Question

一元二次方程.有过程最好.1.已知x1,x2是方程x2+kx+p=0的两根,x3，x4是方程x2+kx+q=0的两根,则

Answer 1

(1)由题意得x1+x2=-k x1*x2=p x3+x4=-k x3*x4=q推出得x1+x2=x3+x4推出得x1-x3=x4-x2 x1-x4=x3-x2题目中式子等于(x1-x3)^2*(x1-x4)^2=[(x1-x3)*(x1-x4)]^2=[x1^2-(x3+x4)x1+x3x4]^2=[x1^2+kx1+q]^2令x1^2+kx1+q=A有因为x1是的X^2+kx+p=0解所以x1^2+Kx1+p=0推出得A=q-p所以答案为(q-p)^2 (2)不会(3)由题意得B^2-4AC>=0推出得a^2+a-2b>=0又因为对任意a都有解所以B^2+4AC