Question

解关于x的方程:mx^2+(4m+1)x+4m+2=0和x^2-(p^2+q^2)x+pq(p+q)(p-q)=0这两个

Answer 1

mx^2+(4m+1)x+4m+2=0

[mx+(2m+1)](x+2)=0

所以

m=0,x=-2

m≠0,x=-(2m+1)/m,x=-2

x^2-(p^2+q^2)x+pq(p+q)(p-q)=0这

[x-q(p+q)][x-p(p-q)]=0

所以x=pq+q^2,x=p^2-pq