## anonymous one year ago Solve 5x2 = -30x - 65.

1. myininaya

$5x^2=-30x-65 \\ 5x^2+30x+65=0$ you could divide both sides by 5 and see if you end up with something factorable on the the left hand side or you can just use quadratic formula

2. anonymous

i tried that and i wound up with $\frac{ 6\pm \sqrt{-16} }{ -2}$

3. myininaya

$\text{ dividing both sides by 5} \\ x^2+6x+13=0$

4. myininaya

$x=\frac{-6 \pm \sqrt{6^2-4(1)(13)}}{2(1)} \\ =\frac{-6 \pm \sqrt{36-52}}{2}=\frac{-6 \pm \sqrt{-16}}{2} =\frac{-6 \pm \sqrt{-1(16)}}{2} \\ =\frac{-6 \pm \sqrt{-1} \sqrt{16}}{2}$ you are right so far

5. myininaya

just need to find sqrt(16) and just recall how we rename sqrt(-1)

6. anonymous

how did you get -1 in radical and 16 ?

7. anonymous

so 4? sqrt -1?

8. anonymous

a. x = -3 ± 2i b.x = -3 ± 4i c.x = -6 ± 2i d.x = -6 ± 4i These are the answers that I got and I'm so confused

9. anonymous

wait would it be d?

10. myininaya

$=\frac{-6 \pm i (4)}{2} \\ =\frac{-6}{2} \pm i \frac{4}{2}$ can you simplify? Are you asking where I got sqrt(-16) didn't you also get that for that one part? That is what your expression said.

11. anonymous

oh so A i got it ! lol

12. anonymous

can you help me with this one as well ? Solve -3x2 + 30x - 90 = 0.

13. myininaya

try dividing both sides by -3 and use quadratic formula again

14. anonymous

lol hold on a baby stole my pen and paper lol

15. myininaya

oh no!

16. anonymous

i got it back lol

17. myininaya

see how far you can get with the problem and if you can't get all the way through it post what you can get through

18. anonymous

ok

19. anonymous

k im stuck on almost the last part im at $\frac{ -10\pm \sqrt{220} }{ 2 }$

20. myininaya

let me check what you have so far one sec

21. myininaya

$-3x^2+30x-90=0 \\ \text{ divide both sides by -3 just \to make things easier \to play with } \\ x^2-10x+30=0 \\ x=\frac{-(-10) \pm \sqrt{(-10)^2-4(1)(30)}}{2(1)}$

22. myininaya

so -(-10)=10 and 100-120=-20

23. myininaya

$x=\frac{10 \pm \sqrt{-20}}{2}$

24. myininaya

now sqrt(-20)=sqrt(-1)*sqrt(20) but there is a perfect square factor in 20 sqrt(-20)=sqrt(-1)*sqrt(4)*sqrt(5)

25. myininaya

$x=\frac{10 \pm \sqrt{-1} \sqrt{4} \sqrt{5}}{2}$ see if you can simplify from there

26. anonymous

is it this?x equals 10 plus or minus 2i square root of 5

27. myininaya

all over 2 right ?

28. anonymous

huh?

29. myininaya

$x=\frac{ 10 \pm i (2)\sqrt{5}}{2}$ you didn't say anything about the 2 on bottom

30. anonymous

ohj yea

31. myininaya

anyways you can separate the fraction and simplify a bit more

32. anonymous

with the 10 and the 2?

33. myininaya

(A+B)/C=A/C+B/C so from previous we have: $x=\frac{10}{2} \pm i \frac{2}{2}\sqrt{5}$

34. anonymous

x equals 5 plus or minus i square root of 5?

35. myininaya

yes since 10/2=5 and 2/2=1

36. anonymous

can you help me with the rest xDDD lol

37. myininaya

i think i can help with one more but I want to see you work it all the way to the end and I will check it and we can make changes if needed

38. anonymous

Solve 5x2 = -30x - 65.

