How do you simplify #sqrt(16r^6)#?

1 Answer
Jan 6, 2017

Answer:

#4r^3#

Explanation:

The trick is to look for squared value and then to take them outside the square root but not in squared form.

Note that #16 = 4^2" and "r^6 = r^(3xx2)= (r^3)^2#

#sqrt(16r^6)" "->" "sqrt(4^2(r^3)^2)" "=" "4r^3#

If you wished to be really fussy consider this example:
#(-2)xx(-2)" " =" " +4" " =" "(+2)xx(+2)#

So we have: #sqrt(16r^6)" "->" "sqrt(4^2(r^3)^2)" "=" "+-4r^3#

I suspect they are just looking for #+4r^3# as the answer