📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程完全掌握
Quadratic equations are a central topic in IGCSE Mathematics, appearing across algebra, graphs, and real-life modelling. They test your ability to manipulate algebraic expressions, choose an efficient solving method, and interpret solutions in context. This guide explains the standard form, factorising, completing the square, the quadratic formula, the discriminant, graphs, disguised quadratics, word problems, and exam-style applications, with worked examples and mark-scheme tips throughout.
二次方程是 IGCSE 数学的核心主题，广泛出现在代数、图像和实际建模中。它考查你化简代数式、选择高效解法以及结合实际情境解释答案的能力。本指南将系统讲解标准形式、因式分解法、配方法、求根公式、判别式、二次函数图像、隐蔽二次方程、应用题建模以及考试常见题型，并配有例题与评分要点提示。
By the end you should be able to look at any quadratic and decide, within a few seconds, which method will get you to the answer fastest with the fewest chances of an arithmetic slip. That decision-making skill is exactly what separates a grade 7 from a grade 9 student.
读完本文，你应当能够在几秒钟内判断出：面对任意一个二次方程，用哪种方法最快、最不容易算错。这种"选择解法"的判断力，正是 7 分与 9 分学生之间的差距所在。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a, b, c are real numbers. The condition a ≠ 0 omes linear, not quadratic. The highest power of the unknown is always 2, and that single fact controls everything that follows.
二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且为实数。条件 a ≠ 0 非常重要；如果 a = 0，方程就变成一次方程，而不是二次方程。未知数的最高次数永远是 2，正是这一点决定了后面所有的性质。
Common examples include 2x² + 3x − 5 = 0 (a = 2, b = 3, c = −5), x² − 4x = 0 (a = 1, b = −4, c = 0), and 3x² − 27 = 0 the equation is called a pure quadratic; when c = 0 it is called an incomplete quadratic.
常见例子包括 2x² + 3x − 5 = 0（a = 2，b = 3，c = −5）、x² − 4x = 0（a = 1，b = −4，c = 0）和 3x² − 27 = 0（a = 3，b = 0，c = −27）。后两个例子说明 b 或 c 可以为零，但 a 不能为零。当 b = 0 时称为纯二次方程；当 c = 0 时称为不完全二次方程。
5) = 2x − 1 expands to x² − 2x − 15 = 2x − 1, and subtracting 2x − 1 from both sides gives x² − 4x − 14 = 0. Get into the habit of writing "= 0" explicitly before you attempt to factorise or substitute into the formula.
很多考题并不会直接给出标准形式，你必须先展开括号并合并同类项。例如 (x + 3)(x − 5) = 2x − 1 展开为 x² − 2x − 15 = 2x − 1，两边同时减去 2x − 1 后得到 x² − 4x − 14 = 0。请养成在因式分解或代入公式之前，先把方程明确写成"= 0"的习惯。
ax² + bx + c = 0, a ≠ 0
2. Solving by Factorising | 因式分解法
If a quadratic expression factorises into a product of two linear factors, we can use the zero product property: if p × q = 0, then p = 0 or q = 0 (or both). This is the fastest method when it works, and examiners clearly reward it because it shows algebraic insight rather than mechanical substitution.
如果二次式可以因式分解为两个一次因式的乘积，我们就可以使用零积性质：若 p × q = 0，则 p = 0 或 q = 0（或两者同时成立）。当它可行时，这是最快的方法，考官也明显青睐这种方法，因为它体现的是代数洞察力而不是机械代入。
For example, x² − 5x + 6 = 0 becomes (x − 2)(x − 3) = 0 dd to give b. Here −2 × −3 = 6 and −2 + (−3) = −5, so the factors are (x − 2) and (x − 3).
例如，x² − 5x + 6 = 0 可变为 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。对于首项系数为 1 的二次式，方法就是找出两个数：它们相乘等于 c，相加等于 b。这里 (−2) × (−3) = 6 且 (−2) + (−3) = −5，因此因式是 (x − 2) 与 (x − 3)。
xample, 6x² + 11x + 3: ac = 18, and 9 + 2 = 11 with 9 × 2 = 18. So 6x² + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1), giving x = −3/2 or x = −1/3.
