## anonymous one year ago 4√(80x^17w^8)

1. CoconutJJ

4 times sqrt of 4th root ?

2. anonymous

?

3. CoconutJJ

|dw:1439531008723:dw|

4. anonymous

oh the second one

5. CoconutJJ

i c.... Assuming you're suppose to simplify, do you remember the exponent laws that apply to radicals

6. anonymous

sorry not really

7. CoconutJJ

|dw:1439531324193:dw|

8. CoconutJJ

|dw:1439531372578:dw|

9. anonymous

so divide x^17 by 4? like 17/4?

10. CoconutJJ

yes...

11. anonymous

would i put it in a decimal form?

12. jtvatsim

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

yes

14. anonymous

i got 2x^4w^4 4th√(5x^13 w^2) @jtvatsim

15. jtvatsim

k let me check what you did. :)

16. jtvatsim

You are close, just a few minor errors, but overall it seems like you have the idea. :)

17. anonymous

can you tell me which part i got wrong?

18. jtvatsim

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

ohhh that is much easier

20. jtvatsim

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

21. anonymous

yes it helped a lot! thank you so much!

22. jtvatsim

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! :)