Question #3f390

2 Answers
Mar 16, 2017

Answer:

#-2/3sqrt(3)-sqrt(2)#

Explanation:

Consider the just the denominator for a moment.

Demonstrating a principle by example:

#a^2-b^2=(a-b)(a+b)#

Now compare to:

#(sqrt(12))^2-(sqrt(18))^2 = (sqrt(12)-sqrt(18))(sqrt(12)+sqrt(18))#

So to remove the roots from the denominator we need to 'force' it to be #(sqrt(12)-sqrt(18))(sqrt(12)+sqrt(18))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Getting rid of the roots in the denominator")#

Multiply by 1 and you do not change the value. However, 1 comes in many forms so you can change the way numbers look without changing their intrinsic value.

#color(green)(2/(sqrt(12)-sqrt(18))color(red)(xx1))#

#color(green)(2/(sqrt(12)-sqrt(18))color(red)( xx(sqrt(12)+sqrt(18))/(sqrt(12+sqrt(18))#

#(2(sqrt(12)+sqrt(18)))/( (sqrt(12))^2-(sqrt(18))^2)#

#(2(sqrt(12)+sqrt(18)))/( 12-18)#

#(2(sqrt(12)+sqrt(18)))/( -6)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Simplifying")#

#(cancel(2)^1(sqrt(12)+sqrt(18)))/( -cancel(6)^3)#

#-1/3(sqrt(12)+sqrt(18))#

But #sqrt(12) = sqrt(3xx2^2)=2sqrt(3)#
And #sqrt(18)=sqrt(2xx3^2)=3sqrt(2)#

#-1/3(2sqrt(3)+3sqrt(2))#

#-2/3sqrt(3)-sqrt(2)#

Mar 16, 2017

Answer:

#-2/3sqrt3-sqrt2#

Explanation:

Multiply the numerator/denominator of the fraction by the #color(blue)"conjugate"# of the denominator.

The #color(blue)"conjugate"" of "sqrt12-sqrt18" is "sqrt12color(red)(+)sqrt18#

Considering the denominator.

#color(orange)"Reminder " (sqrtaxxsqrta)=a#

#"Expand " (sqrt12-sqrt18)(sqrt12color(red)(+)sqrt18)# using the FOIL method.

#=12+cancel(sqrt18xxsqrt12)cancel(-sqrt18xxsqrt12)-18#

#=-6larrcolor(red)" rational denominator"#

#rArr2/(sqrt12-sqrt18)#

#=2/(sqrt12-sqrt18)xx(sqrt12+sqrt18)/(sqrt12+sqrt18)#

#=(2(sqrt12+sqrt18))/(-6)#

#=-1/3(sqrt12+sqrt18)#

Simplifying #sqrt12" and " sqrt18#

#sqrt12=sqrt(4xx3)=sqrt4xxsqrt3=2sqrt3#

#sqrt18=sqrt(9xx2)=sqrt9xxsqrt2=3sqrt2#

#rArr-1/3(sqrt12+sqrt18)#

#=-1/3(2sqrt3+3sqrt2)larr" this is acceptable"#

#=-2/3sqrt3-sqrt2#