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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.

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  1. HorseCrazyGirlForever
    • 3 years ago
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    |dw:1356140871952:dw|

  2. zepdrix
    • 3 years ago
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    \[\huge \sqrt[3]{16}+5\;\sqrt[3]{54}\]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, ...\[\huge 2 \cdot2\cdot2=8 \qquad \qquad 2^3=8\]

  3. zepdrix
    • 3 years ago
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    Since neither of our numbers is larger than 64, we'll be trying to find factors of 27 or 8 under the roots.

  4. zepdrix
    • 3 years ago
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    \[\huge \sqrt[3]{16} \quad \rightarrow \sqrt[3]{\color{orangered}{2}\cdot \color{orangered}{8}}\]Understand how I did the first part? :D

  5. HorseCrazyGirlForever
    • 3 years ago
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    Yes.

  6. zepdrix
    • 3 years ago
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    \[\huge \sqrt[3]{\color{orangered}{2}\cdot\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot \sqrt[3]{\color{orangered}{8}} \qquad = \qquad \sqrt[3]{\color{orangered}{2}}\cdot 2\]

  7. zepdrix
    • 3 years ago
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    Have any ideas for the second one? :)

  8. HorseCrazyGirlForever
    • 3 years ago
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    So you are trying to break it down as much as possible?

  9. zepdrix
    • 3 years ago
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    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.

  10. HorseCrazyGirlForever
    • 3 years ago
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    Ohhh I see! So the factor of 54 would be 27?

  11. zepdrix
    • 3 years ago
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    Yah I think that works! 27*2 = 54

  12. HorseCrazyGirlForever
    • 3 years ago
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    Awesomeness!

  13. HorseCrazyGirlForever
    • 3 years ago
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    So it would just be written like that?

  14. zepdrix
    • 3 years ago
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    \[\huge \sqrt[3]{54}\qquad =\qquad \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\qquad = \qquad \sqrt[3]{2}\cdot 3\]That part make sense? We have a little bit more work to do on this one.

  15. HorseCrazyGirlForever
    • 3 years ago
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    I understand the first part, but I don't under stand how 2*27 turns into 2*3...

  16. zepdrix
    • 3 years ago
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    \[\huge \sqrt[3]{\color{royalblue}{2}\cdot \color{royalblue}{27}}\]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.\[\huge 3^3=27\]

  17. zepdrix
    • 3 years ago
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    So taking the cube root of 27 should give us 3.

  18. HorseCrazyGirlForever
    • 3 years ago
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    Oh okay.

  19. zepdrix
    • 3 years ago
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    So our problem has simplified to this,\[\huge \sqrt[3]{16}+5\;\sqrt[3]{54} \qquad \rightarrow \qquad 2\sqrt[3]{2}+5(3\sqrt[3]{2})\]

  20. HorseCrazyGirlForever
    • 3 years ago
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    Okay I understand all of that. So how would I solve the rest?

  21. zepdrix
    • 3 years ago
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    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.\[\huge 17\; \sqrt[3]{2}\]

  22. zepdrix
    • 3 years ago
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    If this part is confusing you, maybe think of the cuberoot 2 as something simpler, like x.\[\huge 2x+15x=17x\]

  23. HorseCrazyGirlForever
    • 3 years ago
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    Awesome! Thank you so much! I have a few more, could you please help me with those to?

  24. zepdrix
    • 3 years ago
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    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

  25. HorseCrazyGirlForever
    • 3 years ago
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    Okay! Thank you so much!

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