(4a)^2-4* 1*(-4a+3)≥0

(a- 1)^2-4* 1*a^2≥0

(2a)^2-4* 1*(-2a)≥0

Find the range of a separately and take the union.

The final answer should be: A.

2.A∩B≠ empty set

-2∈A

x^2+px+q=0 ( 1)

qx^2+px+ 1=0 (2)

Two equations have the same real root, and -2 is the solution of equation (1).

The same root is X.

x-2=-p，

-2x=q，

qx^2+px+ 1=0

-2x*x^2+(2-x)x+ 1=0

2x^3+x^2-2x- 1=0

x^3- 1+x^3+x^2-2x=0

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

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

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

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

X= 1 or x=- 1/2 or x=- 1

When 1 and x= 1, P = 1 and Q =-2.

2. When x=- 1/2, p = 5/2 and q = 1.

3. when x=- 1, p=3 q=2.

There are two sources.

So this is a quadratic equation.

1-a is not equal to 0, and s is not equal to 1.

There are two positive sources.

x 1+x2 & gt； 0，x 1x 2 & gt； 0

So x1+x2 =-(a+2)/(1-a) > 0.

(a+2)(a- 1)>0

a & lt-2，x & gt 1

x 1x2=-4/( 1-a)>0

a- 1 >0，a & gt 1

So a> 1

Discriminant is greater than 0

(a+2)^2+ 16( 1-a)>； 0

a^2+4a+4- 16a+ 16>； 0

a^2- 12a+20>； 0

(a-2)(a- 10)>0

a & lt2，a & gt 10

To sum it up

1 & lt； a & lt2，a & gt 10