已知一元二次方程ax^2+bx+c=0的两根之和为p,两根的平方根为q,两根的立方和为r,求ar+bq+cp的值.

Question

已知一元二次方程ax^2+bx+c=0的两根之和为p,两根的平方根为q,两根的立方和为r,求ar+bq+cp的值.

Answer 1

如果你的题设似乎这样的话呢
两根之和为p,两根的平方和为q,两根的立方和为r 就会有
x1+x2=-(b/a)=p
x1*x2=c/a
x1^2+x2^2
=(x1+x2)^2-2x1*x2
=p^2-2c/a
=(b/a)^2-2c/a
=(b^2 - 2 a c)/a^2
=q
x1^3+x2^3
=(x1+x2)^3-3x1*x2(x1+x2)
=p^3-3p*(c/a)
=(b/a)^3-3(b/a)(c/a)
=(b^3 - 3 a b c)/a^3
=r
ar+bq+cp
=a*(b^3 - 3 a b c)/a^3+b*(b^2 - 2 a c)/a^2+c*(-b/a)
=(-b^3 + 3 a b c)/a^2+(b^3 - 2 a b c)/a^2-bc/a
=(b c)/a-bc/a
=0