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