設(shè)原方程為ax^2+bx+c=0
則x1+x2=-b/a
x1x2=c/a
新方程的根為y1,y2
y1=x1^2,y2=x2^2
則有y1+y2=(x1+x2)^2-2x1x2=b^2/a^2-2c/a
y1y2=(x1x2)^2=c^2/a^2
所以新方程為;y^2-(b^2/a^2-2c/a)y+c^2/a^2=0
即a^2y^2-(b^2-2ac)y+c^2=0