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∵抛物線y=ax2+2x+c與x軸的兩個不同的交點都在原點右側,∴分兩種情况進行求當a>0時,△=4-4ac>0f(0)=c>0-22a>0,無解;當a<0時,△=4-4ac>0f(0)=c<0-22a>0,則a<0,c<0;則點(a,c)在第三象限.故答案為:三.
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