📚 Quadratic Equations for IGCSE Mathematics | IGCSE数学：二次方程
rown object. Any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0, is a quadratic equation. Mastering them matters because quadratics link algebra, graphs and problem solving together, and they appear regularly in both the Core and the Extended papers.
二次方程贯穿 IGCSE 数学大纲，从简单三项式的因式分解到解释现实中的抛体问题都会遇到。任何可以写成 ax² + bx + c = 0 形式的方程都叫二次方程，其中 a、b、c 是常数且 a ≠ 0。掌握二次方程非常重要，因为它把代数、图像和问题解决联系在一起，并且在核心卷和扩展卷中都经常出现。
1. What is a Quadratic Equation? | 什么是二次方程？
6 = 0 and 3x² + 2x - 1 = 0. The general form is ax² + bx + c = 0, and the leading coefficient a must not be zero; if a = 0 the equation becomes linear, so the condition a ≠ 0 is essential. Every quadratic equation has at most two real solutions, called roots.
二次方程是一个次数为 2 的多项式方程，也就是说未知数的最高次数是 2。例如 x² - 5x + 6 = 0 和 3x² + 2x - 1 = 0 都是二次方程。一般形式为 ax² + bx + c = 0，其中首项系数 a 一定不能为零；如果 a = 0，方程就变成一次方程，因此条件 a ≠ 0 非常关键。每个二次方程最多有两个实数解，我们称为根。
The words below are used constantly in exam questions, so learn them precisely:
以下术语在考题中反复出现，务必准确掌握：
- Root / solution: a value of x that makes the equation true.
- Zero: the same value, described from the point of view of the graph.
- Monic quadratic: one in which a = 1.
- Pure quadratic: one in which b = 0, for example x² - 9 = 0.
- 根 / 解：使方程成立的 x 的取值。
- 零点：同一个值，但从图像的角度来称呼。
- 首一二次式：a = 1 的二次式。
- 纯二次方程：b = 0 的二次方程，例如 x² - 9 = 0。
2. Standard Form and Coefficients | 标准形式与系数
the equation into standard form before you do anything else. For instance, 2x² = 5x - 3 becomes 2x² - 5x + 3 = 0, so a = 2, b = -5 and c = 3.
标准形式为 ax² + bx + c = 0。其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。解题前一定要先把方程整理成这种形式。例如 2x² = 5x - 3 可化为 2x² - 5x + 3 = 0，所以 a = 2、b = -5、c = 3。
The most common mistake here is a sign error. Move every term to the left-hand side by subtracting, then read off the coefficients with their signs attached. A quick table helps:
这里最常见的错误是符号出错。要把所有项都通过减法移到等号左边，然后连同符号一起读出系数。用一张小表格会很有帮助：
| Equation | Standard form | a, b, c |
| 3x² = 12 | 3x² - 12 = 0 | 3, 0, -12 |
| x(x + 4) = 5 | x² + 4x - 5 = 0 | 1, 4, -5 |
| 5 - 2x = x² | x² + 2x - 5 = 0 | 1, 2, -5 |
Rows one and two show that b or c may be zero, and that is perfectly acceptable. Only a is forbidden from being zero.
第一行和第二行说明 b 或 c 可以为零，这是完全允许的。唯独 a 不允许为零。
3. Solving by Factorising | 因式分解法
Factorising is the fastest method when the quadratic breaks into two linear factors with integer terms. It relies on the zero product property: if the product of two numbers is zero, then at least one of them must be zero.
当二次式可以分解成两个整数项的一次因式时，因式分解法是最快的方法。它依靠零积性质：如果两个数的乘积为零，那么其中至少有一个数为零。
3 = 0, giving x = 2 or x = 3. Always check by expanding the brackets to make sure you recover the original expression.
对于 x² - 5x + 6 = 0，要找两个数，它们的乘积是 +6，和是 -5，即 -2 和 -3，于是方程化为 (x - 2)(x - 3) = 0。所以 x - 2 = 0 或 x - 3 = 0，得到 x = 2 或 x = 3。解完后要养成展开括号检验的习惯，确认能还原成原来的式子。
the product a × c = 6 and the two numbers that multiply to 6 and add to 7 are 6 and 1, so:
当 x² 的系数不是 1 时，可以拆分中项。对于 2x² + 7x + 3 = 0，乘积 a × c = 6，而乘积为 6、和为 7 的两个数是 6 和 1，于是：
2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0
f two squares, x² - k² = (x - k)(x + k), and the perfect square, x² + 2kx + k² = (x + k)².
