📚 IGCSE Maths: Quadratic Equations | IGCSE 数学：二次方程
Quadratic equations are one of the most important algebraic topics in the IGCSE Mathematics syllabus. They appear in both Core and Extended papers, and they are almost always linked to graphs, problem solving and real-life modelling, so a confident grasp of this topic pays off across the whole examination.
二次方程是 IGCSE 数学教学大纲中最重要的代数主题之一。它同时出现在核心课程和扩展课程试卷中，并且几乎总是与图像、问题解决和实际建模相结合，因此扎实掌握这一主题会在整份试卷中带来回报。
1. Introduction to Quadratic Equations | 二次方程简介
A quadratic equation in one variable is an equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The name comes from the Latin word for 'square' 0, the equation becomes linear rather than quadratic, so the condition a ≠ 0 is essential.
一个变量的二次方程是能够写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数，且 a ≠ 0。这个名字来自拉丁语中表示"平方"的词，因为未知数 x 的最高次幂是 2。如果 a = 0，该方程就变成了一次方程，而不是二次方程，因此 a ≠ 0 这个条件至关重要。
ax² + bx + c = 0, a ≠ 0
Because the highest power of x is 2, the graph of a quadratic function is a parabola. A quadratic equation can have two distinct real solutions, one repeated solution, or no real solutions at all, and the discriminant tells you which case applies.
由于 x 的最高次幂是 2，二次函数的图像是一条抛物线。二次方程可以有两个不同的实数解、一个重复解，或者没有实数解，而判别式可以告诉你属于哪一种情况。
2. Standard Form and Identifying Coefficients | 标准形式与系数识别
− 5x + 2 = 0, a = 1, b = −5 and c = 2. A hidden coefficient of 1 or −1 is still a coefficient, so never leave it out when substituting into the formula.
在使用任何方法之前，你必须先把方程整理成一边恰好为 0 的形式。系数必须连同符号一起读取：在 x² − 5x + 2 = 0 中，a = 1，b = −5，c = 2。隐藏的系数 1 或 −1 依然是系数，代入公式时千万不要漏掉。
| Given equation | Standard form | a | b | c |
|---|---|---|---|---|
| 3x² = 5x − 2 | 3x² − 5x + 2 = 0 | 3 | −5 | 2 |
| (x + 2)(x − 5) = 0 | x² − 3x − 10 = 0 | 1 | −3 | −10 |
| x²/2 + 3x = 1 | x² + 6x − 2 = 0 | 1 | 6 | −2 |
Fractions and brackets are common in exam questions precisely because they test whether you can rearrange accurately. Multiply or expand first, then collect all terms on one side, and only then start solving.
考试题中经常出现分数和括号，正是为了考查你能否准确整理方程。先乘开或展开，再把所有项移到同一侧，最后才开始求解。
3. Solving by Factorisation | 因式分解法求解
rs is zero, then at least one of the factors must be zero. In symbols, if pq = 0 then p = 0 or q = 0.
当根是整数或简单分数时，因式分解是最快的方法。它依据的是零积法则：如果两个因式的乘积为零，那么至少有一个因式为零。用符号表示就是：若 pq = 0，则 p = 0 或 q = 0。
x² + bx + c = (x + p)(x + q), where p + q = b and pq = c
hen the leading coefficient is not 1, use the same idea but split the middle term: 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3), giving x = −1/2 or x = −3.
对于 x² − 5x + 6 = 0，你需要找到两个数，它们的乘积为 +6，和为 −5。这两个数是 −2 和 −3，所以 (x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。当二次项系数不为 1 时，思路相同，但要把中间项拆分：2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)，得到 x = −1/2 或 x = −3。
ts x = 7 and x = −7. A perfect square such as x² − 10x + 25 = (x − 5)² gives the single repeated root x = 5. Always check for a common factor before anything else: 3x² − 12x = 0 becomes 3x(x − 4) = 0, giving x = 0 or x = 4.
有两个特殊模式值得记住。平方差公式给出 x² − a² = (x + a)(x − a)，所以 x² − 49 = 0 可分解为 (x + 7)(x − 7) = 0，根为 x = 7 和 x = −7。完全平方如 x² − 10x + 25 = (x − 5)²，给出唯一的重复根 x = 5。在做任何事之前都要先检查公因式：3x² − 12x = 0 化为 3x(x − 4) = 0，得到 x = 0 或 x = 4。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q, which reveals the turning point of the graph. The rule is simple: take half of the coefficient of x, square it, add it and subtract it again so the value of the expression does not change.
配方法把二次式改写成 (x + p)² + q 的形式，从而直接显示出图像的转折点。规则很简单：取 x 的系数的一半，平方，先加再减，使表达式的值保持不变。
x² + bx + c = (x + b/2)² − (b/2)² + c
+ 3)² − 9 + 1 = (x + 3)² − 8. Setting this equal to zero gives (x + 3)² = 8, so x + 3 = ±√8 = ±2√2 and therefore x = −3 ± 2√2. If the question asks for the answer to 3 significant figures, x ≈ −0.172 or x ≈ −5.83.
对方程 x² + 6x + 1 = 0 使用该方法时，6 的一半是 3，3² = 9，所以 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。令其等于零得到 (x + 3)² = 8，所以 x + 3 = ±√8 = ±2√2，因此 x = −3 ± 2√2。如果题目要求保留 3 位有效数字，则 x ≈ −0.172 或 x ≈ −5.83。
= 2(x + 2)² − 3. The expression therefore has a minimum value of −3, occurring at x = −2.
