题目内容
设x1,x2是一元二次方程ax2+bx+c=0(a≠0)的两根,
(1)试推导x1+x2=-
,x1•x2=
;
(2)求代数式a(x13+x23)+b(x12+x22)+c(x1+x2)的值.
(1)试推导x1+x2=-
| b |
| a |
| c |
| a |
(2)求代数式a(x13+x23)+b(x12+x22)+c(x1+x2)的值.
(1)∵x1、x2是ax2+bx+c=0(a≠0)的两根,
∴x1=
,x2=
,
∴x1+x2=
=-
,
x1•x2=
•
=
;
(2)∵x1,x2是ax2+bx+c=0的两根,
∴ax12+bx1+c=0,ax22+bx2+c=0,
∴原式=ax13+bx12+cx1+ax23+bx22+cx2,
=x1(ax12+bx1+c)+x2(ax22+bx2+c),
=0.
∴x1=
-b+
| ||
| 2a |
-b-
| ||
| 2a |
∴x1+x2=
-b+
| ||||
| 2a |
| b |
| a |
x1•x2=
-b+
| ||
| 2a |
-b-
| ||
| 2a |
| c |
| a |
(2)∵x1,x2是ax2+bx+c=0的两根,
∴ax12+bx1+c=0,ax22+bx2+c=0,
∴原式=ax13+bx12+cx1+ax23+bx22+cx2,
=x1(ax12+bx1+c)+x2(ax22+bx2+c),
=0.
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