SOLUTION: How would you solve 4x√(9x^2y^4)? in other words: 4x in front of the square root of 9x to the second power and y to the fourth power?

Algebra ->  Square-cubic-other-roots -> SOLUTION: How would you solve 4x√(9x^2y^4)? in other words: 4x in front of the square root of 9x to the second power and y to the fourth power?       Log On


   

Question 173695: How would you solve
4x√(9x^2y^4)?
in other words: 4x in front of the square root of 9x to the second power and y to the fourth power?
Found 2 solutions by stanbon, Mathtut:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
solve
4x√(9x^2y^4)
= (4x)*3xy^2
= 12x^2y^2
==================
Chers,
Stan H.

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
well the way you have written it, 9 is not to the second power only x...and 2 is not to the 4th power only y....when you want to show the whole thing being raised to a power it must all be in parenthesis like %282x%29%5E2
:
%284x%29sqrt%28%289x%29%5E2%282y%29%5E4%29 since the square root of x%5E2is x and the
:
square root of x%5E4is x%5E2.....we are dealing with both of those here
:
%289x%29%5E2comes out as 9x and %282y%29%5E4comes out as %282y%29%5E2
:
so %284x%29%289x%29%282y%29%5E2=72x%5E2y%5E2