How do you simplify #1/{1+sqrt(3)-sqrt(5)}#?
1 Answer
Explanation:
You're going to have to do a little work here to simplify this expression.
The way to go is by rationalizing the denominator. The only problem is the fact that your denominator is a trinomial, and conjugates are only formed for binomials.
More specifically, you get the conjugate of a binomial by changing the sign of its second term.
#a + b -> underbrace(a color(red)(-) b)_(color(blue)("conjugate"))" "# or#" "a - b -> overbrace(a color(red)(+) b)^(color(blue)("conjugate"))" "#
This means that you're going to have to group the denominator as a binomial, for which you can write
#overbrace(1)^(color(red)(a)) + overbrace((sqrt(3) - sqrt(5)))^(color(red)(b)) -> underbrace(1 color(red)(-) (sqrt(3) - sqrt(5)))_(color(blue)("conjugate"))#
So, multiply your expression by
#1/(1 + (sqrt(3) - sqrt(5))) * (1 - (sqrt(3) - sqrt(5)))/(1 - (sqrt(3) - sqrt(5)))#
#(1 - sqrt(3) + sqrt(5))/([1 + (sqrt(3) - sqrt(5))][1 - (sqrt(3) - sqrt(5))]#
The denominator can be rewritten as
#[1 + (sqrt(3) - sqrt(5))][1 - (sqrt(3) - sqrt(5))] = 1^2 - (sqrt(3) - sqrt(5))^2#
This, in turn, will be equal to
#1 - ((sqrt(3))^2 - 2sqrt(3 * 5) + (sqrt(5))^2) = 1 - 3 + 2sqrt(15) - 5#
#=2sqrt(15) - 7#
The expression becomes
#(1 - sqrt(3) + sqrt(5))/(2sqrt(15) - 7)#
Now do the same thing with the new denominator, i.e. find its conjugate
#2sqrt(15) - 7 -> 2sqrt(15) color(red)(+) 7#
and multiply the expression by
#(1 - sqrt(3) + sqrt(5))/(2sqrt(15) - 7) * (2sqrt(15) + 7)/(2sqrt(15) + 7)#
# ((1- sqrt(3) + sqrt(5))(2sqrt(15) + 7))/((2sqrt(15) - 7)(2sqrt(15) + 7))#
The denominator will be equal to
#(2sqrt(15) - 7)(2sqrt(15) + 7) = (2sqrt(15))^2 - 7^2#
# =4 * 15 - 49 = 11#
The numerator will be
#(1 - sqrt(3) + sqrt(5))(2sqrt(15) + 7)#
#2sqrt(15) - 2sqrt(45) + 2sqrt(75) + 7 - 7sqrt(3) + 7sqrt(5)#
#2sqrt(15) - 6sqrt(5) + 10sqrt(3) + 7 - 7sqrt(3) + 7sqrt(5)#
#7 + 3sqrt(3) + sqrt(5) + 2sqrt(15)#
The simplified expression will thus be
#1/(1 + sqrt(3) - sqrt(5)) = color(green)( (7 + 3sqrt(3) + sqrt(5) + 2sqrt(15))/(11))#
