+118608 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?
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
