Question

（1）一元二次不等式ax^2+bx+2>0的解集是（-1/2，-1/3），求a+b的值?（2）若x∈R，不等式ax^2+ax+1>0恒...

（1）一元二次不等式ax^2+bx+2>0的解集是（-1/2，-1/3），求a+b的值?（2）若x∈R，不等式ax^2+ax+1>0恒成立，求a的取值范围

Answer 1

ax^2+bx+2>0的解集为-1/2<x<1/3
所以2/a=-(1/2)*(-1/3)
a=12
因为
-b/a=-1/2-1/3
b/12=5/6
b=10
a+b=12+10=24
2.开口向上a>0
没有交点a²-4a<0
即0＜a ＜4

Answer 2

1,ax平方+bx+2=0的2个根为x1=-1/2,x2=1/3,
韦达定理:x1+x2=-b/a,x1*x2=2/a,
-1/2+1/3=-b/a,-1/2*1/3=2/a,解方程组得:a=-12,b=-2,a+b=-14;
2,
ax^2+ax+1=0无实数解,△=a平方-4*a*1<0,(a-2)平方<4,0<a<4;

Answer 3

解：（1）ax^2+bx+2>0的解集是（-1/2，-1/3），知道-1/2和-1/3恰好是ax^2+bx+2=0的两个根，将x=-1/2和x=-1/3代入ax^2+bx+2=0，解得：a=12，b=10，所以a+b=22。
（2）a>0，且图形恒在X轴上方，a²-4a<0 即0＜a ＜4

Answer 4

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