分类：教育/科学-理工学科-数学
问题：数学若r，s是方程x&sup2;
若r，s是方程x²-px+q=0的两根，求以r²+1/s²，s²+1/r²为根的方程
回答：r+s=p, rs=q
r^2+s^2=(r+s)^2-2rs=p^2-2q
r^2s^2=(rs)^2=q^2
1/r^2+1/s^2=(r^2+s^2)/(r^2s^2)=(p^2-2q)/q^2
r^4+q^4=(r^2+q^2)^2-2r^2q^2=(p^2-2q)^2-2q^2
=p^4-2p^2q+2q^2

r^2+1/s^2+s^2+1/r^2=p^2-2q+(p^2-2q)/q^2
=(p^2-2q)(q^2+1)/q^2
(r^2+1/s^2)(s^2+1/r^2)
=r^2s^2+1/(r^2s^2)+r^2/s^2+s^2/r^2
=r^2s^2+1/(r^2s^2)+(r^4+q^4)/(s^2r^2)
=q^2+1/q^2+p^4-2p^2q+2q^2
=(p^4q^2-2p^2q^3+3q^4)/q^2

所求方程是
x^2-[(p^2-2q)(q^2+1)/q^2]x+(p^4q^2-2p^2q^3+3q^4)/q^2=0
化简为
q^2x^2-(p^2-2q)(q^2+1)x+(p^4q^2-2p^2q^3+3q^4)=0