How do you simplify 1/{1+sqrt(3)-sqrt(5)}11+√3−√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))
