(1)$x_1^2 + x_2^2 - 3x_1x_2 =$;(2)$|x_1 - x_2| =$.

(1)30 (2)$\sqrt{29}$

变式1.已知$m,n$是方程$x^2 - 3x - 4 = 0$的两根，则$(m^2 - 1)(n^2 - 1)$的值是（）

A.0
B.$-6$
C.$-7$
D.6

解:$\because m^2 = 3m + 4,n^2 = 3n + 4$,
$\therefore$原式$=(3m + 3)(3n + 3)=9(mn + m + n + 1)=0$.

变式2.(2025·武昌)已知关于$x$的方程$x^2-(m-2)x-\frac{m^2}{4}=0$.若这个方程的两个实数根满足$x_1=x_2+2$,求$m$的值及相应的两根.

解:$\because x_1 = x_2 + 2,\therefore x_1 - x_2 = 2$,
$\therefore (x_1 - x_2)^2 = 4$,
$(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$,
又$x_1 + x_2 = m - 2,x_1 · x_2 = -\frac{m^2}{4}$,
$\therefore (m - 2)^2 + m^2 = 4$,
解得$m = 0$或$m = 2$,
$\therefore$当$m = 0$时,
解得$x_1 = 0,x_2 = -2$;
当$m = 2$时,
解得$x_1 = 1,x_2 = -1$.

【典例2】关于$x$的一元二次方程$x^2 - mx + 2m - 1 = 0$的两个实数根分别为$x_1$、$x_2$，且$x_1^2 + x_2^2 = 7$。
(1)求$m$的值；
(2)求$x_1^3 + 4x_2 - 6$的值。

解:(1)$x_1 + x_2 = m,x_1 · x_2 = 2m - 1$,
$(x_1 + x_2)^2 - 2x_1x_2 = 7$,
$m^2 - 2(2m - 1) = 7$,
$m^2 - 4m - 5 = 0,m_1 = -1,m_2 = 5$,
又$\because \Delta \ge 0,\therefore m = -1$;
(2)$x_1^2 = 3 - x_1$,
$x_1^3 = 3x_1 - x_1^2 = 4x_1 - 3$,
$\therefore$原式$= 4x_1 - 3 + 4x_2 - 6 = -13$.

变式.关于$x$的一元二次方程$x^2+(2k+1)x+k^2+1=0$有两个不相等的实数根$x_1,x_2$.
(1)求实数$k$的取值范围；
(2)若方程的两实根$x_1,x_2$满足$|x_1+x_2|=x_1· x_2$，求$k$的值.

解:(1)$\because$原方程有两个不相等的实数根,
$\therefore \Delta = (2k + 1)^2 - 4(k^2 + 1) = 4k^2 + 4k + 1 - 4k^2 - 4 = 4k - 3 ＞ 0$,解得$k ＞ \frac{3}{4}$;
(2)$\because x_1 + x_2 = -(2k + 1),x_1x_2 = k^2 + 1$,
$\therefore |2k + 1| = k^2 + 1$,
$\therefore k_1 = 0,k_2 = 2$,
又$\because k ＞ \frac{3}{4},\therefore k = 2$.