Evaluate #sqrt(6)-sqrt(3)# when #sqrt(2)=1.414# ?

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Answer 1 · 2018-03-14T03:38:43

#0.414sqrt3#
or
#0.717#

#sqrt6-sqrt3#

#=sqrt2sqrt3-sqrt3#

#=sqrt3(sqrt2-1)#

#=sqrt3(1.414-1)#

#=0.414sqrt3#

Note: #sqrt3 approx 1.732#
Therefore,

#sqrt6-sqrt3#

#=0.414(1.732)#

#=0.717#

Answer 2 · 2018-03-14T03:38:46

#sqrt6-sqrt3=color(blue)(0.414sqrt3#

Given:

#sqrt6-sqrt3#

Prime factorize #sqrt6#.

#sqrt(2xx3)-sqrt3#

Apply rule: #sqrt(ab)=sqrt(a)sqrt(b)#

#sqrt2sqrt3-sqrt3#

Substitute #1.414# for #sqrt2#.

#1.414sqrt3-sqrt3#

The coefficient of #sqrt3# is understood to be #1#.

#sqrt3=1xxsqrt3#

#1.414sqrt3-1sqrt3#

#0.414sqrt3#