Question

mg(x2-x1)=1/2k(x2^2-x1^2) can you help me rearrange for x1

Answer 1

If you start by dividing by

mg

and multiplying by
\[ (x_2^2-x_1^2)\]

it'll be a good start ^_^

Answer 2

thanks for the help

Answer 3

\[(x_2^2-x_1^2)(x_2-x_1)=\frac{1}{2kmg}\]
\[(x_2^3-x_1^2 \ x_2 -x_2^2 \ x_1-x_1^3)=\frac{1}{2kmg}\]

blegh. This i not a happy polynomial :P

Answer 4

\[(x_2^3-x_1^2 \ x_2 -x_2^2 \ x_1+x_1^3)=\frac{1}{2kmg}\]

Answer 5

I don't think you can get x1 by itself....

Answer 6

maybeee.....

Answer 7

that is what i thought. but this is for ap physics so it has to, i just couldnt get past this step on my own

Answer 8

maybe this
\[(x_2^3-x_1^2 \ x_2 -x_2^2 \ x_1-x_1^3)=\frac{1}{2kmg}\]
\[x_2(x_2^2-x_2 \ x_1-x_1^2 \ )=\frac{1}{2kmg}+x_1^3\]
quadrtic formula for the guy in the middle?
\[x_2=\frac{x_1±\sqrt{x_1^2+4x_1^2}}{2}\]

does that do anything or is even allowed?I forget at this hour :/