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anonymous

  • one year ago

4√(80x^17w^8)

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  1. CoconutJJ
    • one year ago
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    4 times sqrt of 4th root ?

  2. anonymous
    • one year ago
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    ?

  3. CoconutJJ
    • one year ago
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    |dw:1439531008723:dw|

  4. anonymous
    • one year ago
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    oh the second one

  5. CoconutJJ
    • one year ago
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    i c.... Assuming you're suppose to simplify, do you remember the exponent laws that apply to radicals

  6. anonymous
    • one year ago
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    sorry not really

  7. CoconutJJ
    • one year ago
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    |dw:1439531324193:dw|

  8. CoconutJJ
    • one year ago
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    |dw:1439531372578:dw|

  9. anonymous
    • one year ago
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    so divide x^17 by 4? like 17/4?

  10. CoconutJJ
    • one year ago
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    yes...

  11. anonymous
    • one year ago
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    would i put it in a decimal form?

  12. jtvatsim
    • one year ago
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    Here's an alternative way to think through these types of questions. First, think about the normal square roots. When you have square roots you need two factors inside in order to "escape" as in \[\sqrt{x^2} = \sqrt{x \cdot x} = x\] but if we only have one factor we are "trapped" \[\sqrt{2} = stuck\] Does that make sense so far?

  13. anonymous
    • one year ago
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    yes

  14. anonymous
    • one year ago
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    i got 2x^4w^4 4th√(5x^13 w^2) @jtvatsim

  15. jtvatsim
    • one year ago
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    k let me check what you did. :)

  16. jtvatsim
    • one year ago
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    You are close, just a few minor errors, but overall it seems like you have the idea. :)

  17. anonymous
    • one year ago
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    can you tell me which part i got wrong?

  18. jtvatsim
    • one year ago
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    Alright, so effectively we want to "bunch" our factors in groups of 4 whenever possible in order to escape the radical. We need groups of 4 because of the 4th root. \[\sqrt[4]{80x^{17}w^8} = \sqrt[4]{5\cdot2^4 \cdot x^4 \cdot x^4 \cdot x^4 \cdot x^4 \cdot x \cdot w^4 \cdot w^4} \] \[= 2\cdot x\cdot x \cdot x \cdot x \cdot w \cdot w\sqrt[4]{5x} = 2x^4w^2\sqrt[4]{5x}\]

  19. anonymous
    • one year ago
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    ohhh that is much easier

  20. jtvatsim
    • one year ago
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    Does that help you "see" it more clearly? I found that explanation to make the most sense when I was learning. :)

  21. anonymous
    • one year ago
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    yes it helped a lot! thank you so much!

  22. jtvatsim
    • one year ago
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    You are welcome! This same trick works no matter what root you are dealing with! Bring on the 100th roots, just bunch into groups of 100 (maybe not... don't give your teachers any ideas...) Have a good one! :)

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