Here's the question you clicked on:
brittany_lundgren
1.Solve S = 4v2 for v. Show your work
Wel, first isolate the term for v^2
(divide by 4, if you don't know)
Then, take the plus-minus square root of both sides.
You know how to simplify \(\Huge \pm\sqrt\frac{S}{4}\) right?
ok im following you
Let me show you the whole thing. ONe sec, this may take a minute or two.
ok im waiting :)
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?
I can try........ I dont remember how to do any of ths... I'm sooo lost!
Alright, it's ok. Just tell me which steps you don't understand :)
I honestly need help with the whole problem, im soooo confused! :'(
Ok, so, from step one, right? Which parts of step one do you not understand? Or is it the problem itself?
It's the problem itself, I just don't understand :'(
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.
Note that v can ONLY be on one side. If you have v on both sides, it's a mistake.
okay, im following you so far!
i have solved it before - not is clearly there ?
Now, do you remember how we move things around?
Like, combine like terms, move things from one side to the other, and take square roots?
ok well can you show me how? Like the steps ??
Ok. In general, or just for this equation?
just the steps on how to solve this problem... The problem told me to show my work.......
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.
okay so what would I do then?
1.Solve S = 4v2 for v.
I mean, what's in front of "v^2"?
So, how do we move the 4 to the other side?
-4(4-4)=-2(2v-1)
Not that complicated. Just divide both sides by 4 :)
Is the answer to the question v=2?
Well, we can't actually solve for an exact value of v. We can only express it in terms of "s".
okay, I'm sooo confused!!! :'(
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.
OMG!!! dumb right now :(
Ok. If you have the equation x-2y=0 you can't find x, because you don't know y.
okay, I understand that, I just don''t understand on how to solve this problem!
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"
ok,im following so far.
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.
ok so S/4=4v^2/4
Ok. Now simplify the rigt side.
Careful, you can't cancel out the left side...
Now, what do we do to get rid of v^2? (turn it into v)
Careful. You have to take the square root of BOTH sides.
I got confused, just now. I beeen following this whole time and i just now got lost!!! :'(
S/4=4v^2/4 -> \(\huge \frac{S}{4}=\frac{(4v^2)}{4}\)
Then \(\Huge \frac{S}{4}=v^2\)
Now, here's the part you got lost. Let me show it in more detail.
\(\Huge \pm\sqrt{\frac{S}{4}}=\sqrt{v^2}\)
There's a lot going on right here.
Does everything make sense?
ima little lost there
Ok. Do you know why we have the plus minus sign?
nope, im confused whole step.......
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.
damn, what time is it there?
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
ACtually, I'm going to take both right now. No chances.
Time for a cup of coffee. thanks, @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 \[ v=\pm\sqrt{\frac{S}{4}} \]
@nbouscal brb imma go get something to drink
ok, im following so far.
Brittany, I can help too, becaues I just realized I only have one test.
\(\huge v=\pm\frac{\sqrt{S}}{\sqrt{4}}\)
Remember, \(\Huge \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)
Now, what's the square root of 4?
(@nbouscal , I'm going to go ahead and take my scantron before the coffee wears off)
@inkyvoyd @nbouscal
Ok, then, what do we get?
(use draw if you have to)
\(v=\pm\dfrac{\sqrt{S}}{\sqrt{4}}\), \(\sqrt{4}=2\), so \(v=?\)
@brittany_lundgren , do you get it?