## brittany_lundgren 1.Solve S = 4v2 for v. Show your work one year ago one year ago

1. inkyvoyd

Wel, first isolate the term for v^2

2. inkyvoyd

*first isolate v^2

3. inkyvoyd

(divide by 4, if you don't know)

4. brittany_lundgren

ok

5. inkyvoyd

S/4=v^2

6. brittany_lundgren

ok

7. inkyvoyd

Then, take the plus-minus square root of both sides.

8. inkyvoyd

You know how to simplify $$\Huge \pm\sqrt\frac{S}{4}$$ right?

9. brittany_lundgren

ok im following you

10. inkyvoyd

Let me show you the whole thing. ONe sec, this may take a minute or two.

11. brittany_lundgren

ok im waiting :)

12. inkyvoyd

S = 4v2 $$\Large S=4v^2$$ ->Isolate the "$$v^2$$" $$\Large \frac{S}{4}=v^2$$ Take the positive negative square root of both sides $$\huge \pm\sqrt{\frac{S}{4}}=v$$ $$\huge \pm\frac{\sqrt{S}}{\sqrt{4}}=v$$ Can you try from here?

13. inkyvoyd

@brittany_lundgren ?

14. brittany_lundgren

I can try........ I dont remember how to do any of ths... I'm sooo lost!

15. inkyvoyd

Alright, it's ok. Just tell me which steps you don't understand :)

16. brittany_lundgren

I honestly need help with the whole problem, im soooo confused! :'(

17. inkyvoyd

Ok, so, from step one, right? Which parts of step one do you not understand? Or is it the problem itself?

18. brittany_lundgren

It's the problem itself, I just don't understand :'(

19. inkyvoyd

Ok, what they do is give you an equation. What they want you to do is to "solve for v" This means they want you to put v on one side, and everything else on the other.

20. inkyvoyd

Note that v can ONLY be on one side. If you have v on both sides, it's a mistake.

21. brittany_lundgren

okay, im following you so far!

22. inkyvoyd

Good. :)

23. jhonyy9

i have solved it before - not is clearly there ?

24. inkyvoyd

Now, do you remember how we move things around?

25. inkyvoyd

Like, combine like terms, move things from one side to the other, and take square roots?

26. brittany_lundgren

ok well can you show me how? Like the steps ??

27. inkyvoyd

Ok. In general, or just for this equation?

28. brittany_lundgren

just the steps on how to solve this problem... The problem told me to show my work.......

29. inkyvoyd

Okk.

30. inkyvoyd

Remember, we want to put "v" on one side, and everything else on the other. The first step is to get rid of what's in front of v.

31. brittany_lundgren

okay so what would I do then?

32. inkyvoyd

What's in front of v?

33. brittany_lundgren

1.Solve S = 4v2 for v.

34. inkyvoyd

I mean, what's in front of "v^2"?

35. brittany_lundgren

4

36. inkyvoyd

So, how do we move the 4 to the other side?

37. brittany_lundgren

-4(4-4)=-2(2v-1)

38. inkyvoyd

Not that complicated. Just divide both sides by 4 :)

39. brittany_lundgren

Is the answer to the question v=2?

40. inkyvoyd

Well, we can't actually solve for an exact value of v. We can only express it in terms of "s".

41. brittany_lundgren

okay, I'm sooo confused!!! :'(

42. inkyvoyd

Ok. We can't always solve for v (in terms of a number, like 3, or -2). But, we can move everything over to one side and leave v on the other.

43. brittany_lundgren

OMG!!! dumb right now :(

44. inkyvoyd

Ok. If you have the equation x-2y=0 you can't find x, because you don't know y.

45. brittany_lundgren

okay, I understand that, I just don''t understand on how to solve this problem!

46. inkyvoyd

Ok. Solve for v doesn't mean "get a number value for v". It means, "put just v on one side, and everything else without v on the other"

47. brittany_lundgren

ok,im following so far.

48. inkyvoyd

Alright. Since we just want to move everything to the other side, we divide both sides by 4 to get rid of what's in front of v.

49. brittany_lundgren

ok so S/4=4v^2/4

50. inkyvoyd

Ok. Now simplify the rigt side.

51. brittany_lundgren

S=v^2

52. inkyvoyd

Careful, you can't cancel out the left side...

53. brittany_lundgren

S/4=v^2

54. inkyvoyd

yess.

55. inkyvoyd

Now, what do we do to get rid of v^2? (turn it into v)

56. brittany_lundgren

S/4=v

57. inkyvoyd

Careful. You have to take the square root of BOTH sides.

58. brittany_lundgren

I got confused, just now. I beeen following this whole time and i just now got lost!!! :'(

59. inkyvoyd

That's ok.

60. inkyvoyd

S/4=4v^2/4 -> $$\huge \frac{S}{4}=\frac{(4v^2)}{4}$$

61. inkyvoyd

Then $$\Huge \frac{S}{4}=v^2$$

62. inkyvoyd

Now, here's the part you got lost. Let me show it in more detail.

63. inkyvoyd

$$\Huge \pm\sqrt{\frac{S}{4}}=\sqrt{v^2}$$

64. inkyvoyd

There's a lot going on right here.

65. inkyvoyd

Does everything make sense?

66. brittany_lundgren

ima little lost there

67. inkyvoyd

Ok. Do you know why we have the plus minus sign?

68. brittany_lundgren

nope, im confused whole step.......

69. inkyvoyd

I'm really sorry, but I have to go to bed. I'll try to make sure I get people here to help you. (I have to take some tests myself tomorrow, and I can't afford to miss them) @nbouscal , would you be so kind as to help out? THanks.

70. brittany_lundgren

damn, what time is it there?

71. inkyvoyd

2:00 AM. I have to go to sleep and get done before 2:00PM, but I have two tests around 1 hour each, :S

72. brittany_lundgren

dang........

73. inkyvoyd

ACtually, I'm going to take both right now. No chances.

74. inkyvoyd

Time for a cup of coffee. thanks, @nbouscal .

75. nbouscal

Okay, if you have $$v^2=\frac{S}{4}$$, and you want to know what $$v$$ is, you have to take the square root of both sides. That's the only way to get the left side to just have $$v$$ by itself. The problem is, when you square root something, you always have to put a $$\pm$$ sign in front, because you don't know whether it's positive or negative. For example, $$2^2=4, (-2)^2=4$$, so the square root of 4 could be either 2 or -2. So, we take the square root of both sides and we add a plus-or-minus, and we get $v=\pm\sqrt{\frac{S}{4}}$

76. brittany_lundgren

@nbouscal brb imma go get something to drink

77. brittany_lundgren

ok, im following so far.

78. inkyvoyd

Brittany, I can help too, becaues I just realized I only have one test.

79. inkyvoyd

$$\huge v=\pm\frac{\sqrt{S}}{\sqrt{4}}$$

80. inkyvoyd

Remember, $$\Huge \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$

81. inkyvoyd

Now, what's the square root of 4?

82. inkyvoyd

(@nbouscal , I'm going to go ahead and take my scantron before the coffee wears off)

83. brittany_lundgren

2

84. brittany_lundgren

@inkyvoyd @nbouscal

85. inkyvoyd

Ok, then, what do we get?

86. inkyvoyd

(use draw if you have to)

87. nbouscal

$$v=\pm\dfrac{\sqrt{S}}{\sqrt{4}}$$, $$\sqrt{4}=2$$, so $$v=?$$

88. inkyvoyd

@brittany_lundgren , do you get it?