How do you simplify #( 8 + sqrt 6) /sqrt 2#?

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Answer 1 · 2016-03-26T05:37:39

#4sqrt2+sqrt3#

#(8+sqrt6)/sqrt2#

#(8+sqrt6)/sqrt2 xx sqrt2/sqrt2#

#[sqrt2(8+sqrt6)]/2#

#(8sqrt2+sqrt12)/2#

#(8sqrt2+sqrt(4xx3))/2#

#(8sqrt2+2sqrt3)/2#

#[2(4sqrt2+sqrt3)]/2#

#4sqrt2+sqrt3#