How do you rationalize the denominator and simplify #(3+sqrt18) /( 1 + sqrt8 )#?

2 Answers
Jul 15, 2016

Answer:

#3/7(3+sqrt(2)) larr" corrected solution"#

Explanation:

A very useful relationship to remember is #a^2-b^2=(a-b)(a+b)#
In the context of this question we can change
#1+sqrt(8)" to "1^2-(sqrt(8))^2#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply by 1 but in the form of #1= (1-sqrt(8))/(1-sqrt(8))#

#" "color(red)("Error - should be "sqrt(18))#
#" "color(red)(darr)#
#cancel(((3+color(red)(sqrt(8)))(1-sqrt(8)))/((1+sqrt(8))(1-sqrt(8)))) #
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)("Corrected!!!")#

#((3+sqrt(18))(1-sqrt(8)))/((1+sqrt(8))(1-sqrt(8))) = (3-3sqrt(8)+sqrt(18)-sqrt(8xx18))/(1-8)#

#=(3-3sqrt(2xx2^2)+ sqrt(2xx3^2)- sqrt(2xx2^2xx2xx3^2))/(-7)#

#=(3-6sqrt(2)+3sqrt(2)-12)/(-7)#

#=(-9-3sqrt(2))/(-7)" "=" "+3/7(3+sqrt(2))#

Got there in the end! Amazing how much a typo can mess things up!

Jul 15, 2016

Answer:

#3/7(3+sqrt2)#

Explanation:

Rationalisation factor of #a+sqrtb# is #a-sqrtb# and vice-versa.

Given Exp. #=(3+sqrt18)/(1+sqrt8)=(3+sqrt(9*2))/(1+sqrt(4*2))#

#=(3+3sqrt2)/(1+2sqrt2)#

Multiplying, in #Nr. & Dr# by the rationalisation factor of Dr., i.e., by#(1-2sqrt2)#, the Exp.

#={(3+3sqrt2)(1-2sqrt2)}/{(1+2sqrt2)(1-2sqrt2)},#

#={3+3sqrt2-6sqrt2-6*2}/{1^2-(2sqrt2)^2},#

#=(-9-3sqrt2)/(1-8),=3/7(3+sqrt2).#