当 a ≠ 1 时，先计算 a 与 c 的乘积 ac，再找两个数：相乘等于 ac、相加等于 b，然后拆开中间项并分组分解。例如 6x² + 11x + 3：ac = 18，而 9 + 2 = 11 且 9 × 2 = 18。于是 6x² + 9x + 2x + 3 = 3x(2x + 3) + 1(2x + 3) = (2x + 3)(3x + 1)，得到 x = −3/2 或 x = −1/3。
Two special patterns appear constantly. The difference of two squares: a² − b² = (a + b)(a − b), so x² − 16 = 0 gives x = 4 or x = −4. The perfect square: a² ± 2ab + b² = (a ± b)², so x² − 10x + 25 = 0 gives the repeated root x = 5.
有两种特殊模式反复出现。平方差公式：a² − b² = (a + b)(a − b)，所以 x² − 16 = 0 得到 x = 4 或 x = −4。完全平方公式：a² ± 2ab + b² = (a ± b)²，所以 x² − 10x + 25 = 0 得到重根 x = 5。
Finally, always rearrange to equal zero before factorising. If x appears in every term, do not divide both sides by x, because you may lose the valid solution x = 0 x = 4. Losing the zero root is one of the most common single-mark losses in IGCSE papers.
最后，因式分解前一定要将方程整理为等于零的形式。如果每一项都含有 x，不要两边同时除以 x，因为你可能会漏掉 x = 0 这个有效解。正确做法是提取公因式：x² − 4x = 0 变为 x(x − 4) = 0，所以 x = 0 或 x = 4。漏掉零根是 IGCSE 试卷中最常见的丢 1 分原因之一。
3. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. For a monic quadratic x² + bx + c, use (x + b/2)² − (b/2)² + c. The idea is to turn the first two terms into a perfect square and then subtract whatever you added on.
配方法将二次式改写为 a(x + p)² + q 的形式。对于首项系数为 1 的二次式 x² + bx + c，可使用 (x + b/2)² − (b/2)² + c。其思路是把前两项凑成完全平方，再把多加的部分减掉。
Example: x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8 k by expanding: (x + 3)² − 8 = x² + 6x + 9 − 8 = x² + 6x + 1. Always expand back to verify — it takes five seconds and catches nearly every sign error.
例如：x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。6 的一半是 3，而 3² = 9，所以要减去 9。展开检验：(x + 3)² − 8 = x² + 6x + 9 − 8 = x² + 6x + 1。请务必回代展开验证，只需五秒，却能查出几乎所有的符号错误。
Written in the completed-square form y = (x − h)² + k, the vertex of the parabola is (h, k) the function (when the coefficient of x² is positive). This is why completing the square is especially useful when a quadratic does not factorise neatly, or when a question asks for the minimum or maximum value of a function.
写成配方形式 y = (x − h)² + k 后，抛物线的顶点就是 (h, k)，对称轴是竖直线 x = h，而 k 就是函数的最小值（当 x² 的系数为正时）。这正是配方法在二次式无法整齐分解、或题目要求函数最大/最小值时特别好用的原因。
out of the first two terms first. For 2x² − 8x + 5: 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3. The vertex is (2, −3) and the minimum value is −3. A very common error is to forget to multiply the subtracted term by the factor outside the bracket.
当 a ≠ 1 时，先从前两项中提出系数。对 2x² − 8x + 5：2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。顶点为 (2, −3)，最小值为 −3。一个非常常见的错误是忘记把减去的项乘以括号外的系数。
d take the square root of both sides, remembering the ± sign: (x + 3)² = 8 gives x + 3 = ±√8, so x = −3 ± 2√2 or x ≈ −5.83. Leave the answer in exact surd form unless the question asks for a decimal.
用配方法解方程时，把平方括号单独留下，然后两边开平方，切记带上 ± 号：(x + 3)² = 8 得到 x + 3 = ±√8，所以 x = −3 ± 2√2，即 x ≈ −0.17 或 x ≈ −5.83。除非题目要求小数，否则答案应保留精确的根式形式。
4. The Quadratic Formula | 求根公式
rm ax² + bx + c = 0 and rearranging for x. Because it always works, it is your safety net.
