Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

richyw

  • 3 years ago

Having trouble with some algebra

  • This Question is Closed
  1. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    welcome to the club

  2. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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!

  3. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  4. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    without expanding it all out of course...

  5. ganeshie8
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh wow

  7. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i need to sleep haha

  8. ganeshie8
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    haha i agree ;p

  9. richyw
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thanks a lot

  10. ganeshie8
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    np :)

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy