anonymous
  • anonymous
Simplify the following roots. Please click on my question to see the problems because I have to draw it out. Thanks.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1356140871952:dw|
zepdrix
  • zepdrix
\[\huge \sqrt[3]{16}+5\;\sqrt[3]{54}\]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
  • zepdrix
Since neither of our numbers is larger than 64, we'll be trying to find factors of 27 or 8 under the roots.

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zepdrix
  • zepdrix
\[\huge \sqrt[3]{16} \quad \rightarrow \sqrt[3]{\color{orangered}{2}\cdot \color{orangered}{8}}\]Understand how I did the first part? :D
anonymous
  • anonymous
Yes.
zepdrix
  • zepdrix
\[\huge \sqrt[3]{\color{orangered}{2}\cdot\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot \sqrt[3]{\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot 2\]
zepdrix
  • zepdrix
Have any ideas for the second one? :)
anonymous
  • anonymous
So you are trying to break it down as much as possible?
zepdrix
  • zepdrix
Not 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.
anonymous
  • anonymous
Ohhh I see! So the factor of 54 would be 27?
zepdrix
  • zepdrix
Yah I think that works! 27*2 = 54
anonymous
  • anonymous
Awesomeness!
anonymous
  • anonymous
So it would just be written like that?
zepdrix
  • zepdrix
\[\huge \sqrt[3]{54}\qquad =\qquad \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\qquad = \qquad \sqrt[3]{2}\cdot 3\]That part make sense? We have a little bit more work to do on this one.
anonymous
  • anonymous
I understand the first part, but I don't under stand how 2*27 turns into 2*3...
zepdrix
  • zepdrix
\[\huge \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\]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
  • zepdrix
So taking the cube root of 27 should give us 3.
anonymous
  • anonymous
Oh okay.
zepdrix
  • zepdrix
So our problem has simplified to this,\[\huge \sqrt[3]{16}+5\;\sqrt[3]{54} \qquad \rightarrow \qquad 2\sqrt[3]{2}+5(3\sqrt[3]{2})\]
anonymous
  • anonymous
Okay I understand all of that. So how would I solve the rest?
zepdrix
  • zepdrix
Let'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
  • zepdrix
If this part is confusing you, maybe think of the cuberoot 2 as something simpler, like x.\[\huge 2x+15x=17x\]
anonymous
  • anonymous
Awesome! Thank you so much! I have a few more, could you please help me with those to?
zepdrix
  • zepdrix
Close 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
anonymous
  • anonymous
Okay! Thank you so much!

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