Question

设方程x^2+px+q=0的两根之差与方程x^2+qx+p=0的两根之差相等,求p+q的值

Answer 1

设x^2+px+q=0的两根为x1,x2,

x^2+qx+p=0的两根为x3,x4,

由韦达定理：x1+x2=-p, x1*x2=q

x3+x4=-q,x3*x4=p

又∵x1-x2＝x3-x4

(x1-x2)2＝(x3-x4)2

(x1+x2)2-4x1*x2＝(x3+x4)2-4x3*x4

p2-4q=q2-4p

(p-q)(p+q+4)=0

p+q=0或p+q= -4