《2025年启东中学作业本九年级数学上册人教版》

12. 已知方程$4x^{2}-2x - 1 = 0的两个根分别为x_{1}$,$x_{2}$,不解方程,求下列代数式的值:
(1)$\frac{1}{x_{1}}+\frac{1}{x_{2}}$;
(2)$x_{1}^{2}+x_{2}^{2}$;
(3)$\frac{x_{2}}{x_{1}}+\frac{x_{1}}{x_{2}}$;
(4)$(x_{1}-x_{2})^{2}$.

答案： 解:
∵方程$4x^{2}-2x-1=0$的两个根分别为$x_{1},x_{2}$,
$\therefore x_{1}+x_{2}=\frac{1}{2}$,$x_{1}x_{2}=-\frac{1}{4}$.
(1)原式$=\frac{x_{1}+x_{2}}{x_{1}x_{2}}=\frac{\frac{1}{2}}{-\frac{1}{4}}=-2$.
(2)原式$=(x_{1}+x_{2})^{2}-2x_{1}x_{2}=\frac{1}{4}-2×(-\frac{1}{4})=\frac{3}{4}$.
(3)原式$=\frac{(x_{1}+x_{2})^{2}-2x_{1}x_{2}}{x_{1}x_{2}}=\frac{\frac{3}{4}}{-\frac{1}{4}}=-3$.
(4)原式$=(x_{1}+x_{2})^{2}-4x_{1}x_{2}=\frac{1}{4}-4×(-\frac{1}{4})=\frac{5}{4}$.

13. (2024·钱塘区期末)已知关于$x的一元二次方程x^{2}-2(k - 1)x + k^{2}+3 = 0$.
(1)若该方程有一个根是$-2$,求$k$的值;
(2)若该方程有两个实数根,求$k$的取值范围;
(3)若该方程的两个实数根$x_{1}$,$x_{2}满足(x_{1}-1)(x_{2}-1)= 14$,求$k$的值.

答案： 解:
(1)当$x=-2$时,$4-2(k-1)×(-2)+k^{2}+3=0$,
整理得$k^{2}+4k+3=0$,解得$k=-1$或$-3$.
(2)根据题意得$\Delta=(2k-2)^{2}-4(k^{2}+3)\geq0$,
解得$k\leq-1$.
(3)根据题意得$x_{1}+x_{2}=2k-2$,$x_{1}x_{2}=k^{2}+3$.
$\because (x_{1}-1)(x_{2}-1)=14$,$\therefore x_{1}x_{2}-(x_{1}+x_{2})+1=14$,
即$k^{2}+3-(2k-2)+1=14$,
整理得$k^{2}-2k-8=0$,解得$k_{1}=-2$,$k_{2}=4$,
$\because k\leq-1$,$\therefore k=-2$.

14. (2024·内江)已知关于$x的一元二次方程x^{2}-px + 1 = 0$($p$为常数)有两个不相等的实数根$x_{1}和x_{2}$.
(1)填空:$x_{1}+x_{2}=$p,$x_{1}x_{2}=$1;
(2)求$\frac{1}{x_{1}}+\frac{1}{x_{2}}$,$x_{1}+\frac{1}{x_{1}}$的值;
解:$\because x_{1}+x_{2}=p$,$x_{1}x_{2}=1$,
$\therefore \frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{2}+x_{1}}{x_{1}x_{2}}=\frac{p}{1}=p$.
$\because$关于$x$的一元二次方程$x^{2}-px+1=0$（p为常数）有两个不相等的实数根$x_{1}$和$x_{2}$,
$\therefore x_{1}^{2}-px_{1}+1=0$,$\therefore x_{1}-p+\frac{1}{x_{1}}=0$,即$x_{1}+\frac{1}{x_{1}}=p$.
(3)已知$x_{1}^{2}+x_{2}^{2}= 2p + 1$,求$p$的值.
解:由根与系数的关系得$x_{1}+x_{2}=p$,$x_{1}x_{2}=1$,
$\because x_{1}^{2}+x_{2}^{2}=2p+1$,$\therefore (x_{1}+x_{2})^{2}-2x_{1}x_{2}=2p+1$,
$\therefore p^{2}-2=2p+1$,解得$p_{1}=3$,$p_{2}=-1$,
当$p=3$时,$\Delta=p^{2}-4=9-4=5＞0$;
当$p=-1$时,$\Delta=p^{2}-4=-3＜0$,$\therefore p=3$.

答案：
(1)p 1
(2)解:$\because x_{1}+x_{2}=p$,$x_{1}x_{2}=1$,
$\therefore \frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{2}+x_{1}}{x_{1}x_{2}}=\frac{p}{1}=p$.
$\because$关于$x$的一元二次方程$x^{2}-px+1=0$（p为常数）有两个不相等的实数根$x_{1}$和$x_{2}$,
$\therefore x_{1}^{2}-px_{1}+1=0$,$\therefore x_{1}-p+\frac{1}{x_{1}}=0$,即$x_{1}+\frac{1}{x_{1}}=p$.
(3)解:由根与系数的关系得$x_{1}+x_{2}=p$,$x_{1}x_{2}=1$,
$\because x_{1}^{2}+x_{2}^{2}=2p+1$,$\therefore (x_{1}+x_{2})^{2}-2x_{1}x_{2}=2p+1$,
$\therefore p^{2}-2=2p+1$,解得$p_{1}=3$,$p_{2}=-1$,
当$p=3$时,$\Delta=p^{2}-4=9-4=5＞0$;
当$p=-1$时,$\Delta=p^{2}-4=-3＜0$,$\therefore p=3$.