## anonymous one year ago SOMEONE PLEASE HELP ME WITH THIS ALGEBRA 2 . its hard and im really stressed

1. Jhannybean

Well, whats the question? :D

2. anonymous

Match the complex expression to its correct simplified form. 1. i^22 2. i&7 3. 5 + 8i - 4 + 2i 4. (2 - 3i)(3 + 5i) 5. (2 - i)(2 + i)

3. Astrophysics

i^22 = -1

4. UsukiDoll

hmm... well for the first one , we know that $i^2= -1$ so we can split the $i^{22}$ with the exponent rule but that can take long... wow @Astrophysics so fast lol

5. anonymous

lol . i just dont understand it period

6. UsukiDoll

well I know why it was done fast.. the exponent was an even number XD it's like writing i^2 11 times

7. Astrophysics

Main thing is you understand that i^2 = -1

8. Astrophysics

Which can be proved from $i^2 = (0+i)^2 \implies (0+i)(0+i) \implies ...-1+0i = -1$

9. UsukiDoll

long way... not recommended lol $(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)(i^2)$ $(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)(-1)$ $(1)(1)(1)(1)(1)(-1) \rightarrow (1)(-1)= -1$

10. Astrophysics

lool

11. Astrophysics

Nice haha

12. UsukiDoll

i^{100} is gonna hurt by hand XD

13. UsukiDoll

so next question For 3. 5 + 8i - 4 + 2i we can combine like terms so our answer is in the form of ai+b or a+bi

14. Astrophysics

$i^7 \implies (i^2)(i^2)(i^2)i$

15. anonymous

so number 2 would be i^7 (-1)(-1)(-1)(-1)(-1)(-1)(-1) = -1

16. Astrophysics

Naw, $i = \sqrt{-1}$

17. UsukiDoll

oh I read that second problem wrong. I thought it was i and 7 XD

18. Astrophysics

$i^7 = i^6 i = -1 \times i = -i$

19. anonymous

okay so problem 3 would be i^3 ?

20. UsukiDoll

problem 3 has 3. 5 + 8i - 4 + 2i we can combine like terms so our answer is in the form of ai+b or a+bi

21. Astrophysics

Yeah just combine like terms haha

22. anonymous

-1+10i

23. Astrophysics

close but should be +1

24. UsukiDoll

we can rearrange it to make it a bit easier 5-4+8i+2i -> 1+10i

25. UsukiDoll

well the attempt wasn't that bad.... just 5-4=1

26. UsukiDoll

the last 2 problems is just using FOIL and using the fact that$i^2 = -1$ again the final form of these questions is a+bi or ai+b

27. anonymous

(2 - i)(2 + i) 4 + 2i - 2i - i^2 4+i^2

28. anonymous

did i do that right

29. Astrophysics

$(2-i)(2+i) \implies (2)(2)+2i+(2)(-i)-i^2 \implies 4+2i-2i-i^2 = 4-i^2 = 4-(-1) = 4+1 = 5$

30. Astrophysics

$4-i^2 = 4-(-1) \implies 4+1 = 5$

31. UsukiDoll

(2-i)(2+i) 4+2i-2i-i^2 4-i^2 since i^2 = -1 4-(-1) = 4+1 = 5

32. Astrophysics

Sorry it got cut off there

33. Astrophysics

You're on the right path though, just a sign error :-)

34. UsukiDoll

it was a good attempt though .. just when we see i^2, we need to use substitution Replace i^2 with -1

35. anonymous

(2 - 3i)(3 + 5i) 6 + 10i - 9i - 15i 6 - 14i

36. Astrophysics

Hmm I think not

37. UsukiDoll

(2-3i)(3+5i) 6+10i-9i-15i^2

38. UsukiDoll

remember... i^2 = -1

39. anonymous

how do you apply -15 and i^2

40. Astrophysics

$(2-3i)(3+5i) = 6+10i-9i-15i^2 \implies 6+i-15i^2$ $6+i-15(-1) \implies 6+i+15 \implies 21 +i$

41. Astrophysics

Do you see your error now?

42. UsukiDoll

i^2 = -1 replace the i^2 with -1 6+10i-9i-15i^2 since i^2 = -1 6+10i-9i-15(-1)

43. UsukiDoll

6+i+15 21+i

44. Astrophysics

$\huge \checkmark$

45. anonymous

omg thank yall . !!!!!! can you help me with something else ?

