How do you simplify # (64+sqrt(128) )/ 8#?

1 Answer
Meave60
Jun 27, 2015

Simplify the numerator of #(64+sqrt 128)/8# to #(64+sqrt (2xx64))/8#. Simplify #(64+sqrt(2xx64))/8# to #(64+8sqrt 2)/8#. Factor out 8 from #(64+8sqrt 2)/8# to get #(8(8+sqrt 2))/8#. Cancel the #8# in the numerator and denominator to get #8+sqrt 2#.

Explanation:

#(64+sqrt(128))/8#

Simplify.

#(64+sqrt(2xx64))/8# =

#(64+sqrt 2sqrt 64)/8# =

#(64+8sqrt 2)/8# =

Factor out #8# in the numerator.

#(8(8+sqrt 2))/8#

Cancel the #8# in the numerator and denominator.

#(cancel 8(8+sqrt 2))/cancel8# =

#(8+sqrt 2)#

Related questions
See all questions in Simplification of Radical Expressions
Impact of this question
1733 views around the world
You can reuse this answer
Creative Commons License