A community for students.
Here's the question you clicked on:
 0 viewing
HorseCrazyGirlForever
 3 years ago
Simplify the following roots. Please click on my question to see the problems because I have to draw it out. Thanks.
HorseCrazyGirlForever
 3 years ago
Simplify the following roots. Please click on my question to see the problems because I have to draw it out. Thanks.

This Question is Closed

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356140871952:dw

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \sqrt[3]{16}+5\;\sqrt[3]{54}\]Hmm so this is a bit tricky. We want to simplify the roots. So we need to find FACTORS of those numbers that are PERFECT CUBES. Examples of Perfect Cubes > 8, 27, 64, ...\[\huge 2 \cdot2\cdot2=8 \qquad \qquad 2^3=8\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Since neither of our numbers is larger than 64, we'll be trying to find factors of 27 or 8 under the roots.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \sqrt[3]{16} \quad \rightarrow \sqrt[3]{\color{orangered}{2}\cdot \color{orangered}{8}}\]Understand how I did the first part? :D

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \sqrt[3]{\color{orangered}{2}\cdot\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot \sqrt[3]{\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot 2\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Have any ideas for the second one? :)

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0So you are trying to break it down as much as possible?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Not AS MUCH AS POSSIBLE. We certainly could have broken the 8 down a little further, into 4*2, and then into another 2*2*2.. But that wouldn't have helped us. We're trying to break down the number into a perfect cube. 8 is a perfect cube, so we were able to break it down to 8 nicely.

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0Ohhh I see! So the factor of 54 would be 27?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yah I think that works! 27*2 = 54

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0Awesomeness!

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0So it would just be written like that?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \sqrt[3]{54}\qquad =\qquad \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\qquad = \qquad \sqrt[3]{2}\cdot 3\]That part make sense? We have a little bit more work to do on this one.

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0I understand the first part, but I don't under stand how 2*27 turns into 2*3...

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\]From here, remember that you can split up the root if you want. So we'll write it like so,\[\huge \sqrt[3]{\color{royalblue}{2}}\cdot \sqrt[3]{\color{royalblue}{27}}\] From here, we just need to remember that 27 is a perfect square.\[\huge 3^3=27\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So taking the cube root of 27 should give us 3.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So our problem has simplified to this,\[\huge \sqrt[3]{16}+5\;\sqrt[3]{54} \qquad \rightarrow \qquad 2\sqrt[3]{2}+5(3\sqrt[3]{2})\]

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0Okay I understand all of that. So how would I solve the rest?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Let's multiply the 3 and 5 out.\[\huge 2\sqrt[3]{2}+5(3\sqrt[3]{2}) \qquad =\qquad 2\;\sqrt[3]{2}+15\;\sqrt[3]{2}\] From here, you might notice that we have similar terms. Sure, they're kinda ugly but they should combine nonetheless.\[\huge 17\; \sqrt[3]{2}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1If this part is confusing you, maybe think of the cuberoot 2 as something simpler, like x.\[\huge 2x+15x=17x\]

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0Awesome! Thank you so much! I have a few more, could you please help me with those to?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Close this thread, open a new one. In the description somewhere, type @zepdrix It will give me a little popup, so I can easily find your question. There are lots of good helpers on here. So if I can't get to your question I'm sure someone can. But yah I'll come take a look ^^ heh

HorseCrazyGirlForever
 3 years ago
Best ResponseYou've already chosen the best response.0Okay! Thank you so much!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.