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1.关于$x$的一元二次方程$x^2 + bx + 4 = 0$有一根为1，则另一根为
4
.答案
4
2.已知一元二次方程$x^2 -4x -c=0$的一个根是$2+\sqrt{3}$，则另一根为（
A.$-2-\sqrt{3}$
B.$\sqrt{3}-2$
C.$2-\sqrt{3}$
D.$-6-\sqrt{3}$
C
）A.$-2-\sqrt{3}$
B.$\sqrt{3}-2$
C.$2-\sqrt{3}$
D.$-6-\sqrt{3}$
答案
C
3.已知关于$x$的方程$x^2 -6x +k=0$的两根分别是$x_1,x_2$，且满足$\frac{1}{x_1}+\frac{1}{x_2}=3$，求$k$的值.
答案
解:$\because x_1 + x_2 = 6, x_1 x_2 = k$,
$\therefore \frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = 3, \therefore k = 2$.
$\therefore \frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = 3, \therefore k = 2$.
4.关于$x$的方程$x^2 - (m^2 - 4)x + m = 0$的两根互为相反数，求$m$的值.
答案
解:$x_1 + x_2 = m^2 - 4 = 0, \therefore m = \pm 2$,
当$m=2$时,$x^2 + 2 = 0, \Delta ＜ 0$,舍去,
当$m=-2$时,$x^2 - 2 = 0, \therefore m = -2$.
当$m=2$时,$x^2 + 2 = 0, \Delta ＜ 0$,舍去,
当$m=-2$时,$x^2 - 2 = 0, \therefore m = -2$.
5.若$m,n$是一元二次方程$x^2+x=4$的两个实数根,求多项式$2n^2 - mn - 2m$的值.
答案
解:$\because n^2 = 4 - n$,
$\therefore$原式$=2(4-n) - mn - 2m$
$=8 - 2(m + n) - mn$,
又$\because m + n = -1$,
$mn = -4$,
$\therefore$原式$=14$.
$\therefore$原式$=2(4-n) - mn - 2m$
$=8 - 2(m + n) - mn$,
又$\because m + n = -1$,
$mn = -4$,
$\therefore$原式$=14$.
6.关于$x$的一元二次方程$x^2 -4x -2m +5=0$的两根为$a,b$.
(1)若$a-b=2$,求$m$的值;
(2)若$a,b$均为正数,直接写出$m$的范围.
(1)若$a-b=2$,求$m$的值;
(2)若$a,b$均为正数,直接写出$m$的范围.
答案
解:(1)$\because a + b = 4, a - b = 2$,
$\therefore a = 3, b = 1$,
而$ab = 5 - 2m = 3, \therefore m = 1$;
(2)$\begin{cases} \Delta ≥ 0, \\ x_1 · x_2 ＞ 0, \end{cases} \therefore \frac{1}{2} ≤ m ＜ \frac{5}{2}$.
$\therefore a = 3, b = 1$,
而$ab = 5 - 2m = 3, \therefore m = 1$;
(2)$\begin{cases} \Delta ≥ 0, \\ x_1 · x_2 ＞ 0, \end{cases} \therefore \frac{1}{2} ≤ m ＜ \frac{5}{2}$.
7.已知$x_1,x_2$是关于$x$的一元二次方程$x^2 -6x +k=0$的两个实数根，且$x_1^2x_2^2 -x_1 -x_2=115$，求$k$的值.
答案
解:$\because x_1 + x_2 = 6, x_1 · x_2 = k$,
$\therefore k^2 - 6 = 115, k = \pm 11$,
又$\because \Delta = 36 - 4k ≥ 0$,
$\therefore k ≤ 9, \therefore k = -11$.
$\therefore k^2 - 6 = 115, k = \pm 11$,
又$\because \Delta = 36 - 4k ≥ 0$,
$\therefore k ≤ 9, \therefore k = -11$.
