Question

You want to show that
V^2 - Vo^2 = 2 a (X - Xo)
first rewrite that as:
(V-Vo)(V+Vo) = 2a (X-Xo)
now
your average velocity during the interval = (V+Vo)/2
so
X = Xo + t(V+Vo)/2
but
V = Vo + a t
so
t = (V-Vo)/a
so
X = Xo +(V-Vo)(V+Vo)/2a
we are there.

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but wouldn't you have to rearange it and end up subtracting Xo

Answer 1

This is what i did

x = 2^-1(v0 + v)a^-1(v-v0) + x0
simplfy
x = (a2)^-1(v^2 -v0^2)+ x0
(x = (a2)^-1(v^2 -v0^2)+ x0)2a
(2a)x = (v^2-v0^2) + x0
(2a)x - x0 = (v^2-v0^2) + x0 - x0
(2a)x - x0 + v0^2= (v^2 - v0^2) + v0^2
(2a)x - x0 + v0^2 = v^2

i do not see were i went wrong

Answer 2

Please check your previous posts. Here is my answer from several hours ago:
You want to show that
V^2 - Vo^2 = 2 a (X - Xo)
first rewrite that as:
(V-Vo)(V+Vo) = 2a (X-Xo)
now
your average velocity during the interval = (V+Vo)/2
so
X = Xo + t(V+Vo)/2
but
V = Vo + a t
so
t = (V-Vo)/a
so
X = Xo +(V-Vo)(V+Vo)/2a
we are there.

Answer 3

Of course you have to rearrange it
X -Xo = (V-Vo)(V+Vo)/2a
(X-Xo)2a = (V-Vo)(V+Vo)
2 a (X-Xo) = V^2-Vo^2