Question

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

Answer 1

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`

Answer 2

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..