若关于x轴的不等式 ax＾2+bx+c＜0的解集是空集,则： A.a>0,且b＾2-4ac>0 B.a0,且b＾2-4ac>0 B.a0,且b＾2-4ac小于等于0D.a0怎么做,写出过程.

若关于x轴的不等式 ax＾2+bx+c＜0的解集是空集,则： A.a>0,且b＾2-4ac>0 B.a0,且b＾2-4ac>0 B.a0,且b＾2-4ac小于等于0D.a0怎么做,写出过程.
若关于x轴的不等式 ax＾2+bx+c＜0的解集是空集,则： A.a>0,且b＾2-4ac>0 B.a<0,且b＾2-4ac小于等...
若关于x轴的不等式
ax＾2+bx+c＜0的解集是空集,则：
A.a>0,且b＾2-4ac>0 B.a<0,且b＾2-4ac小于等于0
C.a>0,且b＾2-4ac小于等于0
D.a<0,且b＾2-4ac>0
怎么做,写出过程.

若关于x轴的不等式 ax＾2+bx+c＜0的解集是空集,则： A.a>0,且b＾2-4ac>0 B.a0,且b＾2-4ac>0 B.a0,且b＾2-4ac小于等于0D.a0怎么做,写出过程.
本题选C.不妨令f(x)=ax^2+bx+c.由题知：f(x)>或=0在R上恒成立.故可据此画出f(x)的图像,则可知：只有a>0,且b^2-4ac<或=0,即不让其图像与X轴有两个交点.

说明ax＾2+bx+c大于等于0是恒成立的，则开口向上，a>0,与X轴的只有一个交点或者没有交点，则b＾2-4ac小于等于0 所以答案为C

利用数形结合的思想：ax＾2+bx+c＜0的解集是空集===>y=ax＾2+bx+c的图像在x轴上方（可以相切）===>函数图象开口向上，和x轴至多一个交点===>方程ax＾2+bx+c=0无解或有重根===>C.a>0,且b＾2-4ac小于等于0

Create an informal type of a gentleman
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