《2025年拔尖特训九年级数学上册人教版》

14. (1)若$p = -4,q = 3$,求方程$x^2 + px + q = 0$的两根.
(2)已知实数a,b满足$a^2 - 15a - 5 = 0,b^2 - 15b - 5 = 0$,求$\frac{a}{b} + \frac{b}{a}$的值.
(3)已知关于x的方程$x^2 + mx + n = 0(n \neq 0)$,求出一个一元二次方程,使它的两个根分别是已知方程两根的倒数.

答案：
(1)$\because p=-4,q=3,$$\therefore$方程为$x^{2}-4x+3=0$，解得$x_{1}=3,x_{2}=1.$（2）$\because a,b$满足$a^{2}-15a-5=0,b^{2}-15b-5=0,$$\therefore a,b$是方程$x^{2}-15x-5=0$的解.当$a≠b$时，$a+b=15,ab=-5.$$\therefore \frac{a}{b}+\frac{b}{a}=\frac{a^{2}+b^{2}}{ab}=\frac{(a+b)^{2}-2ab}{ab}=\frac{15^{2}-2×(-5)}{-5}=-47.$当$a=b$时，原式$=2.$$\therefore \frac{a}{b}+\frac{b}{a}$的值为-47或2.（3）设方程$x^{2}+mx+n=0(n≠0)$的两个根分别是$x_{1},x_{2}.$$\therefore \frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1}x_{2}}=-\frac{m}{n},\frac{1}{x_{1}}\cdot\frac{1}{x_{2}}=\frac{1}{x_{1}x_{2}}=\frac{1}{n}.$$\therefore$方程$x^{2}+\frac{m}{n}x+\frac{1}{n}=0$的两个根分别是已知方程两根的倒数（方程不唯一）.

15. 若$x_1,x_2$是关于x的一元二次方程$x^2 - 2(m + 1)x + m^2 + 2 = 0$的两个实数根,且$(x_1 + 1)(x_2 + 1) = 8$,则$m = $1.

答案： 1 解析：$\because x_{1},x_{2}$是关于$x$的一元二次方程$x^{2}-2(m+1)x+m^{2}+2=0$的两个实数根，$\therefore x_{1}+x_{2}=2(m+1),x_{1}x_{2}=m^{2}+2.\because (x_{1}+1)(x_{2}+1)=8$，即$x_{1}x_{2}+x_{1}+x_{2}+1=8,\therefore m^{2}+2+2(m+1)+1=8$，解得$m=1$或$m=-3.\because \Delta=[-2(m+1)]^{2}-4(m^{2}+2)=8m-4\geq0$，解得$m\geq\frac{1}{2},\therefore m=1.$

16. 阅读材料:
材料1:为了解方程$(x^2)^2 - 13x^2 + 36 = 0$,若把$x^2$看成一个整体,设$y = x^2$,则原方程可化为$y^2 - 13y + 36 = 0$,经过运算,原方程的解为$x_1 = 2,x_2 = -2,x_3 = 3,x_4 = -3$.我们把以上这种解决问题的方法叫做换元法.
材料2:已知实数m,n满足$m^2 - m - 1 = 0,n^2 - n - 1 = 0$,且$m \neq n$,显然m,n是方程$x^2 - x - 1 = 0$的两个不等的实数根,由一元二次方程根与系数的关系,可知$m + n = 1,mn = -1$.
根据材料,解答下列问题:
(1)直接应用:方程$x^4 - 5x^2 + 6 = 0$的解为$x_{1}=\sqrt{2},x_{2}=-\sqrt{2},x_{3}=\sqrt{3},x_{4}=-\sqrt{3}$.
(2)间接应用:已知实数a,b满足$2a^4 - 7a^2 + 1 = 0,2b^4 - 7b^2 + 1 = 0$,且$a \neq b$,求$a^4 + b^4$的值.$\because a≠b,$$\therefore a^{2}≠b^{2}$或$a^{2}=b^{2}(a=-b).$① 当$a^{2}≠b^{2}$时，令$a^{2}=m,b^{2}=n.$$\therefore m≠n$，则$2m^{2}-7m+1=0,2n^{2}-7n+1=0.$$\therefore m,n$是方程$2x^{2}-7x+1=0$的两个不等的实数根.$\therefore m+n=\frac{7}{2},mn=\frac{1}{2}.$$\therefore a^{4}+b^{4}=m^{2}+n^{2}=(m+n)^{2}-2mn=\frac{45}{4}.$② 当$a^{2}=b^{2}(a=-b)$时，易得$a^{2}=b^{2}=\frac{7\pm\sqrt{41}}{4}$，此时$a^{4}+b^{4}=2a^{4}=2(a^{2})^{2}=\frac{45\pm7\sqrt{41}}{4}.$综上所述，$a^{4}+b^{4}$的值为$\frac{45}{4}$或$\frac{45+7\sqrt{41}}{4}$或$\frac{45-7\sqrt{41}}{4}.$
(3)拓展应用:已知实数m,n满足$\frac{1}{m^4} + \frac{1}{m^2} = 7,n^2 - n = 7$,且$n ＞ 0$,求$\frac{1}{m^4} + n^2$的值.令$\frac{1}{m^{2}}=a,-n=b$，则$a^{2}+a-7=0,b^{2}+b-7=0.$$\because n＞0,$$\therefore \frac{1}{m^{2}}≠-n$，即$a≠b.$$\therefore a,b$是方程$x^{2}+x-7=0$的两个不等的实数根.$\therefore a+b=-1,ab=-7.$$\therefore \frac{1}{m^{4}}+n^{2}=a^{2}+b^{2}=(a+b)^{2}-2ab=15.$

答案：
(1)$x_{1}=\sqrt{2},x_{2}=-\sqrt{2},x_{3}=\sqrt{3},x_{4}=-\sqrt{3}.$（2）$\because a≠b,$$\therefore a^{2}≠b^{2}$或$a^{2}=b^{2}(a=-b).$① 当$a^{2}≠b^{2}$时，令$a^{2}=m,b^{2}=n.$$\therefore m≠n$，则$2m^{2}-7m+1=0,2n^{2}-7n+1=0.$$\therefore m,n$是方程$2x^{2}-7x+1=0$的两个不等的实数根.$\therefore m+n=\frac{7}{2},mn=\frac{1}{2}.$$\therefore a^{4}+b^{4}=m^{2}+n^{2}=(m+n)^{2}-2mn=\frac{45}{4}.$② 当$a^{2}=b^{2}(a=-b)$时，易得$a^{2}=b^{2}=\frac{7\pm\sqrt{41}}{4}$，此时$a^{4}+b^{4}=2a^{4}=2(a^{2})^{2}=\frac{45\pm7\sqrt{41}}{4}.$综上所述，$a^{4}+b^{4}$的值为$\frac{45}{4}$或$\frac{45+7\sqrt{41}}{4}$或$\frac{45-7\sqrt{41}}{4}.$（3）令$\frac{1}{m^{2}}=a,-n=b$，则$a^{2}+a-7=0,b^{2}+b-7=0.$$\because n＞0,$$\therefore \frac{1}{m^{2}}≠-n$，即$a≠b.$$\therefore a,b$是方程$x^{2}+x-7=0$的两个不等的实数根.$\therefore a+b=-1,ab=-7.$$\therefore \frac{1}{m^{4}}+n^{2}=a^{2}+b^{2}=(a+b)^{2}-2ab=15.$