求根公式可以解出任何二次方程，无论它能否因式分解。它的推导过程就是对一般形式 ax² + bx + c = 0 使用配方法，再解出 x。由于它永远有效，它是你的"保底"方法。
x = (−b ± √b² − 4ac) / (2a)
Worked example: solve x² − 5x + 3 = 0. Here a = 1, b = −5, c = 3, so b² − 4ac = 25 − 12 = 13. Therefore x = (5 ± √13 ≈ 0.70 (to 2 decimal places). Note how the negative value of b turns the numerator into +5 — mishandling minus signs here is the single biggest source of lost marks with this formula.
例题：解 x² − 5x + 3 = 0。这里 a = 1，b = −5，c = 3，所以 b² − 4ac = 25 − 12 = 13。因此 x = (5 ± √13) / 2，得到 x ≈ 4.30 或 x ≈ 0.70（保留两位小数）。注意 b 为负值时分子变成 +5——在这里处理不好负号是使用该公式时最大的丢分点。
Useful habits that examiners look for: write the formula down before substituting; substitute inside brackets, for example (−5)² rather than −5²; keep the whole calculation visible; and only round at the very end. If the question says "give your answers to 3 significant figures", do not round the intermediate value of √13.
考官方看重的几个好习惯：先写出公式再代入；代入时加括号，例如写 (−5)² 而不是 −5²；把完整计算过程写出来；只在最后一步取近似。如果题目要求"答案保留三位有效数字"，不要在中间步骤就把 √13 取近似。
- Step 1: Rearrange into the form ax² + bx + c = 0 and identify a, b, c. / 第一步：整理成 ax² + bx + c = 0，写出 a、b、c。
- Step 2: Compute b² − 4ac separately, with brackets. / 第二步：单独计算 b² − 4ac，注意加括号。
- Step 3: Substitute into the formula and simplify the surd if possible. / 第三步：代入公式，能化简根式就化简。
- Step 4: Give both roots, in exact form or to the required accuracy. / 第四步：给出两个根，精确形式或按要求精度。
If the discriminant is a perfect square, the surd simplifies and the answer is rational — a useful self-check that often tells you the question could have been factorised instead. If the discriminant is negative, stop and write that there are no real solutions.
如果判别式是完全平方数，根式可以化简，答案就是有理数——这是一个很好的自检信号，往往说明题目本来可以用因式分解来做。如果判别式为负数，就应停下并写出"没有实数解"。
5. The Discriminant | 判别式
The discriminant is the expression under the square root: Δ = b² − 4ac. It tells you how many real roots a quadratic has, without solving it. Questions about the discriminant are extremely common in IGCSE and are usually worth 2 to 4 marks.
判别式就是根号下的表达式：Δ = b² − 4ac。它能在不解方程的情况下告诉你二次方程有多少个实数根。判别式的题目在 IGCSE 中极其常见，通常值 2 到 4 分。
| Discriminant value / 判别式取值 | Nature of roots / 根的性质 | Graph / 图像 |
| Δ > 0 | Two distinct real roots / 两个不同实根 | Curve cuts x-axis twice / 曲线与 x 轴相交两次 |
| Δ = 0 | One repeated real root / 一个重根 | Curve touches x-axis at the vertex / 曲线在顶点处与 x 轴相切 |
| Δ < 0 | No real roots / 无实根 | Curve never meets x-axis / 曲线与 x 轴不相交 |
of k for which the equation has equal roots. Equal roots require Δ = 0, so k² − 36 = 0 and k = ±6. For "two distinct real roots" you solve an inequality instead: k² − 36 > 0 gives k < −6 or k > 6.
例：对 x² + kx + 9 = 0，求使方程有等根的 k 值。等根要求 Δ = 0，即 k² − 36 = 0，得 k = ±6。若要求"两个不同实根"，则要解不等式：k² − 36 > 0，得 k < −6 或 k > 6。
Two extra facts worth remembering. First, if a, b and c are rational and Δ is a perfect square, both roots are rational. Second, if Δ is negative, the quadratic cannot be factorised over the real numbers, so do not waste time trying to find factors — this is exactly the kind of efficiency that saves minutes in a timed paper.
