How do you simplify #sqrt(196y^10)#?

2 Answers
Sep 10, 2015

Answer:

#sqrt(196y^10) = 14y^5#

Explanation:

#14xx14 = 196#
#y^5xxy^5=y^10#

So #sqrt(196y^10) = sqrt(14^2(y^5)^2)#

#color(white)("XXXXXX")= abs(14y^5)#
...we take the absolute value to ensure the root extracted is the principal root

Sep 10, 2015

Answer:

#sqrt(196y^10) = 14abs(y^5)#

Explanation:

First, note that in general #sqrt(x^2) = abs(x)# rather than #x#,

since:

#sqrt(x^2) = { (x, "if x >= 0"), (-x, "if x < 0") :}#

Also, if #a, b >= 0# then #sqrt(ab) = sqrt(a)sqrt(b)#

So:

#sqrt(196y^10) = sqrt(14^2 (y^5)^2) = sqrt(14^2)sqrt((y^5)^2) = 14abs(y^5)#