Let a, b, c, and d be distinct real numbers such that

a = (sqrt4 + (sqrt(5+a)))

b = (sqrt4 - (sqrt(5+b)))

c = (sqrt4 + (sqrt(5-c)))

d = (sqrt4 - (sqrt(5-d)))

Compute abcd.

Destryox Jan 27, 2021

#1**0 **

Solve the following system:

{a = -a + 2 + sqrt(5) | (equation 1)

b = -b + 2 - sqrt(5) | (equation 2)

c = -c + 2 + sqrt(5) | (equation 3)

d = -d + 2 - sqrt(5) | (equation 4)

Express the system in standard form:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+2 b+0 c+0 d = 2 - sqrt(5) | (equation 2)

0 a+0 b+2 c+0 d = 2 + sqrt(5) | (equation 3)

0 a+0 b+0 c+2 d = 2 - sqrt(5) | (equation 4)

Divide equation 4 by 2:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+2 b+0 c+0 d = 2 - sqrt(5) | (equation 2)

0 a+0 b+2 c+0 d = 2 + sqrt(5) | (equation 3)

0 a+0 b+0 c+d = 1 - (sqrt(5))/(2) | (equation 4)

Divide equation 3 by 2:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+2 b+0 c+0 d = 2 - sqrt(5) | (equation 2)

0 a+0 b+c+0 d = (sqrt(5) + 2)/(2) | (equation 3)

0 a+0 b+0 c+d = 1 - sqrt(5)/2 | (equation 4)

Divide equation 2 by 2:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+b+0 c+0 d = 1 - (sqrt(5))/(2) | (equation 2)

0 a+0 b+c+0 d = 1/2 (2 + sqrt(5)) | (equation 3)

0 a+0 b+0 c+d = 1 - sqrt(5)/2 | (equation 4)

Divide equation 1 by 2:

{a+0 b+0 c+0 d = (sqrt(5) + 2)/(2) | (equation 1)

0 a+b+0 c+0 d = 1 - sqrt(5)/2 | (equation 2)

0 a+0 b+c+0 d = 1/2 (2 + sqrt(5)) | (equation 3)

0 a+0 b+0 c+d = 1 - sqrt(5)/2 | (equation 4)

Collect results:

a = 1/2 (2 + sqrt(5))

b = 1 - sqrt(5)/2

c = 1/2 (2 + sqrt(5))

d = 1 - sqrt(5)/2 If you multiply them together, you get:abcd =1/16

Guest Jan 27, 2021