4√(80x^17w^8)

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4√(80x^17w^8)

Mathematics
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4 times sqrt of 4th root ?
?
|dw:1439531008723:dw|

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oh the second one
i c.... Assuming you're suppose to simplify, do you remember the exponent laws that apply to radicals
sorry not really
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so divide x^17 by 4? like 17/4?
yes...
would i put it in a decimal form?
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?
yes
i got 2x^4w^4 4th√(5x^13 w^2) @jtvatsim
k let me check what you did. :)
You are close, just a few minor errors, but overall it seems like you have the idea. :)
can you tell me which part i got wrong?
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. \[\sqrt[4]{80x^{17}w^8} = \sqrt[4]{5\cdot2^4 \cdot x^4 \cdot x^4 \cdot x^4 \cdot x^4 \cdot x \cdot w^4 \cdot w^4} \] \[= 2\cdot x\cdot x \cdot x \cdot x \cdot w \cdot w\sqrt[4]{5x} = 2x^4w^2\sqrt[4]{5x}\]
ohhh that is much easier
Does that help you "see" it more clearly? I found that explanation to make the most sense when I was learning. :)
yes it helped a lot! thank you so much!
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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