#"Solve the characteristic quartic equation without the y's first :"#
#x^4 + 2 x^3 - 3 x^2 - 4x - 1 = 0#
#=> (x^2-x-1)(x^2+3x+1) = 0" (*)"#
#"1) "x^2+3x+1=0 => x = (-3 pm sqrt(5))/2#
#"2) "x^2-x-1 = 0 => x = (1 pm sqrt(5))/2#
#"If we apply this on the given polynomial we get"#
#(x^2 - x y - y^2)(x^2 + 3 xy + y^2) = 0#
#=> (x-(1+sqrt(5))y/2)(x-(1-sqrt(5))y/2)#
#(x+(3+sqrt(5))y/2)(x-(sqrt(5)-3)y/2)=0#
#"(*) With the substitution "x = y-1/2" we get : "#
#y^4 - (9/2) y^2 + 1/16 = 0#
#"Now put "z = y^2" and multiply with 16 : "#
#16 z^2 - 72 z + 1 = 0#
#"disc : "72^2 - 4*16 = 5120 = 32^2 * 5#
#=> z = (72 pm 32 sqrt(5))/32 = 9/4 pm sqrt(5)#
#=> y = pm sqrt(9/4 pm sqrt(5))#
#=> y = pm sqrt(9 pm 4 sqrt(5))/2#
#=> y = pm sqrt((2 pm sqrt(5))^2)/2#
#=> y = pm(1 pm sqrt(5)/2)#
#=> x = (1 pm sqrt(5))/2 " or "(-3 pm sqrt(5))/2#
