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aroub

  • 3 years ago

sqrt[4]{6} times sqrt[4]{36} times sqrt[4]{192}

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  1. aroub
    • 3 years ago
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    lol, more of a fail -.- So, this is the question: \[\sqrt[4]{6} \times \sqrt[4]{36} \times \sqrt[4]{192}\]

  2. aroub
    • 3 years ago
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    I know it looks easy, I just keep on getting different answer :/

  3. aroub
    • 3 years ago
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    O.o

  4. aroub
    • 3 years ago
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    I got \[2\sqrt[4]{2592}\] Yeah -.- It looks wrong -.-

  5. Hero
    • 3 years ago
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    If it has the same root....

  6. Hero
    • 3 years ago
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    rewrite it like this: \[\large \sqrt[4]{6 \times 36 \times 192}\]

  7. aroub
    • 3 years ago
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    Yes, this is what I did..

  8. phi
    • 3 years ago
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    I would never multiply these numbers together. on the contrary, you want to factor them all are under the root sign. so you have \[ 6 \cdot 6^2 \cdot 6\cdot 2^5 \] or \[ 6^4 \cdot 2^4 \cdot 2\] pull out the powers of 4

  9. RaphaelFilgueiras
    • 3 years ago
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    |dw:1347737858173:dw|

  10. Hero
    • 3 years ago
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    I didn't multiply them together. I just simplified them under one root @phi. I agree that what you did would be the next step AFTER putting them under one root

  11. helder_edwin
    • 3 years ago
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    sorry. i got \[ \large 12\sqrt[4]{2} \]

  12. phi
    • 3 years ago
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    @Hero that was to @aroub not you. I know you know how to do it!

  13. aroub
    • 3 years ago
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    I didn't multiply them together at first. This is was what I did (it's stupid though compared to what you did): \[\sqrt[4]{3\times2 \times 6^2\times2^6\times3} = \sqrt[4]{3^2\times2^4\times6^2\times2^3}\] Don't ask me how or why :P And then after realizing I cant factor them more I multiplied them.

  14. aroub
    • 3 years ago
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    Thank you phi and everyone!! =D

  15. phi
    • 3 years ago
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    doing that is fine. The only mistake is not factor the 6^2 into 2^2 * 3^2 after combining bases, you would have gotten 3^4 2^9 or 3^4 * 2^4 * 2^4 * 2 take the 4th root, you would get 3*2*2 * 2^(1/4) or 12* 2^(1/4)

  16. aroub
    • 3 years ago
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    If i still want the square root that would be equal to \[12\sqrt[4]{2}?\]

  17. phi
    • 3 years ago
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    yes, but it is the 4th root (not the square root)

  18. aroub
    • 3 years ago
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    Oh yeah sorry -.- I'm used to square roots :P

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