【答案】
分析:(1)由一元二次方程的二次項(xiàng)系數(shù)不為0和根的判別式求出a的取值范圍.設(shè)拋物線y=x
2-(2a+1)x+2a-5與x軸的兩個(gè)交點(diǎn)的坐標(biāo)分別為(α,0)、(β,0),且α<β,∴α、β是關(guān)于x的方程x
2-(2a+1)x+2a-5=0的兩個(gè)不相等的實(shí)數(shù)根,再利用方程x
2-(2a+1)x+2a-5=0的根的判別式求a的取值范圍,又∵拋物線y=x
2-(2a+1)x+2a-5與x軸的兩個(gè)交點(diǎn)分別位于點(diǎn)(2,0)的兩旁,利用根與系數(shù)的關(guān)系確定;
(2)把代數(shù)式變形后,利用根與系數(shù)的關(guān)系求出a的值.
解答:解:(1)∵關(guān)于x的方程(a+2)x
2-2ax+a=0有兩個(gè)不相等的實(shí)數(shù)根
∴
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解得:a<0,且a≠-2 ①
設(shè)拋物線y=x
2-(2a+1)x+2a-5與x軸的兩個(gè)交點(diǎn)的坐標(biāo)分別為(α,0)、(β,0),且α<β
∴α、β是關(guān)于x的方程x
2-(2a+1)x+2a-5=0的兩個(gè)不相等的實(shí)數(shù)根
∵△=[-(2a+1)]
2-4×1×(2a-5)=(2a-1)
2+21>0
∴a為任意實(shí)數(shù)②
由根與系數(shù)關(guān)系得:α+β=2a+1,αβ=2a-5
∵拋物線y=x
2-(2a+1)x+2a-5與x軸的兩個(gè)交點(diǎn)分別位于點(diǎn)(2,0)的兩旁
∴α<2,β>2
∴(α-2)(β-2)<0
∴αβ-2(α+β)+4<0
∴2a-5-2(2a+1)+4<0
解得:a>-
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③
由①、②、③得a的取值范圍是-
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<a<0;
(2)∵x
1和x
2是關(guān)于x的方程(a+2)x
2-2ax+a=0的兩個(gè)不相等的實(shí)數(shù)根
∴x
1+x
2=
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,x
1x
2=
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∵-
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<a<0,∴a+2>0
∴x
1x
2=
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<0不妨設(shè)x
1>0,x
2<0
∴|x
1|+|x
2|=x
1-x
2=2
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∴x
12-2x
1x
2+x
22=8,即(x
1+x
2)
2-4x
1x
2=8
∴(
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)
2-
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=8
解這個(gè)方程,得:a
1=-4,a
2=-1(16分)
經(jīng)檢驗(yàn),a
1=-4,a
2=-1都是方程(
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)
2-
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=8的根
∵a=-4<-
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,舍去
∴a=-1為所求.
點(diǎn)評(píng):本題綜合性強(qiáng),考查了一元二次方程中的根與系數(shù)的關(guān)系和根的判別式的綜合利用.