How do you rationalize the denominator and simplify #(2-3sqrt5) / (3+2sqrt5)#?

1 Answer
Apr 7, 2016

Answer:

#(-36 +-13sqrt(5))/11#

Explanation:

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#