Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

leya454

  • 3 years ago

olve this radical expression (show work) http://i1052.photobucket.com/albums/s441/leya454/mathhel_zps1ebfb197.png

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

    There is no solving here, you can't solve an expression, only simplify/reduce one...$$\frac{k+1}{k^2+6k+9}+\frac4{k+3}-\frac{-6}{k-1}=\frac{(k+1)(k-1)}{(k+3)^2(k-1)}+\frac{4(k+3)(k-1)}{(k+3)^2(k-1)}-\frac{-6(k+3)^2}{(k+3)^2(k-1)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{(k+1)(k-1)+4(k+3)(k-1)+6(k+3)^2}{(k+3)^2(k-1)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{k^2-1+4k^2+8k-12+6k^2+36k+54}{(k+3)^2(k-1)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{11k^2+44k-41}{k^3+5k^2+3k-9}$$

  2. leya454
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks

  3. 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