39. myininaya

that is the same as the first equation you put

40. myininaya

but we can do it again if you didn't understand it

41. anonymous

oh lol i put the wrong one cx

42. anonymous

Solve x2 + 4x + 8 = 0.

43. myininaya

since we already did this one I will be a bit more detailed with the explanation: $5x^2=-30x-65 \\ \text{ I first added } 30x \text{ and } 65 \text{ on both sides } \\ 5x^2+30x+65=0 \\ \text{ I then notice all terms are divisible by 5} \\ \text{ So I divided both sides by 5} \\ x^2+6x+13=0 \\ \text{ comparing this to } ax^2+bx+c=0 \\ \text{ I identified } a=1, b=6,c=13 \\ \text{ so using the quadratic formula we have } \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(6)\pm \sqrt{(6)^2-4(1)(13)}}{2(1)} \\ x=\frac{-6 \pm \sqrt{36-52}}{2} \\ x=\frac{-6 \pm \sqrt{-16}}{2} \\ \text{ now } \sqrt{-16}=\sqrt{-1(16)}=\sqrt{-1} \sqrt{16}=i (4)=4i \\ \text{ so we have } \\ x=\frac{-6 \pm 4i}{2} \\ \text{ now seperating the fraction } \\ x=\frac{-6}{2}\pm\frac{4i}{2} \\ x=\frac{-6}{2}\pm\frac{4}{2}i \\ x=-3 \pm 2i$ now I see you have another equation there see if you can follow what I did here to solve your equation

44. myininaya

45. myininaya

which I think your problem isn't identifying a,b, and c

46. myininaya

your problem is playing with the square root and separating the fraction

47. myininaya

$x^2+4x+8=0 \\ \text{ is the same as } 1x^2+4x+8=0 \\ \\ a=1,b=4,c=8 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(4) \pm \sqrt{4^2-4(1)(8)}}{2(1)} \\$ what do you get when you simplify this monster (and he isn't a bad looking monster; just a few teeth)

48. anonymous

$\frac{ -4\pm \sqrt{16-32} }{ 2 }$ so $\frac{ -4\pm \sqrt{-16} }{ 2 }$ but im stuck now :/

49. myininaya

beautiful so far

50. myininaya

I actually simplify sqrt(-16) for you above

51. anonymous

would i divide the -16 by 4?

52. myininaya

sqrt(16)=4 sqrt(-1)=i you have sqrt(-16) though sqrt(-16) remember -16 is same as -1*16 so sqrt(-16)=sqrt(-1*16) =sqrt(-1)*sqrt(16) you should be able to identify sqrt(-1) as i and you should be able to simplify sqrt(16) since you know 4^2=16 (Where 4>0) then sqrt(16)=?

53. anonymous

so would it be x equals negative 4 plus or minus 2 I

54. myininaya

all over ?

55. anonymous

2

56. myininaya

oh wait and why 2?

57. myininaya

sqrt(16)=4 not 2

58. anonymous

cause it was all over two

59. anonymous

minus what i said

60. myininaya

$x^2+4x+8=0 \\ \text{ is the same as } 1x^2+4x+8=0 \\ \\ a=1,b=4,c=8 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(4) \pm \sqrt{4^2-4(1)(8)}}{2(1)} \\ \\ x=\frac{-4 \pm \sqrt{16-32}}{2} =\frac{-4 \pm \sqrt{-16}}{2} \\ x=\frac{-4 \pm 4i }{2} \\ \text{ now separate the fraction and simplify a bit more }$

61. anonymous

so $-2\pm2i$

62. myininaya

yes

63. myininaya

since -4/2=-2 and 4/2=2 so yep $x=\frac{-4}{2} \pm \frac{4}{2}i =-2 \pm 2i$

64. anonymous

i think i got the rest <3 thank you . now i gotta fill out some more job apps :/ ugh growing up sucks wish there was such thing as never land

65. myininaya

with peterpan remember those kids never get pizza

66. anonymous

that is true but i would find a way to create it xD

67. anonymous

thank you again!!

68. myininaya

np :)

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