Question

Having trouble with some algebra

Answer 1

welcome to the club

Answer 2

my solutions manual goes from \[\vec{E}=k_eq\left( \frac{1}{(x-a)^2}-\frac{1}{(x+a)^2}\right)\hat{i}\]to\[\vec{E}=k_eq\frac{4ax}{(x^2-a^2)^2}\hat{i}\]
and I can't quite remember how to get that!

Answer 3

I can get to \[\vec{E}=k_eq\left(\frac{4ax}{(x-a)^2(x+a)^2}\right)\hat{i}\]
so basically I have forgotten how to deal with the denominator :O

Answer 4

without expanding it all out of course...

Answer 5

\(\vec{E}=k_eq\left(\frac{4ax}{(x-a)^2(x+a)^2}\right)\hat{i} \)

\(\vec{E}=k_eq\left(\frac{4ax}{((x-a)(x+a))^2}\right)\hat{i} \)

Answer 6

oh wow

Answer 7

i need to sleep haha

Answer 8

haha i agree ;p

Answer 9

thanks a lot

Answer 10

np :)