How do you show that #(3+sqrt2)/(5+sqrt8)# can be written to #(11-sqrt2)/17#?

2 Answers
Aug 12, 2018

Answer:

Please see below.

Explanation:

We know that ,

#color(red)((1)a^2-b^2=(a-b)(a+b)#

Let ,

#X=(3+sqrt2)/(5+sqrt8)#

Multiplying numerator and denominator by # (5-sqrt8)#

#X=(3+sqrt2)/(5+sqrt8)xx (5-sqrt8)/(5-sqrt8)#

Simplifying we get

#X=(3(5-sqrt8)+sqrt2(5-sqrt8))/((5)^2-(sqrt8)^2)tocolor(red)([because(1)])#

#:.X=(15-3sqrt8+5sqrt2-sqrt16)/(25-8)#

#:.X#=#(15-3*2sqrt2+5sqrt2- 4)/17to[becausesqrt8=sqrt(4xx2)=2sqrt2]#

#:.X=(11-6sqrt2+5sqrt2)/17#

#:.X=(11-sqrt2)/17#

Aug 13, 2018

Answer:

By Rationalization

Explanation:

#(3 + sqrt2)/(5 + sqrt8)#

By Rationalization, Note: #a/(x + sqrty) = a/(x + sqrty) xx (x - sqrty)/(x - sqrty)# or vice versa..

Hence;

#(3 + sqrt2)/(5 + sqrt8) xx (5 - sqrt8)/(5 - sqrt8)#

Rationalizing..

#[(3 + sqrt2)(5 - sqrt8)]/[(5 + sqrt8)(5 - sqrt8)]#

Expanding..

#[(15 - 3sqrt8 + 5sqrt2 - sqrt16)]/[(25 - 8)]#

Simplifying..

#[(15 - 3sqrt8 + 5sqrt2 - 4)]/17#

Further simplifying..

#[(15 - 3sqrt8 + 5sqrt2 - 4)]/17#

Collecting like terms..

#[(15 - 4 - 3sqrt8 + 5sqrt2)]/17#

Simplifying..

#[(11 - 3sqrt(2 xx 4) + 5sqrt2)]/17#

Further simplifying..

#[(11 - 3(sqrt2 xx sqrt4)+ 5sqrt2)]/17#

#[(11 - 3(sqrt2 xx 2)+ 5sqrt2)]/17#

#[(11 - 3 xx 2(sqrt2)+ 5sqrt2)]/17#

#[(11 - 6sqrt2+ 5sqrt2)]/17#

#[(11 - sqrt2)]/17#

Therefore;

#(3 + sqrt2)/(5 + sqrt8) = (11 - sqrt2)/17#

As required..