Question

已知：a平方-5a+1=0,求a的4次+a平方+1/a平方的值.

Answer 1

a^2-5a+1=0,a+1/a=5 a^4+a^2+1/a =a(a^3+a+1/a+1/a^3)-1 =a((a+1/a)^3-2(a+1/a))-1 =a(5^3-2*5)-1 =115*a-1 a^2-5a+1=0,a=(5+sqrt(21))/2或a=(5-sqrt(21))/2 a^4+a^2+1/a=115*a-1 =115*(5+sqrt(21))/2-1或115*(5-sqrt(21))/2-1