若方程2x^2-3x-1=0的两根为x1x2 则x1^4+x2^4
问题描述：
若方程2x^2-3x-1=0的两根为x1x2 则x1^4+x2^4
答
2x^2-3x-1=0的两根为x1,x2
x1+x2=3/2
x1*x2=-1/2
x1^4+x2^4
=(x1²+x2²)²-2x1²x2²
=[(x1+x2)²-2x1x2]²-2x1²x2²
=[(3/2)²-2*(-1/2)]²-2*(-1/2)²
=161/16