anonymous one year ago okay my questions are #32. (-8-√-2)^2 #33. (2-3i)(3-i)-(4-i)(4+i) then i will have one more but i will have to write it after working these too.

1. jim_thompson5910

let's focus on one at a time (one per post to devote the whole post to the question) like before, $\Large \sqrt{-2} = i\sqrt{2}$ do you agree?

2. anonymous

yes i agree

3. jim_thompson5910

|dw:1441069958552:dw|

4. jim_thompson5910

go ahead and fill out the table (or what you can)

5. anonymous

|dw:1441069842863:dw|

6. jim_thompson5910

the root 2 in the lower right corner should be root 4

7. jim_thompson5910

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8. anonymous

then i cross out the -8i√2

9. jim_thompson5910

why cross that out?

10. anonymous

i dont know im not sure how this is gonna help me to solve this problem

11. jim_thompson5910

those terms would go away IF you had +8i*sqrt(2) and -8i*sqrt(2)

12. jim_thompson5910

but instead we have two copies of -8i*sqrt(2)

13. jim_thompson5910

$\large i^2*\sqrt{4} = -1*2 = -2$

14. jim_thompson5910

|dw:1441070689367:dw|

15. anonymous

okay

16. jim_thompson5910

now add up all the terms inside the boxes $\Large 64 + (-8i\sqrt{2})+(-8i\sqrt{2})+(-2) = 62 - 16i\sqrt{2}$

17. jim_thompson5910

make sense?

18. anonymous

how do you get rid of the sqrt on the two or does it need to stay.

19. jim_thompson5910

there's nothing we can do to simplify further, so we leave as is

20. anonymous

21. jim_thompson5910

yes, $\Large \left(-8-\sqrt{-2}\right)^2 = 62 - 16i\sqrt{2}$

22. anonymous

actually i will have one more additional problem but i will post each into a new window. thank you thoughtfor the help bc that was the correct problem.

23. jim_thompson5910

yeah one per post is a good idea to avoid clutter

24. jim_thompson5910

and avoid lag too