题目内容
关于x的方程x2+2(k+1)x+k-2=0
(1)试说明:不论k取何值时,方程总有实数根;
(2)若方程有一根为x=1,求k的值并求出方程的另一根.
(1)试说明:不论k取何值时,方程总有实数根;
(2)若方程有一根为x=1,求k的值并求出方程的另一根.
(1)证明:∵△=[2(k+1)]2-4(k-2)
=4k2+8k+2-4k+8
=4k2+4k+10
=4(k2+k)+10
=4(k2+k+
-
)+10
=4(k+
)2-1+10
=4(k+
)2+9>0,
∴不论k取何值时,方程总有实数根;
(2)将x=1代入x2+2(k+1)x+k-2=0得,
1+2(k+1)+k-2=0,
解得,k=-
,
则k-2=-
-2=-
;
∴x•1=-
,
解得k=-
,
故k=-
,另一个根为x=-
.
=4k2+8k+2-4k+8
=4k2+4k+10
=4(k2+k)+10
=4(k2+k+
1 |
4 |
1 |
4 |
=4(k+
1 |
2 |
=4(k+
1 |
2 |
∴不论k取何值时,方程总有实数根;
(2)将x=1代入x2+2(k+1)x+k-2=0得,
1+2(k+1)+k-2=0,
解得,k=-
1 |
3 |
则k-2=-
1 |
3 |
7 |
3 |
∴x•1=-
7 |
3 |
解得k=-
7 |
3 |
故k=-
1 |
3 |
7 |
3 |
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