Algebra II Simplifying Radicals Using Product and Quotient Properties

Click For Summary
SUMMARY

The discussion centers on simplifying the expression \(\sqrt[3]{\frac{5}{4}}\) using the product and quotient properties of cube roots. The correct simplification is \(\frac{\sqrt[3]{10}}{2}\), achieved by rewriting the fraction as \(\frac{5}{4} \cdot \frac{2}{2}\) to form \(\frac{10}{8}\). Participants emphasized the importance of recognizing perfect cubes and suggested multiplying by \(\sqrt[3]{16}\) instead of \(\sqrt[3]{4}\) to facilitate the simplification process. The user expressed gratitude for the hints provided, which clarified the solution method.

PREREQUISITES
  • Understanding of cube roots and their properties
  • Familiarity with simplifying fractions
  • Knowledge of perfect cubes and their significance in algebra
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the properties of cube roots in detail
  • Practice simplifying various radical expressions
  • Learn techniques for identifying perfect cubes in algebraic fractions
  • Explore advanced topics in radical expressions and their applications
USEFUL FOR

Students studying Algebra II, educators teaching radical simplification techniques, and anyone looking to improve their skills in manipulating cube roots and algebraic expressions.

velox_xox
Messages
34
Reaction score
0

Homework Statement


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

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

Homework Equations


--


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!
 
Physics news on Phys.org
...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!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
5K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K