is there a way to solve this by hand (1/2)x^4-x^2-1=0
Apr 16, 2009 #3 nameVoid 241 0 (1/2)x^4-x^2-1=(1/2)p^2-p-1=0 quadratic in p [1+||- (1+2)^(1/2) ] = 1+||-3^(1/2) = p = x^2 x=+||-[1+||-3^(1/2) ]^(1/2) =+||-[1+3^(1/2) ]^(1/2)
(1/2)x^4-x^2-1=(1/2)p^2-p-1=0 quadratic in p [1+||- (1+2)^(1/2) ] = 1+||-3^(1/2) = p = x^2 x=+||-[1+||-3^(1/2) ]^(1/2) =+||-[1+3^(1/2) ]^(1/2)
