When we rationalize the denominator of \(\frac{1}{1 + \sqrt{3} + \sqrt{5}},\)

we obtain an expression of the form \(a + b \sqrt{3} + c \sqrt{5} + d \sqrt{15},\)

where a, b, c, and d are rational numbers. Find a + b + c + d.

Guest Jul 26, 2019

#2**+1 **

Simplify the following:

1/(1 + sqrt(3) + sqrt(5)

Multiply numerator and denominator of 1/(1 + sqrt(3) + sqrt(5)) by 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)) = 1 (1 - sqrt(3) - sqrt(5)) + sqrt(3) (1 - sqrt(3) - sqrt(5)) + sqrt(5) (1 - sqrt(3) - sqrt(5)) = (1 - sqrt(3) - sqrt(5)) + (sqrt(3) - 3 - sqrt(15)) + (sqrt(5) - sqrt(15) - 5) = -2 sqrt(15) - 7:

(1 - sqrt(3) - sqrt(5))/(-2 sqrt(15) - 7)

Factor -1 out of -7 - 2 sqrt(15) giving -(2 sqrt(15) + 7):

(1 - sqrt(3) - sqrt(5))/(-(2 sqrt(15) + 7))

Multiply numerator and denominator of (1 - sqrt(3) - sqrt(5))/(-(2 sqrt(15) + 7)) by -1:

-(1 - sqrt(3) - sqrt(5))/(2 sqrt(15) + 7)

-(1 - sqrt(3) - sqrt(5)) = -1 + sqrt(3) + sqrt(5):

(-1 + sqrt(3) + sqrt(5))/(2 sqrt(15) + 7)

Multiply numerator and denominator of (-1 + sqrt(3) + sqrt(5))/(2 sqrt(15) + 7) by 2 sqrt(15) - 7:

((-1 + sqrt(3) + sqrt(5)) (2 sqrt(15) - 7))/((2 sqrt(15) + 7) (2 sqrt(15) - 7))

(2 sqrt(15) + 7) (2 sqrt(15) - 7) = 7 (-7) + 7×2 sqrt(15) + 2 sqrt(15) (-7) + 2 sqrt(15)×2 sqrt(15) = -49 + 14 sqrt(15) - 14 sqrt(15) + 60 = 11:

**((-1 + sqrt(3) + sqrt(5)) (2 sqrt(15) - 7))/11 = 1/11 (7 + 3 sqrt(3) - sqrt(5) - 2 sqrt(15)) =a + b + c + d =7 + 3 - 1 - 2 = 7**

Guest Jul 27, 2019