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

3 Answers
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

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

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)