x1 не равно x2 x1, x2 > 1/2 (a+3)x^2+3(a+1)x-2(a+1)=0

Question

x1 не равно x2

x1, x2 > 1/2

(a+3)x^2+3(a+1)x-2(a+1)=0

Answer 1

D=9(a+1)^2+8(a+3)(a+1)=(a+1)(9a+9+8a+24)=(a+1)(17a+33)>0,

(a+1)(17a+33)=0,

a1=-1 16/17, a2=-1,

a∈(-∞;-1 16/17)U(-1;+∞),

 

x1+x2=-3(a+1)/(a+3)>1/2+1/2,

-3a-3>a+3,

-4a>6,

a<-1 1/2,</p>

 

x1*x2=-2(a+1)/(a+3)>1/2*1/2,

-8a-8>a+3,

-9a>11,

a<-1 2/9,</p>

 

a<-1 16/17,</p>

a∈(-∞;-1 16/17).