已知抛物线y=x 2 +(2a-1)x+a 2 +3a+ 17 4 与x轴交于点A(x 1 ,0),B(x 2 ,0).
1个回答

(1)∵抛物线y=x 2+(2a-1)x+a 2+3a+

17

4 与x轴交于点A(x 1,0),B(x 2,0).

∴b 2-4ac>0,

即(2a-1) 2-4(a 2+3a+

17

4 )>0,

解得a<-1.

(2)设方程x 2+(2a-1)x+a 2+3a+

17

4 =0的两根为x 1,x 2

∴x 1+x 2=1-2a,x 1•x 2=a 2+3a+

17

4 ,

∵x 1 2+x 2 2=(x 1+x 2 2-2x 1•x 2=(1-2a) 2-2(a 2+3a+

17

4 )=2(a-

5

2 ) 2-20,

∵a<-1,

∴(a-

5

2 ) 2

49

4 ,

∴2(a-

5

2 ) 2-20>

9

2 ,

即S>

9

2 .