SOLUTION: Thank you in advance for helping me with this problem: {{{ sqrt (p^2q^4)/ sqrt(pq) }}} I truly appreciate your help in helping me understand this.

Algebra ->  Square-cubic-other-roots -> SOLUTION: Thank you in advance for helping me with this problem: {{{ sqrt (p^2q^4)/ sqrt(pq) }}} I truly appreciate your help in helping me understand this.       Log On


   

Question 271883: Thank you in advance for helping me with this problem:
+sqrt+%28p%5E2q%5E4%29%2F+sqrt%28pq%29+
I truly appreciate your help in helping me understand this.

Found 2 solutions by ankor@dixie-net.com, Alan3354:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt+%28p%5E2q%5E4%29%2F+sqrt%28pq%29+
is the same as
+sqrt+%28%28p%5E2q%5E4%29%2F%28pq%29%29+
cancel out the denominator
sqrt%28p%2Aq%5E3%29

Answer by Alan3354(69443) About Me  (Show Source):