Algebra II Simplifying Radicals Using Product and Quotient Properties

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Homework Help Overview

The discussion revolves around simplifying the expression \(\sqrt[3]{\frac{5}{4}}\) using properties of radicals in the context of Algebra II. Participants are exploring various methods to arrive at the textbook answer of \(\frac{\sqrt[3]{10}}{2}\).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to separate the numerator and denominator into individual cube roots and questions the correctness of their approach. Some participants suggest alternative methods, such as multiplying by different forms of 1 or changing the fraction to an equivalent form to facilitate simplification.

Discussion Status

Participants are actively engaging with the problem, offering hints and alternative strategies without reaching a consensus. The original poster expresses appreciation for the guidance received and indicates a growing understanding of the concepts involved.

Contextual Notes

The original poster requests minimal hints to encourage independent problem-solving, indicating a desire to learn and grasp the material thoroughly. There is a mention of the challenge of receiving direct answers from peers, which influences their approach to seeking help.

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Homework Statement


Simplify. \sqrt[3]{\frac{5}{4}}

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

Homework Equations


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The Attempt at a Solution



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

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!
 
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...but \sqrt[3]{4}^2 \neq 4
\sqrt[3]{4}=\sqrt[3]{2^2} = \sqrt[3]{2}^2
 
Instead of multiplying by the cube root of 4 over itself, multiply by the cube root of 16 over itself.
 
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 \sqrt[3]{10} in the numerator and 2 in the denominator.

Hint : \sqrt[3]{2^{3}}
 
Or (pretty much the same) since 4= 2^2, you need another "2" to get a perfect cube:
\sqrt[3]{\frac{5}{4}}= \sqrt[3]{\frac{5}{4}\frac{2}{2}}= \sqrt[3]{\frac{10}{8}}
= \frac{\sqrt[3]{10}}{2}
 
... 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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