还有两个值得记住的结论。第一，若 a、b、c 都是有理数且 Δ 是完全平方数，则两个根都是有理数。第二，若 Δ 为负数，该二次式在实数范围内无法因式分解，因此不要浪费时间去找因式——这种效率判断能在限时考试中为你省下好几分钟。
6. Quadratic Graphs and the Parabola | 二次函数图像与抛物线
The graph of y = ax² + bx + c is a parabola. If a > 0 it opens upwards (a "u" shape) and has a minimum point; if a < 0 it opens downwards (an "n" shape) and has a maximum point. The larger the value of |a|, the narrower the curve.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，开口向上（"U"形），有最低点；若 a < 0，开口向下（"n"形），有最高点。|a| 越大，曲线越窄。
Key features you should be able to read off or label: the y-intercept is at (0, c); the x-intercepts (if any) are the roots of ax² + bx + c = 0; the axis of symmetry is x = −b / (2a) und by substituting back. In completed-square form y = a(x − h)² + k, the vertex is simply (h, k).
你应当能够读出或标出的关键特征：y 轴截距为 (0, c)；x 轴截距（若有）就是 ax² + bx + c = 0 的根；对称轴为 x = −b / (2a)；顶点在该对称轴上，因此其横坐标为 −b/(2a)，纵坐标把横坐标代回即可。若写成配方形式 y = a(x − h)² + k，顶点就是 (h, k)。
= 2 (equivalently −(−4)/(2 × 1) = 2). Substituting x = 2 gives y = 4 − 8 + 3 = −1, so the vertex is (2, −1).
例题：y = x² − 4x + 3。y 轴截距为 (0, 3)。因式分解得 (x − 1)(x − 3)，所以根为 x = 1 与 x = 3。对称轴为 x = (1 + 3)/2 = 2（等价地 −(−4)/(2 × 1) = 2）。代入 x = 2 得 y = 4 − 8 + 3 = −1，因此顶点为 (2, −1)。
When sketching, use the correct shape, mark the intercepts and vertex, draw a smooth curve rather than a series of straight lines, and label the curve with its equation. Marks are given for these features, not for artistic quality, but a mislabelled vertex or a straight-sided "parabola" will cost you.
作图时，要用正确的开口方向，标出截距与顶点，画出光滑曲线而不是一连串直线段，并在曲线旁标注方程。得分点是这些特征，而不是画得漂不漂亮；但顶点标错、或者画出"直边抛物线"都会丢分。
7. Solving Equations Graphically | 用图像解方程
s x = 1 and x = 3. Reading roots from a graph this way is a standard 1 or 2 mark question.
如果题目给出 y = x² − 4x + 3 的图像，那么 x² − 4x + 3 = 0 的解就是曲线与 x 轴交点的横坐标，即 x = 1 和 x = 3。这种从图像上读根的方法是标准的 1 到 2 分题。
A harder version asks you to solve a related equation using the same graph, by drawing a suitable straight line. To solve x² − 4x + 3 = x ch is not the graph you were given. Instead, rewrite the target equation so that the graph's expression appears on one side: x² − 4x + 3 = x, so draw y = x and read off the two intersection points. The general rule is: rearrange until the left-hand side matches the graph exactly, then draw whatever is on the right-hand side.
较难的版本会要求你用同一张图、通过画一条合适的直线来解一个相关方程。要解 x² − 4x + 3 = x，两边减去 x 得 x² − 5x + 3 = 0，这并非你所给的图像。正确做法是把目标方程改写成图像表达式出现在一边的形式：x² − 4x + 3 = x，所以画直线 y = x，读出两个交点即可。通用规则是：整理到左边与给定图像完全一致，然后画出右边那条直线。
wo solutions; tangency means one solution (Δ = 0); no intersection means no real solution (Δ < 0). This links Section 5 and Section 10 together neatly.
图像也是判断联立方程解的个数最快的方法。若一条直线与抛物线相交两次，相应二次方程有两个解；相切意味着一个解（Δ = 0）；不相交意味着无实数解（Δ < 0）。这把第 5 节与第 10 节很好地联系了起来。
8. Disguised Quadratics and Substitution | 隐蔽二次方程与换元法
Some equations are quadratic in a hidden variable. Recognising the pattern x⁴, x², constant — or 22x, 2x, constant — lets you substitute u for the squared term and solve a simple quadratic instead of a
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