46. anonymous

1. X^2 + 9 = 0 2. 3x^2 + 11 = -181

47. UsukiDoll

are these questions solve for x?

48. Astrophysics

You can set x^2 = -9 this has to do with imaginary numbers as well

49. anonymous

Solve the following: x^2 + 9 = 0 Select one: a. 2 + 4i b. + 3i c. not possible d. + 9i

50. Astrophysics

Remember as I mentioned before $i = \sqrt{-1}$

51. anonymous

Solve: 3x^2 + 11 = -181 Select one: a. not possible b. + 8i c. 1 + 13i d. + 4i

52. Astrophysics

$x^2 = - 9$ what do you get

53. UsukiDoll

ah yes once we square root both sides on x^2=-9 we're gonna have an imaginary case. since negative signs in radicals aren't allowed, the result is imaginary.. denoted by i

54. anonymous

3

55. anonymous

no

56. anonymous

3i ?

57. Astrophysics

That's one solution remember when we take a square root we get a positive and negative value :-)

58. UsukiDoll

hmm I guess for this multiple choice, it's only considering the positive version when taking the square root |dw:1439181642696:dw|

59. anonymous

so positive and negative 3i

60. Astrophysics

Yup

61. UsukiDoll

yeah

62. Astrophysics

So here is what's going on really

63. Astrophysics

$x^2 = -9 \implies x = \pm \sqrt{-9} \implies \pm \sqrt{-1} \sqrt{9} \implies \pm 3i$

64. anonymous

the next one is positive or negative 8i

65. Astrophysics

We just factor out the $\sqrt{-1}$

66. Astrophysics

Yea thats right

67. UsukiDoll

3x^2 +11=-181 3x^2 = -181-11 3x^2=-192 x^2 = -64 X = \sqrt{-64} x = -8i, 8i

68. anonymous

(2x^2-8x +10=0) im going to go this one tell me if its correct

69. Astrophysics

?

70. anonymous

|dw:1439181899563:dw|

71. anonymous

im not sure what to do next

72. Astrophysics

Why would you do that, just factor

73. UsukiDoll

I wouldn't do that at all

74. Astrophysics

or complete the square i guess

75. UsukiDoll

2x^2-8x+10=0 all of these numbers are even .. .they have one even number in common

76. Astrophysics

Here lets make it easier on you $2(x^2-4x+5)=0$

77. Astrophysics

Oh sorry Usuki xD

78. UsukiDoll

2 x 1 = 2 2 x 4 = 8 2 x 5 = 10 all of them have 2 in common . so yank that out and we should have 2(x^2-4x+5) = 0

79. anonymous

then factor x^2 - 4x + 5 ?

80. UsukiDoll

then factor (x^2-4x+5) since 5 is a prime number the only combination is 1 x 5 = 5 we need a negative number to produce -4 so 1 -5 = -4 (x+1)(x-5) ... only that produces a different sign patern but I'm wondering if this is written right. by discriminant formula (-4)^2-4(1)(5) = 16-20 = -4 oh great we need quadratics -_-! .. complex roots.. mhm

81. Astrophysics

Yup :p

82. UsukiDoll

$\frac{ -b \pm \sqrt{b^2-4ac}}{2a}$

83. Astrophysics

Quadratic is the easiest way to approach this

84. UsukiDoll

since a = 1, b = -4, and c = 5 but we got our b^2-4ac

85. Astrophysics

You'll have two complex solutions

86. UsukiDoll

$\frac{ 4 \pm \sqrt{-4}}{2}$ $\frac{ 4 \pm \sqrt{4}i}{2}$

87. UsukiDoll

take the square root of 4 and then reduce these fractions.

88. anonymous

2 plus or minus i

89. Astrophysics

Nice!

90. anonymous

Yay !!! :)

91. UsukiDoll

$\frac{ 4 \pm 2i}{2} \rightarrow 2 \pm i$ correct :)

92. anonymous

I have other work to complete if I need help will you all be on ?

93. UsukiDoll

depends on the time zone... I'm a bit worn out from last night.

94. Astrophysics

Just open a new question, and someone will get to it eventually :)

95. anonymous

ok thanks for yall help & eastern