How do you solve #(25y^2)^(1/2) #?

1 Answer
Jul 17, 2016

#5y, -5y#

Explanation:

Anything to the #1/2# power is the same as the square root of that value.

#(25y^2)^(1/2) = sqrt(25y^2)#

Find the square root. Note that #sqrt(ab) = sqrtaxxsqrtb#

#sqrt25 times sqrt(y^2)#

#5 times 5# is #25#, to the square root of #25# is #5#. Using the rule from the very top, #sqrt(y^2) = (y^2)^(1/2) = y^(2*1/2) = y^1 = y#.

#5 times y#

#5y#

There can also be a negative answer. Every value has a negative root, because when you multiply two negative numbers you end up with the same positive number.

Example: #sqrt16 = 4, -4# since #4*4 = 16# and #-4*-4 = 16#.

So, the square root of #25# would actually be #5# and #-5#.

The other possible solution: #-5y#