《2025年知识大通关九年级数学全一册人教版》

例3 用公式法解下列方程：
(1)$3x^2 + x - 1 = 0$；
(2)$6x^2 + 4 = 2\sqrt{5}x$.

答案： 解：
(1)$\because a = 3$，$b = 1$，$c = -1$，
$\Delta = b^2 - 4ac = 1^2 - 4 × 3 × (-1) = 13 ＞ 0$，
$\therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{13}}{2 × 3} = \frac{-1 \pm \sqrt{13}}{6}$，
即$x_1 = \frac{-1 + \sqrt{13}}{6}$，$x_2 = \frac{-1 - \sqrt{13}}{6}$.
(2)原方程可化为$3x^2 - \sqrt{5}x + 2 = 0$.
$\because a = 3$，$b = -\sqrt{5}$，$c = 2$，
$\Delta = (-\sqrt{5})^2 - 4 × 3 × 2 = -19 ＜ 0$，
$\therefore$原方程无实数根.

例4 用因式分解法解下列方程：
(1)$x^2 - 14x = 0$；
(2)$x^2 - 6x = -9$；
(3)$x(5x + 4) = 5x + 4$；
(4)$(3x - 1)^2 = (x - 1)^2$.

答案： 解：
(1)因式分解，得$x(x - 14) = 0$，
$\therefore x = 0$或$x - 14 = 0$.
$\therefore x_1 = 0$，$x_2 = 14$.
(2)移项，得$x^2 - 6x + 9 = 0$.
因式分解，得$(x - 3)^2 = 0$.
$\therefore x_1 = x_2 = 3$.
(3)移项，得$x(5x + 4) - (5x + 4) = 0$.
因式分解，得$(x - 1)(5x + 4) = 0$.
$\therefore x - 1 = 0$或$5x + 4 = 0$.
$\therefore x_1 = 1$，$x_2 = -\frac{4}{5}$.
(4)移项，得$(3x - 1)^2 - (x - 1)^2 = 0$.
因式分解，得$(3x - 1 + x - 1)(3x - 1 - x + 1) = 0$，
即$2x(4x - 2) = 0$.
$\therefore 2x = 0$或$4x - 2 = 0$.
$\therefore x_1 = 0$，$x_2 = \frac{1}{2}$.

例5 不解方程，判断下列方程根的情况：
(1)$7x^2 + 2x + 3 = 0$；
(2)$4(x^2 - x) = -1$.

答案： 解：
(1)$\because \Delta = 2^2 - 4 × 7 × 3 = 4 - 84 = -80 ＜ 0$，
$\therefore$原方程没有实数根.
(2)原方程化为一般形式是$4x^2 - 4x + 1 = 0$.
$\because \Delta = (-4)^2 - 4 × 4 × 1 = 16 - 16 = 0$，
$\therefore$原方程有两个相等的实数根.

例6 已知关于$x$的一元二次方程$x^2 + 4x + m - 1 = 0$有两个实数根，分别记为$x_1$，$x_2$.
(1)求$m$的取值范围；
(2)若$2(x_1 + x_2) + x_1x_2 + 10 = 0$，求$m$的值.

答案： 解：
(1)$\because$关于$x$的一元二次方程$x^2 + 4x + m - 1 = 0$有两个实数根，
$\therefore \Delta = b^2 - 4ac = 4^2 - 4(m - 1) \geq 0$，
解得$m \leq 5$.
(2)由题意，得$x_1 + x_2 = -4$，$x_1x_2 = m - 1$.
$\because 2(x_1 + x_2) + x_1x_2 + 10 = 0$，
$\therefore 2 × (-4) + m - 1 + 10 = 0$.
解得$m = -1$.