当 x² 的系数不为 1 时，先把前两项中的该系数提出来。对于 2x² + 8x + 5，可写成 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。因此该表达式的最小值为 −3，在 x = −2 处取得。
5. The Quadratic Formula | 求根公式
eting the square on the general form ax² + bx + c = 0, and it should be memorised exactly.
求根公式可以解任何二次方程，包括那些无法因式分解的方程。它是通过对一般形式 ax² + bx + c = 0 配方推导出来的，必须一字不差地记住。
x = (−b ± √b² − 4ac) / 2a
a = 2, b = −5 and c = −3, so b² − 4ac = (−5)² − 4 × 2 × (−3) = 25 + 24 = 49. Then x = (5 ± √49 otice how the negative value of b turns into +5 in the numerator: keeping the signs straight is where most marks are lost.
对于 2x² − 5x − 3 = 0，有 a = 2，b = −5，c = −3，所以 b² − 4ac = (−5)² − 4 × 2 × (−3) = 25 + 24 = 49。于是 x = (5 ± √49) / 4 = (5 ± 7) / 4，得到 x = 3 或 x = −1/2。注意 b 为负值时，在分子中变成了 +5：把符号处理正确，是大多数失分所在。
r or a decimal answer correct to 2 or 3 significant figures. For x² − 4x + 1 = 0, x = (4 ± √12) / 2 = (4 ± 2√3) / 2 = 2 ± √3, which is approximately 3.73 or 0.27. Always simplify the surd fully before writing your final answer.
当判别式不是完全平方数时，题目通常要求给出根式答案，或保留 2 至 3 位有效数字的小数答案。对于 x² − 4x + 1 = 0，x = (4 ± √12) / 2 = (4 ± 2√3) / 2 = 2 ± √3，约等于 3.73 或 0.27。写出最终答案前，一定要把根式完全化简。
6. The Discriminant and the Number of Roots | 判别式与根的个数
The expression b² − 4ac is called the discriminant, often written as Δ. It sits under the square root in the formula, so its sign decides how many real solutions the equation has.
表达式 b² − 4ac 称为判别式，通常写作 Δ。它位于公式的根号之下，因此它的符号决定了方程有多少个实数解。
| Discriminant | Nature of roots | Graph |
|---|---|---|
| Δ > 0 | Two distinct real roots | Crosses the x-axis twice |
| Δ = 0 | One repeated real root | Touches the x-axis at the vertex |
| Δ < 0 | No real roots | Does not meet the x-axis |
x-axis. By contrast, x² − 8x + 16 = 0 has Δ = 64 − 64 = 0, so it is a perfect square with the repeated root x = 4.
例如，x² + 4x + 5 = 0 的 Δ = 4² − 4 × 1 × 5 = 16 − 20 = −4，为负值，所以该方程没有实数解，抛物线完全位于 x 轴上方。相反，x² − 8x + 16 = 0 的 Δ = 64 − 64 = 0，因此它是完全平方式，有重复根 x = 4。
an unknown coefficient and asks for the values that produce a repeated root, for instance 'find k if x² + kx + 9 = 0 has equal roots'. Set the discriminant to zero: k² − 36 = 0, so k = 6 or k = −6.
扩展试卷中常见的题型是给出一个未知系数，要求找出使方程有重复根的取值，例如"若 x² + kx + 9 = 0 有等根，求 k"。令判别式等于零：k² − 36 = 0，所以 k = 6 或 k = −6。
7. Graphs of Quadratic Functions | 二次函数图像
. If a > 0 the curve opens upwards and has a minimum point; if a < 0 it opens downwards and has a maximum point. The larger the value of |a|, the narrower the parabola.
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。|a| 的值越大，抛物线越窄。
y-intercept: (0, c) | axis of symmetry: x = −b / 2a
where the curve crosses the x-axis, which is why solving an equation and sketching a graph are two views of the same idea. The axis of symmetry is the vertical line x = −b / 2a, which passes through the vertex and lies exactly halfway between the two roots.
从方程中可以直接读出三个特征。y 轴截距总在 (0, c)。方程 ax² + bx + c = 0 的解正是曲线与 x 轴交点的横坐标，这正是"解方程"与"画图像"是同一件事的两种视角的原因。对称轴是竖直线 x = −b / 2a，它经过顶点，并且恰好位于两个根的正中间。
re on that line. Examiners frequently ask you to use a given graph to write down the equation of the axis of symmetry or to estimate the roots from a sketch.
这个中点关系是一个有用的检验方法。如果二次方程的两个根是 x = 1 和 x = 7，那么对称轴一定是 x = 4，顶点就位于这条直线上。考官经常要求你利用给定的图像写出对称轴的方程，或从草图中估算方程的根。
8. Sketching Parabolas and Finding Turning Points | 抛物线作图与转折点
A good sketch needs four pieces of information: the shape (up or down, decided by the sign of a), the y-intercept, the x-intercepts if they exist, and the coordinates of the turning point. Completing the square is the fastest way to obtain the turning point.
一幅合格的草图需要四项信息：开口方向（由 a 的符号决定向上或向下）、y 轴截
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