How do I simplify #(3sqrt3 - 1)^2 / (2sqrt3 - 3)# ?

3 Answers
iceman
Apr 14, 2018

#(2(19sqrt3+24))/3#

Explanation:

#(3sqrt3-1)^2/(2sqrt3-3)#

#=(3sqrt3-1)^2/(2sqrt3-3)*(2sqrt3+3)/(2sqrt3+3)#

#=((27-6sqrt3+1)(2sqrt3+3))/((2sqrt3)^2-(3)^2)#

#=(2(14-3sqrt3)(2sqrt3+3))/(12 - 9)#

#=(2(28sqrt3+42-18-9sqrt3))/3#

#=(2(19sqrt3+24))/3#

Jim G.
Apr 14, 2018

#16+38/3sqrt3#

Explanation:

#"the first thing to do is expand the numerator using FOIL"#

#"note that "sqrtaxxsqrta=a#

#rArr(3sqrt3-1)^2=(3sqrt3-1)(3sqrt3-1)#

#=27-3sqrt3-3sqrt3+1#

#=28-6sqrt3#

#rArr((3sqrt3-1)^2)/(2sqrt3-3)=(28-6sqrt3)/(2sqrt3-3)#

#"the next step is to "color(blue)"rationalise the denominator"#
#"that is, eliminate the radical"#

#"to do this multiply numerator/denominator by the"#
#color(blue)"conjugate ""of the denominator"#

#"the conjugate of "2sqrt3-3" is "2sqrt3color(red)(+)3#

#rArr((28-6sqrt3)(2sqrt3+3))/((2sqrt3-3)(2sqrt3+3)#

#"expand numerator/denominator using FOIL"#

#=(56sqrt3+84-36-18sqrt3)/(12cancel(+6sqrt3)cancel(-6sqrt3)-9)#

#=(48+38sqrt3)/3#

#=16+38/3sqrt3#

A08
Apr 14, 2018

Given expression

#((3sqrt3-1)^2)/(2sqrt3-3)#

Expanding the numerator using the identity

#(a-b)^2=a^2-2ab+b^2#, we get

#((3sqrt3)^2-2xx3sqrt3xx1+1^2)/(2sqrt3-3)#
#=>(27-6sqrt3+1)/(2sqrt3-3)#
#=>(28-6sqrt3)/(2sqrt3-3)#

Rationalizing the denominator and using the identity #(a+b)(a-b)=a^2-b^2#, we get

#(28-6sqrt3)/(2sqrt3-3)xx(2sqrt3+3)/(2sqrt3+3)#
#=>((28-6sqrt3)(2sqrt3+3))/((2sqrt3)^2-(3)^2)#
#=>((56sqrt3+84-36-18sqrt3))/((2sqrt3)^2-(3)^2)#
#=>((38sqrt3+48))/((12-9)#
#=>2/3(19sqrt3+24)#

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