## anonymous one year ago evaluate the radical expression and express the result in a+bi form

1. anonymous

$(3-\sqrt{-5}) (1+\sqrt{-1})$

2. campbell_st

well you need to distribute so it is $3(1 + \sqrt{-1})-\sqrt{-5}(1 + \sqrt{-1})$ what do you think the next line of working is..?

3. anonymous

$3(1+i)-i \sqrt{5}(1+i)$

4. anonymous

?

5. UnkleRhaukus

good, now distribute the 3, and the -i√5

6. anonymous

$(3+3i)(-i \sqrt{5}+1\sqrt{5}) ?$

7. UsukiDoll

$(3-\sqrt{-5}) (1+\sqrt{-1})$ $(3-\sqrt{5}i) (1+i)$ now expand.

8. UsukiDoll

all negatives in the radical should be pulled out first.. so that square root of -5 should be square root of 5 i

9. UnkleRhaukus

$3(1+i)-i \sqrt{5}(1+i)\\=(3+3i)+(-i \sqrt{5}-\sqrt{5}i\times i)$

10. UsukiDoll

$(3-\sqrt{5}i) (1+i)$ $3+3i-\sqrt{5}i-\sqrt{5}(i)(i)$ note $i^2 = -1$

11. UsukiDoll

typed too fast... -1 inside the square root is just an i

12. anonymous

$(3+\sqrt{5})+(3-\sqrt{5})i$

13. anonymous

is that right?

14. UsukiDoll

hold on.

15. UsukiDoll

$3+3i-\sqrt{5}i-\sqrt{5}(-1)$ $3+3i-\sqrt{5}i+\sqrt{5}$ $3+\sqrt{5}+3i-\sqrt{5}i$ $\[3+\sqrt{5}+i(3-\sqrt{5})$\] yeah it's correct

16. UsukiDoll

my i is placed differently, but it shouldn't matter because we still have a+bi only our a =$3+\sqrt{5}$ and b = $3-\sqrt{5}$

17. anonymous

thank you so much!

18. UsukiDoll

it's best to convert all negatives in the square root to i's first and if it's a perfect square like $\sqrt{-1}$ just take the square root and add an i $\sqrt{-1} \rightarrow i$ similarly for $\sqrt{-5} \rightarrow \sqrt{5}i$

19. UsukiDoll

but 5 isn't a perfect square so leave it in the radical and only the negative pops out of the radical and becomes i

20. UsukiDoll

negatives inside the radical produce imaginary results.

21. UsukiDoll

then use foil and i^2 = -1 ... simplify until a+bi or ai+b form is achieved.