因此 x = -1/2 或 x = -3。有两个特殊模式值得记牢：平方差 x² - k² = (x - k)(x + k)，以及完全平方 x² + 2kx + k² = (x + k)²。
4. Solving by Completing the Square | 配方法
mum value, or an exact surd answer. Start with x² + 6x + 2 = 0. Halve the coefficient of x and square it: half of 6 is 3, and 3² = 9.
配方法把二次式改写成 (x + p)² + q 的形式。当题目要求求顶点、最大值或最小值，或者要求带根号的精确解时，这是首选方法。以 x² + 6x + 2 = 0 为例，先取 x 的系数的一半再平方：6 的一半是 3，而 3² = 9。
x² + 6x + 9 - 9 + 2 = 0, so (x + 3)² - 7 = 0
Therefore (x + 3)² = 7, and x = -3 ± √7. Written to 3 significant figures these are x ≈ -0.354 and x ≈ -5.65.
因此 (x + 3)² = 7，x = -3 ± √7。保留 3 位有效数字即 x ≈ -0.354 和 x ≈ -5.65。
² + 8x + 5 = 0. Write it as 2(x² + 4x) + 5 = 0 and complete the square inside the bracket:
如果 a ≠ 1，要先提取首项系数。以 2x² + 8x + 5 = 0 为例，写成 2(x² + 4x) + 5 = 0，然后在括号内配方：
2[(x + 2)² - 4] + 5 = 0, giving 2(x + 2)² - 3 = 0
So (x + 2)² = 1.5, and x = -2 ± √1 tells you immediately that the minimum value of the expression is -3, reached when x = -2.
所以 (x + 2)² = 1.5，x = -2 ± √1.5，即 x ≈ -0.775 或 x ≈ -3.22。注意，配方后的形式立刻告诉你：该表达式的最小值是 -3，在 x = -2 时取得。
5. The Quadratic Formula | 求根公式
atic formula solves any quadratic equation ax² + bx + c = 0, whether or not it factorises:
求根公式可以解出任何二次方程 ax² + bx + c = 0，不管它能否因式分解：
x = ( -b ± √b² - 4ac ) ÷ (2a)
mplify the surd if possible. For x² - 5x + 6 = 0 we have a = 1, b = -5, c = 6, so x = (5 ± √25 - 24) ÷ 2 = (5 ± 1) ÷ 2, giving x = 3 or x = 2, which agrees with factorising.
代入系数时要连同符号一起小心代入，然后尽可能化简根式。对于 x² - 5x + 6 = 0，a = 1、b = -5、c = 6，所以 x = (5 ± √25 - 24) ÷ 2 = (5 ± 1) ÷ 2，得到 x = 3 或 x = 2，与因式分解的结果一致。
Now take 2x² - 4x - 3 = 0. Then a = 2, b = -4, c = -3, so b² - 4ac = 16 + 24 = 40 and √40 = 2√10 .3246. Hence x = (4 ± 6.3246) ÷ 4, giving x ≈ 2.58 or x ≈ -0.581 to 3 significant figures.
再看 2x² - 4x - 3 = 0。此时 a = 2、b = -4、c = -3，所以 b² - 4ac = 16 + 24 = 40，√40 = 2√10 ≈ 6.3246。因此 x = (4 ± 6.3246) ÷ 4，保留 3 位有效数字得 x ≈ 2.58 或 x ≈ -0.581。
Three practical rules apply in the exam. Write the formula down before substituting, keep the negative sign inside the bracket when b is negative, and do not round until the final step. Extended candidates should also be able to leave the answer in exact surd form rather than as a decimal.
考试中有三条实用规则：先写出公式再代入；当 b 为负数时，负号要放在括号内；不到最后一步不要四舍五入。扩展卷考生还应当能够把答案保留为精确的根式形式，而不是写成小数。
6. The Discriminant | 判别式
The expression under the square root, b² - 4ac, is called the discriminant and is usually written as Δ. It tells you how many real roots the equation has before you solve it, and it is a very common exam question in its own right.
根号下的表达式 b² - 4ac 叫做判别式，通常记作 Δ。它能在求解之前告诉你方程有几个实根，本身也是考试中非常常见的考点。
| Value of Δ | Nature of the roots | Graph |
| Δ > 0 | Two distinct real roots | Crosses the x-axis twice |
| Δ = 0 | One repeated real root | Touches the x-axis once |
| Δ < 0 | No real roots | Never meets the x-axis |
6 - 20 = -4, which is negative, so the equation has no real solutions. Questions often say "find the value of k for which the equation has equal roots"; set Δ = 0 and solve the resulting linear equation in k.
