Given:" "(2-3sqrt(5))/(3+2sqrt(5))
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The value of 1 can be written in many forms:
1"; "2/2"; "(-5)/(-5)"; "(3-2sqrt(5))/(3-2sqrt(5))
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the principle that " "a^2-b^2=(a-b)(a+b)
Multiply by 1 but in the form 1=(3-2sqrt(5))/(3-2sqrt(5))
(2-3sqrt(5))/(3+2sqrt(5)) xx (3-2sqrt(5))/(3-2sqrt(5))" "The denominator is now of form (a-b)(a+b)
((2-3sqrt(5))(3-2sqrt(5)))/(3^2-[(2sqrt(5))^2])
(6-4sqrt(5)-9sqrt(5)+6(sqrt(5))^2)/(9-20)
(6-13sqrt(5)+30)/(9-20)
(36-13sqrt(5))/(-11)" Multiply by 1:"->(-1)/(-1) xx(36-13sqrt(5))/(-11)
(13sqrt(5)-36)/11
But sqrt(5) can be either positive or negative.
In that (-5)xx(-5)" "=" "(+5)xx(+5)" "=" "+5
(-36 +-13sqrt(5))/11
