Find z (floor). Please? Add work and explanation.

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Answer 1 · 2017-07-14T14:09:42

#z=1/5(2sqrt2+sqrt3)#

#z= (sqrt3^2 - 2sqrt2^2)/(sqrt3-2sqrt2)#

#= (3-2xx2)/(sqrt3-2sqrt2)#

#= (-1)/(sqrt3-2sqrt2)#

Rationalise the denominator:

#= ((-1)xx(sqrt3+2sqrt2))/((sqrt3-2sqrt2)xx(sqrt3+2sqrt2))#

#= (-(sqrt3+2sqrt2))/(3-4xx2)#

#= (-(sqrt3+2sqrt2))/(3-8)#

#= (sqrt3+2sqrt2)/5#

#=1/5(2sqrt2+sqrt3)#