三、解答题
12.根据一元二次方程的根与系数的关系,求下列方程两个根$x_1,x_2$的和与积.
(1)$x^2 -4x +1=0$;
(2)$-3x^2 -2x +6=0$;
(3)$2x^2 -10=x^2 -8x$.

12.解:(1)$x_1+x_2=-(-4)=4,x_1x_2=1$.
(2)$x_1+x_2=-\frac{-2}{-3}=-\frac{2}{3},x_1x_2=\frac{6}{-3}=-2$. (3)方程化成一般形式为$x^2+8x-10=0,x_1+x_2=-8,x_1x_2=-10$.

13.设$x_1,x_2$是关于$x$的方程$(x-1)(x-2)=m^2$的两根.
(1)当$x_1=-1$时,求$x_2$及$m$的值;
(2)求证:$(x_1-1)(x_2-1)≤0$.

13.(1)解:把$x_1=-1$代入方程$(x-1)(x-2)=m^2$,得$m^2=6$,
$\therefore m=\pm\sqrt{6},\therefore (x-1)(x-2)=6$,即$x^2-3x-4=0.\therefore (x-4)(x+1)=0.\therefore x_1=-1,x_2=4$.
(2)证明:方程$(x-1)(x-2)=m^2$可化为$x^2-3x+2-m^2=0.\because \Delta=9-4(2-m^2)=4m^2+1＞0$,$\therefore$方程有两个不相等的实数根.$\because$方程$(x-1)(x-2)=m^2$即$x^2-3x+2-m^2=0$的两根为$x_1$、$x_2$,
$\therefore x_1+x_2=3,x_1x_2=2-m^2.\therefore (x_1-1)(x_2-1)=x_1x_2-(x_1+x_2)+1=2-m^2-3+1=-m^2.\because m^2≥0,\therefore -m^2≤0$,即$(x_1-1)(x_2-1)≤0$.

14.已知关于$x$的一元二次方程$x^2 - 2(a - 1)x + a^2 - a - 2 = 0$有两个不相等的实数根$x_1,x_2$。
(1)若$a$为正整数,求$a$的值;
(2)若$x_1,x_2$满足$x_1^2 + x_2^2 - x_1x_2 = 16$,求$a$的值.

14.解:(1)由题意,得$\Delta=[-2(a-1)]^2-4(a^2-a-2)＞0$,解得$a＜3$.$\because a$为正整数,$\therefore a=1$或$a=2$. (2)$\because x_1+x_2=2(a-1)$,$x_1x_2=a^2-a-2$,$x_1^2+x_2^2-x_1x_2=16$,$\therefore (x_1+x_2)^2-3x_1x_2=16$,即$[2(a-1)]^2-3(a^2-a-2)=16$,解得$a_1=-1,a_2=6$.$\because a＜3$,$\therefore a=-1$.