Question

olve this radical expression (show work)

http://i1052.photobucket.com/albums/s441/leya454/mathhel_zps1ebfb197.png

Answer 1

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

Answer 2

thanks