How do you simplify #(8+sqrt48)/4#?

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Answer 1 · 2017-04-12T07:22:29

#2+sqrt(3)#

Simplify your radical (square root):

To simplify a square root, use perfect squares such as
#2^2 = 4#
#3^2 = 9#
#4^2 = 16#
#sqrt(48) = sqrt(16*3)#

Use the rule that says #sqrt(m*n) = sqrt(m)sqrt(n)#

#sqrt(48) = sqrt(16) sqrt(3) = 4 sqrt(3)#

So #(8 + sqrt(48))/4 = (8 + 4 sqrt(3))/4#

Factor a 4 from each number in the numerator:

#(8 + 4 sqrt(3))/4 = (4(2+sqrt(3)))/4#

Cancel the 4's because #4/4 = 1#

So #(8 + sqrt(48))/4 = (8 + 4 sqrt(3))/4 = (4(2+sqrt(3)))/4 = 2+sqrt(3)#