Question

已知实系数一元二次方程ax^2+bx+c=0有两个实数根x1,x2,a>b>c,求d=|x1-x2|的取值范围.

Answer 1

∵原方程有两个实数根,∴判别时△=b^2-4ac≥0.

由韦达定理得：x1+x2=-b/a,x1*x2=c/a.

(x1+x2)^2=(-b/a)^2=b^2/a^2.

(x1-x2)^2=(x1+x2)^2-4x1x2.

=b^2/a^2-4c/a.

=(b^2-4ac)/a^2

|x1-x2|^2=(x1-x2)^2.

|x1-x2|=√(b^2-4ac)/a.

∵b^2-4ac≥0,a＞0.

∴0＜|x1-x2|≤√(b^2-4ac)/a.