#1**0 **

If

\(a=\sqrt4+\sqrt5+a\\ then\\ 0=\sqrt4+\sqrt5\\ 0=2+\sqrt5\\ 2=-\sqrt5 \)

This is obviously nonsense.

so

This equation has no real solutions. In fact I do not think it has any solutions at all.

**Perhaps you wrote the question incorrrectly?**

Melody Jan 25, 2018

#3**0 **

There is STILL no solution the way you have the equations written down. However, if you modify the signs, + or - on the first and the fourth, then you have this solution:

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 25, 2018