## anonymous 4 years ago Can someone please check my work?? Wanna make sure I'm doing this right! Simply each number by using the imaginary number i. thanks!

1. anonymous

$\sqrt{-7}$ = 7i or $i \sqrt{7}$

2. anonymous

$\sqrt{-81}$ = 3i

3. anonymous

$\sqrt{-16}$ = $2i \sqrt{4}$

4. anonymous

$3\sqrt{-9}$ = $3i$

5. anonymous

The last one is wrong.

6. anonymous

@dido525 Oh wow!! Thank you. I can't believe I'm doing these somewhat right lol. Would it be $3i \sqrt{3}$ ?? And there was also a problem that was $\sqrt{-72}$ I got: $2i \sqrt{3}$

7. anonymous

The last one should be 9i.

8. anonymous

Well think about it WITHOUT the negative.

9. anonymous

|dw:1352593737003:dw|

10. anonymous

|dw:1352593754329:dw|

11. anonymous

Now what would you do?

12. anonymous

is that a 4 and 18? I would think to factor them to 2 x 2 and 9 x 2 maybe

13. anonymous

14. anonymous

|dw:1352593862710:dw|

15. anonymous

@Dido525

16. anonymous

No No.

17. anonymous

|dw:1352594047004:dw|

18. anonymous

@Dido525 Oh ok! That made since. I was using 8, 9, and i to make -72

19. anonymous

Okay :) .

20. anonymous

@Dido525 Thank you so much for all your help :) Makes me feel more confident about simplying them!

21. anonymous

*simplifying whoops

22. Zarkon

$\sqrt{-72}=\sqrt{(-1)\cdot2\cdot 36}$ $=6i\sqrt{2}$

23. anonymous

Okay that works too.

24. anonymous

Getting the i right, but need to look at $\sqrt{81} and \sqrt{16}$

25. anonymous

@chaguanas Would I do $3i \sqrt{3}$ for -81? and 4i or $4i \sqrt{4}$ for -16?

26. Zarkon

do you know what $$\sqrt{81}$$is?

27. anonymous

@Zarkon 9

28. Zarkon

well...$\sqrt{-81}=\sqrt{(-1)81}=i9=9i$

29. anonymous

@Zarkon Ohh so that's how you do it. You wouldn't factor the 9 out to a 3?

30. Zarkon

no

31. Zarkon

for any $$a>0$$ $\sqrt{-a^2}=a\cdot i$ so $\sqrt{-81}=\sqrt{-9^2}=9i$

32. Zarkon

$\sqrt{-16}=\sqrt{-4^2}=4i$

33. anonymous

@Zarkon Oh alright! I'm not sure if my teacher taught me about the a>0 thing but that makes sense. Thank you very much :) Explained a lot.