Quadratic Equation

[Baked0420]

I need to solve this problem using the quadratic formula, but it's not in the quadratic equation yet. I don't need you to solve this problem, would just like you to show me it in the quadratic equation form and how you got it there.

What I have: 4/v = v-6/v-4

Need it in: ax^2 + bx + c form

thanks for any help, you will get rep+ if you make an honest attempt at trying to help.

[Brain]

multiply every term by v

add/subtract to get it all on one side

[Markus]

Do you really need to get this in the quadratic equitation, or do you only need to solve this?

[barbarianl3t]

4v -16 =v^2 -6v In quadratic mode is v^2 -10v + 16
to get there you start with (assuming you just forgot the paranthesis)
4/v = (v-6)/v-4

multiply both sides by v and you get

4 = (v^2 -6v)/v-4 which gets rid of the v in the denomineter on the left side

next mulitply both sides by v-4 to get the v out of the denomenter on the right side

4v - 16 = v^2 - 6v

next put all terms on one side

0 = v^2 - 10v + 16 hence you get you're equation

[Brain]

I think the parenthesis go:

4/v = v-(6/v)-4

That's how you would do it without the parenthesis according to the order of operations

[Baked0420]

Do you really need to get this in the quadratic equitation, or do you only need to solve this?

I need to solve it, but I need to solve it using the quadratic formula. The way we learned it is x:= -b sqrt(b^2 - 4ac) all over 2a. :p lol at my bits of scar in there (:= and sqrt) but anyway, yea I just needed to get this into ax^2 + bx + c so that I could solve it the way the teacher wanted.'

edit: I don't think multiplying by v gets rid of the v on the right where it's v-6 over v-4, I think to get rid of that you have to multiply by the conjugate, no? which is why I posted here to see if anyone could confirm this. and then wouldn't that mess up the v on the left side?

EDIT: I think barbarianl3t is right. I'll just go with that and if it's wrong my teacher can tell me how I was supposed to do it, thanks for your help everyone.

[Method]

You just need to find the common denominator for both sides, which is (v)(v - 4), so you multiply both sides by those two values, which is what barbarianl3t mentioned.

[Baked0420]

yea, for some reason before I posted I thought I'd need to multiply by a conjugate, no clue why, but after barb's post, I realized when I thought maybe just multiplying by v-4 was right, but I thought that'd mess up the other side of the equation but it won't change the denominator :p shoulda just tried it, well thanks for the help Method and barbarianl3t, and thanks for confirming it Method.