## anonymous one year ago Can someone help me with this? √45n^5

1. SolomonZelman

$$\large\sqrt{45n^5}$$ like this?

2. anonymous

Yeah like that.

3. SolomonZelman

$$\color{black}{ \displaystyle \sqrt{45n^5}=\sqrt{9\cdot 5 \cdot n^4\cdot n} =\sqrt{9\cdot n^4}~\cdot ~\sqrt{5\cdot n}=\sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }$$

4. SolomonZelman

that is how i brake them down.

5. SolomonZelman

what is √9, can you tell me?

6. anonymous

3?

7. SolomonZelman

yes

8. anonymous

How did we lose an n from n^5?

9. SolomonZelman

we didn't, all I did is used the fact that $$n^5=n^{4+1}=n^4\cdot n^1=n^4\cdot n$$

10. SolomonZelman

and of course, I rearranged them putting the terms that can be simplified in the beginning and then the terms that can't be simplified

11. SolomonZelman

$$\sqrt{n^4}$$ is same as $$\sqrt[2]{n^4}$$ and that is =$$n^{4/2}={\rm to~what?}$$

12. anonymous

Oh alright..

13. SolomonZelman

ok, did you get up to finding the $$\sqrt{n^4}$$ ?

14. anonymous

Nope don't understand what's going on this time.

15. SolomonZelman

$$\sqrt{n^4}=\sqrt[2]{n^4}$$ this is because whenever the root is not specificed, it is 2 (i.e. SQUARE root)

16. SolomonZelman

now, applying the rule: $$\sqrt[c]{a^b}=a^{b/c}$$ we get that $$\sqrt[2]{n^4}=n^{4/2}=n^{2/1}=n^2$$

17. anonymous

Okay, didn't know that.

18. SolomonZelman

We had: $$\color{black}{ \displaystyle \sqrt{9} \cdot \sqrt{ n^4}~\cdot ~\sqrt{5\cdot n} }$$ and now we had: $$\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }$$

19. SolomonZelman

oh, now you know.... :0

20. anonymous

Thanks.

21. SolomonZelman

$$\color{black}{ \displaystyle3 \cdot n^2\cdot ~\sqrt{5\cdot n} }$$ is just same as $$\color{black}{ \displaystyle3 n^2~\sqrt{5 n} }$$ $$\bf ...$$yw