例如 x² - 4x + 5 = 0 的 Δ = 16 - 20 = -4，为负数，所以该方程没有实数解。题目常问"求使方程有两个相等实根的 k 值"，这时令 Δ = 0，解出关于 k 的一次方程即可。
7. Graphs of Quadratic Functions | 二次函数的图像
and has a maximum point. The y-intercept is always c, because setting x = 0 leaves y = c.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，曲线开口向上，有最低点；若 a < 0，曲线开口向下，有最高点。y 轴截距永远是 c，因为令 x = 0 就得到 y = c。
h a parabola in the exam you should mark the intercepts, the vertex and the line of symmetry, then draw a smooth curve through them.
用因式分解、配方或公式求出的根，正是曲线与 x 轴的交点，也就是 y = 0 的位置。考试中画抛物线时，要标出截距、顶点和对称轴，然后画出光滑的曲线。
. You can also find it by averaging the two roots, since the vertex lies halfway between them for a symmetric parabola.
对称轴经过顶点，方程为 x = -b ÷ (2a)。你也可以通过求两个根的平均值得到它，因为对于对称的抛物线，顶点正好位于两根的正中间。
8. Vertex Form and Maximum or Minimum Values | 顶点式与最值
, from which the vertex is read off as (-p, q). This is the standard method for optimisation questions in IGCSE.
配方法把 y = ax² + bx + c 转化为顶点式 y = a(x + p)² + q，由此可直接读出顶点为 (-p, q)。这是 IGCSE 中解决最值问题的标准方法。
d term can never be negative, the smallest possible value of y is -7, occurring at x = -3.
考虑 y = x² + 6x + 2。配方得 y = (x + 3)² - 7，所以顶点是 (-3, -7)，对称轴是 x = -3。由于平方项永远非负，y 的最小值就是 -7，在 x = -3 时取得。
s negative the same working gives a maximum instead of a minimum. A typical question reads "show that x² - 8x + 20 is always positive" ete the square as (x - 4)² + 4, and since (x - 4)² ≥ 0 the whole expression is at least 4.
如果 a 为负数，同样的过程得到的是最大值而不是最小值。常见题型是"证明 x² - 8x + 20 恒为正"；做法是配方成 (x - 4)² + 4，由于 (x - 4)² ≥ 0，整个表达式至少为 4。
9. Forming and Solving Quadratics from Word Problems | 由文字题建立并求解二次方程
Word problems usually hide a quadratic inside a geometric or physical situation. The routine is always the same: define a variable, write an equation from the given information, rearrange to standard form, solve, then reject any solution that does not fit the context.
文字题通常把一个二次方程藏在几何或物理情境中。解题步骤总是一样的：设未知数，根据条件列出方程，整理成标准形式，求解，然后舍去不符合实际情况的解。
w² + 3w - 40 = 0, which factorises as (w + 8)(w - 5) = 0, giving w = 5 or w = -8. A length cannot be negative, so w = 5 cm and the rectangle is 5 cm by 8 cm.
例一，面积问题。一个矩形比它的宽长 3 厘米，面积是 40 平方厘米。设宽为 w，则长为 w + 3，可得 w(w + 3) = 40，即 w² + 3w - 40 = 0，因式分解为 (w + 8)(w - 5) = 0，得到 w = 5 或 w = -8。长度不可能为负，所以 w = 5 厘米，矩形尺寸为 5 厘米 × 8 厘米。
ll is 15 m high. Setting h = 15 gives 20t - 5t² = 15, that is 5t² - 20t + 15 = 0, or t² - 4t + 3 = 0, which factorises as (t - 1)(t - 3) = 0. The ball is 15 m high at t = 1 s on the way up and t = 3 s on the way down.
例二，抛体问题。一个球的离地高度与时间的关系为 h = 20t - 5t²（米），t 以秒计。求球高度为 15 米的时刻。令 h = 15，得 20t - 5t² = 15，即 5t² - 20t + 15 = 0，或 t² - 4t + 3 = 0，因式分解为 (t - 1)(t - 3) = 0。球在上升过程中 t = 1 秒和下降过程中 t = 3 秒时高度为 15 米。
ctor such as (x + 2), because that root is a genuine solution the examiner expects to see.
所有文字题还有两条规则：绝不要用变量去除两边，因为那样会丢掉 x = 0 这个根；也绝不要用 (x + 2) 这样的因式去除，因为那个根是真实存在的解，考官期望看到它。
10. Quadratic Inequalities (Extended) | 二次不等式（扩展内容）
The Extended paper sometimes asks you to solve an inequality such as x² - 5x + 6 > 0. The safe method is to find the critical values first, then test the regions they create.
扩展卷有时会要求解 x² - 5x + 6 > 0 这样的不等式。稳妥的做法是先求出临界值，再对它们划分出的区间进行检验。
The
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