How do you simplify #(6+sqrt128)/2#?

1 Answer
Don't Memorise
Jul 10, 2015

#=color(blue)(3 +sqrt32)#

Explanation:

#sqrt128 = color(green)(sqrt(4 * 32)#
The expression can be written as:

#((6+ color(green)(sqrt(4 * 32)))/2)#

#=((6+ 2(sqrt 32))/2)#

#2# is common to both terms of the numerator:

#=cancel2((3+ (sqrt( 32)))/cancel2)#

#=color(blue)(3 +sqrt32)#

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