How do you simplify # (5-3sqrt6)/(3+4sqrt2)#?

2 Answers
NickTheTurtle
Apr 2, 2017

#(9sqrt(6)+20sqrt(2)-24sqrt(3)-15)/23#

Explanation:

In order to rationalize the denominator, we multiply it by its conjugate (the conjugate of #a+bsqrt(c)# is #a-bsqrt(c)#). The reason why will be apparent as we solve the problem.

The conjugate of the denominator #3+4sqrt(2)# is #3-4sqrt(2)#.

Since we are going to multiply its denominator by #3-4sqrt(2)#, we also have to multiply the numerator by #3-4sqrt(2)#.

Then, we have #(5-3sqrt(6))/(3+4sqrt(2))=(5-3sqrt(6))/(3+4sqrt(2))*(3-4sqrt(2))/(3-4sqrt(2))=((5-3sqrt(6))(3-4sqrt(2)))/((3+4sqrt(2))(3-4sqrt(2)))#.

We can simplify the denominator by using the identity #(a+b)(a-b)=a^2-b^2#.

#((5-3sqrt(6))(3-4sqrt(2)))/((3+4sqrt(2))(3-4sqrt(2)))=(15-9sqrt(6)-20sqrt(2)+12sqrt(12))/(9-32)=-(15-9sqrt(6)-20sqrt(2)+24sqrt(3))/23=(9sqrt(6)+20sqrt(2)-24sqrt(3)-15)/23#

mason m
Apr 2, 2017

#-(15-9sqrt6-20sqrt2+24sqrt3)/(23)#

Explanation:

We can simplify the denominator by turning it into a single term by multiplying the fraction by the conjugate of the denominator.

The conjugate of #3+4sqrt2# is found by simply reversing the sign of the second term, which is #3-4sqrt2#.

So, we multiply the numerator and denominator by #3-4sqrt2#.

#(5-3sqrt6)/(3+4sqrt2)*(3-4sqrt2)/(3-4sqrt2)=((5-3sqrt6)(3-4sqrt2))/((3+4sqrt2)(3-4sqrt2))#

Expand both of these by FOILing:

#=(15-(3sqrt6)3+5(-4sqrt2)-3sqrt6(-4sqrt2))/(9+(4sqrt2)3+3(-4sqrt2)+(4sqrt2)(-4sqrt2))#

Simplifying these:

#=(15-9sqrt6-20sqrt2+12sqrt12)/(9+12sqrt2-12sqrt2-16sqrt4)#

Note that #sqrt4=2# and #sqrt12=sqrt(4*3)=sqrt4sqrt3=2sqrt3#.

#=(15-9sqrt6-20sqrt2+12(2sqrt3))/(9-16(2))#

#=(15-9sqrt6-20sqrt2+24sqrt3)/(9-32)#

#=(15-9sqrt6-20sqrt2+24sqrt3)/(-23)#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
2911 views around the world
You can reuse this answer
Creative Commons License