【例1】设α、β是一元二次方程$x^2 + 2x - 1 = 0$的两个根,则αβ的值是:(
A.2
B.1
C.−2
D.−1
D
)A.2
B.1
C.−2
D.−1
答案
D
练习1.若$x_1$、$x_2$是一元二次方程$x^2 -5x +6=0$的两个根,则$x_1 +x_2$的值是(
A.1
B.5
C.−5
D.6
B
)A.1
B.5
C.−5
D.6
答案
B
练习2.下列一元二次方程有两个互为倒数的实数根的是(
A.$2x^2 - 3x + 1 = 0$
B.$x^2 - x + 1 = 0$
C.$x^2 + x - 1 = 0$
D.$x^2 - 3x + 1 = 0$
D
)A.$2x^2 - 3x + 1 = 0$
B.$x^2 - x + 1 = 0$
C.$x^2 + x - 1 = 0$
D.$x^2 - 3x + 1 = 0$
答案
D
练习3.关于$x$的方程$x^2 + px + q = 0$的两个根分别为-3和-1,则$p=$
4
,$q=$3
.答案
p=4,q=3
练习4.已知一元二次方程$x^2 + 4x - 1 = 0$的两根分别为m,n,则$mn - m - n$的值是(
A.5
B.3
C.-3
D.-5
B
)A.5
B.3
C.-3
D.-5
答案
B
【例2】方程$x^2 + mx - 3 = 0$的一根为3,另一根为
方程$ax^2 + bx + c = 0$的两根为$x_1,x_2$,
$x_1 + x_2 = \frac{$
$}{a}$,
-1
。方程$ax^2 + bx + c = 0$的两根为$x_1,x_2$,
$x_1 + x_2 = \frac{$
答案
-1
解:$\because x_1 · x_2 = -3, \therefore$另一根为$-1$.(抓两根之积)
解:$\because x_1 · x_2 = -3, \therefore$另一根为$-1$.(抓两根之积)
练习.方程$x^2 -4x -n=0$的一个根为1,则另一个根为
$x_1 · x_2 = \frac{c}{a}$
3
。$x_1 · x_2 = \frac{c}{a}$
答案
3
解:$\because x_1 + x_2 = 4, \therefore$另一根为3.(抓两根之和)
解:$\because x_1 + x_2 = 4, \therefore$另一根为3.(抓两根之和)
【例3】若$x_1$、$x_2$是一元二次方程$2x^2 - 3x -1=0$的两个根,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)$(x_1 - x_2)^2$.
(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:(1)$\because x_1 + x_2 = \frac{3}{2}, x_1 x_2 = -\frac{1}{2}$,
$\therefore \frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = -3$;
(2)$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = \frac{9}{4} + 1 = \frac{13}{4}$;
(3)$\frac{x_2}{x_1} + \frac{x_1}{x_2} = \frac{x_2^2 + x_1^2}{x_1 x_2} = -\frac{13}{2}$;
(4)$(x_1 - x_2)^2 = x_1^2 + x_2^2 - 2x_1 x_2 = (x_1 + x_2)^2 - 4x_1 x_2 = \frac{17}{4}$.
$\therefore \frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = -3$;
(2)$x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = \frac{9}{4} + 1 = \frac{13}{4}$;
(3)$\frac{x_2}{x_1} + \frac{x_1}{x_2} = \frac{x_2^2 + x_1^2}{x_1 x_2} = -\frac{13}{2}$;
(4)$(x_1 - x_2)^2 = x_1^2 + x_2^2 - 2x_1 x_2 = (x_1 + x_2)^2 - 4x_1 x_2 = \frac{17}{4}$.
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