (pretty much the same) since [itex]4= 2^2[/itex], you need another "2" to get a perfect cube:
    [tex]\sqrt[3]{\frac{5}{4}}= \sqrt[3]{\frac{5}{4}\frac{2}{2}}= \sqrt[3]{\frac{10}{8}}[/tex]
    [tex]= \frac{\sqrt[3]{10}}{2}[/tex]
     
  7. May 23, 2012 #6
    ... Wow. That should have been obvious to me; I was looking at a cube root as a square root. *sigh*

    Thank you everyone for your help. I just now logged in and was able to decipher your hints very quickly. I tried it for myself, and it made sense. Sorry, that was an obvious one for everybody!

    Thank you Joffan, Mark44, BloodyFrozen, and HallsofIvy. I appreciate your speedy and helpful replies!
     
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