1. 如果关于 $ x $ 的一元二次方程 $ ax^2 + bx + c = 0(a≠0) $有实数根,那么必须满足的条件是（）

A.$ b^2 -4ac ≥ 0 $
B.$ b^2 -4ac ≤ 0 $
C.$ b^2 -4ac ＞ 0 $
D.$ b^2 -4ac ＜ 0 $

1.A

2. [2025 南通启秀中学月考] 在用求根公式 $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ 求一元二次方程的根时，小珺正确地代入了 $a, b, c$ 的值，得到 $x = $$$，则她求解的一元二次方程是（）

A.$2x^2 - 3x - 1 = 0$
B.$2x^2 + 4x - 1 = 0$
C.$-x^2 - 3x + 2 = 0$
D.$3x^2 - 2x + 1 = 0$

2.A 由题意,得 $a=2,b=-3,c=-1$.

3. 已知一元二次方程$2x^2 - 3x = 1$可化为$ax^2 + bx + c = 0$的形式，则$b^2 - 4ac =$。

3.17 将原方程化为一般形式,得 $2x^2 - 3x - 1 = 0,\therefore a = 2,b = -3,c = -1,\therefore b^2 - 4ac = (-3)^2 -4 × 2 × (-1) = 17.$

4. 解下列方程：
(1) $-2x^2 +5x -3 =0$；(2) $5x^2 +2x -1 =0$；
(3) $4x^2 +1 =5x$；(4) $\frac{1}{2}x^2 +\frac{1}{8} =\frac{1}{2}x$；
(5) 一题多解 $x^2 +2\sqrt{5}x -2 =0$；
(6) $3x(3 -x)=3x -1$。

4. 解:(1)$\because a = -2,b = 5,c = -3$,
$b^2 -4ac =5^2 -4 × (-2) × (-3) =1＞0$,
$\therefore x = \frac{-5 \pm \sqrt{1}}{2 × (-2)} = \frac{-5 \pm 1}{-4}$,
$\therefore x_1 = 1,x_2 = \frac{3}{2}.$
(2)$\because a = 5,b = 2,c = -1$,
$b^2 -4ac =2^2 -4 × 5 × (-1) =24＞0$,
$\therefore x = \frac{-2 \pm \sqrt{24}}{2 × 5} = \frac{-1 \pm \sqrt{6}}{5}$,
$\therefore x_1 = \frac{-1 + \sqrt{6}}{5},x_2 = \frac{-1 - \sqrt{6}}{5}.$
(3)把方程 $4x^2 + 1 =5x$ 化为一般形式,得 $4x^2 -5x +1 =0.$
$\because a =4,b = -5,c =1$,
$b^2 -4ac = (-5)^2 -4 × 4 × 1 =9＞0$,
$\therefore x = \frac{-(-5) \pm \sqrt{9}}{2 × 4} = \frac{5 \pm 3}{8}$,
$\therefore x_1 =1,x_2 = \frac{1}{4}.$
(4)将$\frac{1}{2}x^2 + \frac{1}{8} = \frac{1}{2}x$ 化为一般形式,得 $4x^2 -4x +1 =0.$
$\because a =4,b = -4,c =1$,
$b^2 -4ac = (-4)^2 -4 × 4 × 1 =0$,
$\therefore x = \frac{-(-4) \pm \sqrt{0}}{2 × 4} = \frac{1}{2}$,
$\therefore x_1 = x_2 = \frac{1}{2}.$
(当 $b^2 -4ac =0$ 时,方程的根要写成“$x_1 = x_2 = ···$”的形式)
(5) 通解 $\because a =1,b = 2\sqrt{5},c = -2$,
$b^2 -4ac = (2\sqrt{5})^2 -4 × 1 × (-2) =28＞0$,
$\therefore x = \frac{-2\sqrt{5} \pm \sqrt{28}}{2 × 1} = -\sqrt{5} \pm \sqrt{7}$,
$\therefore x_1 = -\sqrt{5} + \sqrt{7},x_2 = -\sqrt{5} - \sqrt{7}.$
另解 移项,得 $x^2 + 2\sqrt{5}x =2.$
配方,得 $x^2 + 2\sqrt{5}x + (\sqrt{5})^2 = 2 + (\sqrt{5})^2$,
即$(x + \sqrt{5})^2 =7.$
解这个方程,得 $x + \sqrt{5} = \pm \sqrt{7}$,
$\therefore x_1 = -\sqrt{5} + \sqrt{7},x_2 = -\sqrt{5} - \sqrt{7}.$
(6)把方程 $3x(3 - x) = 3x - 1$ 化成一般形式,得 $3x^2 -6x -1 =0.$
$\because a =3,b = -6,c = -1$,
$b^2 -4ac = (-6)^2 -4 × 3 × (-1) =48＞0$,
$\therefore x = \frac{6 \pm \sqrt{48}}{2 × 3} = \frac{3 \pm 2\sqrt{3}}{3}$,
$\therefore x_1 = \frac{3 + 2\sqrt{3}}{3},x_2 = \frac{3 - 2\sqrt{3}}{3}.$

5. 当正数$x$为何值时，代数式$7x(x+5)$与代数式$-6x^2 -37x -9$的值互为相反数？

5. 解:由题意,得 $7x(x + 5) -6x^2 -37x -9 =0.$
化简,得 $x^2 -2x -9 =0.$
$\because a =1,b = -2,c = -9$,
$b^2 -4ac = (-2)^2 -4 × 1 × (-9) =40＞0$,
$\therefore x = \frac{-(-2) \pm \sqrt{40}}{2 × 1} = \frac{2 \pm 2\sqrt{10}}{2} = 1 \pm \sqrt{10}.$
$\because 1 + \sqrt{10} ＞0,1 - \sqrt{10} ＜0,\therefore$ 正数 $x = 1 + \sqrt{10}.$
故当正数 $x = 1 + \sqrt{10}$ 时,代数式 $7x(x + 5)$ 与代数式 $-6x^2 -37x -9$ 的值互为相反数.

6. 在解方程$\sqrt{2}x^2 + 4\sqrt{3}x = 2\sqrt{2}$时，一位同学的解答过程如下：
$\because a = \sqrt{2}, b = 4\sqrt{3}, c = 2\sqrt{2},$
$\therefore b^2 - 4ac = (4\sqrt{3})^2 - 4 × \sqrt{2} × 2\sqrt{2} = 32 ＞ 0,$
$\therefore x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4\sqrt{3} \pm \sqrt{32}}{2 × \sqrt{2}} = -\sqrt{6} \pm 2,$
$\therefore x_1 = -\sqrt{6} + 2, x_2 = -\sqrt{6} - 2.$
请你分析以上解答有无错误，如有错误，请写出正确的解答过程.

6. 解:有错误,正确的解答过程如下:
移项,得$\sqrt{2}x^2 +4\sqrt{3}x -2\sqrt{2} =0.$
$\because a = \sqrt{2},b =4\sqrt{3},c = -2\sqrt{2}$,
$\therefore b^2 -4ac = (4\sqrt{3})^2 -4 × \sqrt{2} × (-2\sqrt{2}) =64＞0$,
$\therefore x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} = \frac{-4\sqrt{3} \pm \sqrt{64}}{2 × \sqrt{2}} = -\sqrt{6} \pm 2\sqrt{2}$,
$\therefore x_1 = -\sqrt{6} + 2\sqrt{2},x_2 = -\sqrt{6} - 2\sqrt{2}.$