题目内容
已知方程x2+3x-5=0的两根为x1、x2,求
+
值.
| x2 |
| x1 |
| x1 |
| x2 |
分析:根据根与系数的关系得到x1+x2=-3,x1x2=-5,再变形
+
=
=
,然后利用整体思想进行计算.
| x2 |
| x1 |
| x2 |
| x2 |
| ||||
| x1x2 |
| (x1+x2)2-2x1x2 |
| x1x2 |
解答:解:根据题意得x1+x2=-3,x1x2=-5,
+
=
=
=
=-
.
| x2 |
| x1 |
| x2 |
| x2 |
| ||||
| x1x2 |
=
| (x1+x2)2-2x1x2 |
| x1x2 |
=
| (-3)2-2×(-5) |
| -5 |
=-
| 19 |
| 5 |
点评:本题考查了一元二次方程ax2+bx+c=0(a≠0)的根与系数的关系:若方程的两根为x1,x2,则x1+x2=-
,x1•x2=
.
| b |
| a |
| c |
| a |
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