📚 Mastering Quadratic Equations | 掌握一元二次方程
Quadratic equations appear in nearly every IGCSE Mathematics paper, from simple factorisation to complex word problems. This article gives you the complete toolkit: standard form, factorisation, completing the square, the quadratic formula, the discriminant, graphs, inequalities, and exam-ready tips.
一元二次方程几乎出现在每一份 IGCSE 数学试卷中，从简单的因式分解到复杂的应用题。本文将为你提供完整的工具箱：标准形式、因式分解法、配方法、求根公式、判别式、图像、不等式以及考试必备技巧。
1. The Standard Form of a Quadratic Equation | 一元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 real numbers and a ≠ 0. The term ax² is the quadratic term, bx is the linear term, and c is the constant term.
一元二次方程是指任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是实数，且 a ≠ 0。ax² 称为二次项，bx 称为一次项，c 称为常数项。
ition a ≠ 0 is crucial: if a = 0, the equation becomes linear (bx + c = 0), not quadratic.
例如，2x² − 5x + 1 = 0 是一个一元二次方程，其中 a = 2，b = −5，c = 1。条件 a ≠ 0 至关重要：如果 a = 0，方程就变成一次方程 bx + c = 0，而不是二次方程。
rranging. For instance, 3x² = 7x − 2 must be rewritten as 3x² − 7x + 2 = 0 before solving.
有些方程需要经过整理才会变成二次方程。例如，3x² = 7x − 2 必须先改写为 3x² − 7x + 2 = 0 才能开始求解。
2. Solving by Factorisation | 因式分解法
to 6 and add to 5. Those numbers are 2 and 3, so the equation becomes (x + 2)(x + 3) = 0.
当二次表达式可以因式分解时，这种方法通常最快。要解 x² + 5x + 6 = 0，需要找到两个数，使它们的乘积为 6，和为 5。这两个数是 2 和 3，因此方程化为 (x + 2)(x + 3) = 0。
operty, if (x + 2)(x + 3) = 0, then either x + 2 = 0 or x + 3 = 0. Hence x = −2 or x = −3.
根据零乘积性质，若 (x + 2)(x + 3) = 0，则必有 x + 2 = 0 或 x + 3 = 0。因此 x = −2 或 x = −3。
Follow these steps when factorising:
因式分解时请遵循以下步骤：
- Move all terms to one side so the equation is in the form ax² + bx + c = 0. / 将所有项移到一边，使方程成为 ax² + bx + c = 0 的形式。
- Factorise the quadratic expression completely. / 将二次表达式完整地分解因式。
- Set each bracket equal to zero and solve the two linear equations. / 令每个括号等于零，并解出两个一次方程。
and factorise: x(2x + 1) + 3(2x + 1) = 0, giving (2x + 1)(x + 3) = 0, so x = −½ or x = −3.
当 a ≠ 1 时，例如 2x² + 7x + 3 = 0，先将 a 和 c 相乘：2 × 3 = 6。找到 6 的两个因数和为 7，即 1 和 6，然后拆开中间项：2x² + x + 6x + 3 = 0。分组并因式分解：x(2x + 1) + 3(2x + 1) = 0，得到 (2x + 1)(x + 3) = 0，所以 x = −½ 或 x = −3。
3. Solving by Completing the Square | 配方法
tant. The key identity is (x + b/2)² = x² + bx + (b/2)², so x² + bx = (x + b/2)² − (b/2)².
配方法将 x² + bx + c 化为一个完全平方加上一个常数。关键恒等式是 (x + b/2)² = x² + bx + (b/2)²，因此 x² + bx = (x + b/2)² − (b/2)²。
Solve x² + 6x + 1 = 0 by completing the square:
用配方法解 x² + 6x + 1 = 0：
- Rewrite the first two terms: x² + 6x = (x + 3)² − 9. / 重写前两项：x² + 6x = (x + 3)² − 9。
- Insert the constant: (x + 3)² − 9 + 1 = 0, so (x + 3)² − 8 = 0. / 代入常数项：(x + 3)² − 9 + 1 = 0，即 (x + 3)² − 8 = 0。
- Isolate the square: (x + 3)² = 8. / 分离平方项：(x + 3)² = 8。
- Take square roots: x + 3 = ±√8 = ±2√2. / 两边开平方：x + 3 = ±√8 = ±2√2。
- Solve: x = −3 ± 2√2. / 求解：x = −3 ± 2√2。
The answers are x = −3 + 2√2 or x = −3 − 2√2. This method also helps when finding the turning point of a parabola, as we will see later.
答案为 x = −3 + 2√2 或 x = −3 − 2√2。这种方法在后续求抛物线的顶点时也非常有用。
4. The Quadratic Formula | 求根公式
dratic formula. This formula works for every quadratic, even when factorisation is difficult or impossible.
对于任意一元二次方程 ax² + bx + c = 0，解可由求根公式给出。这个公式适用于所有二次方程，即使因式分解很难或根本无法进行。
x = (−b ± √b² − 4ac) / 2a
Solve 2x² − 4x − 3 = 0 using the formula, where a = 2, b = −4 and c = −3:
用求根公式解 2x² − 4x − 3 = 0，其中 a = 2，b = −4，c = −3：
- Substitute into the formula: x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2). / 代入公式：x = (4 ± √((−4)² − 4 × 2 × (−3))) / (2 × 2)。
- Simplify inside the root: 16 + 24 = 40. / 化简根号内：16 + 24 = 40。
- x = (4 ± √40) / 4 = (4 ± 2√10) / 4. / x = (4 ± √40) / 4 = (4 ± 2√10) / 4。
- So x = (2 + √10) / 2 ≈ 2.58 or x = (2 − √10) / 2 ≈ −0.58. / 因此 x = (2 + √10) / 2 ≈ 2.58 或 x = (2 − √10) / 2 ≈ −0.58。
Always check whether the question asks for exact values or decimal answers rounded to a given accuracy, such as 1 decimal place or 3 significant figures.
务必看清题目要求的是精确值还是保留指定精度的小数答案，例如保留 1 位小数或 3 位有效数字。
5. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant, written as Δ. It tells us how many real roots a quadratic equation has without solving it.
根号内的表达式 b² − 4ac 称为判别式，记为 Δ。它不经过求解就能告诉我们一个二次方程有多少个实数根。
| Value of Δ = b² − 4ac | Nature of Roots / 根的情况 |
| Δ > 0 | Two distinct real roots / 两个不相等的实数根 |
| Δ = 0 | One repeated real root / 一个重根（两个相等实数根） |
| Δ < 0 | No real roots / 没有实数根 |
0 has Δ = 4² − 4 × 1 × 5 = 16 − 20 = −4. Since Δ < 0, this equation has no real solutions.
例如，方程 x² + 4x + 5 = 0 的判别式为 Δ = 4² − 4 × 1 × 5 = 16 − 20 = −4。由于 Δ < 0，该方程没有实数解。
6. Solving Quadratic Equations by Graphing | 利用图像解二次方程
-coordinates where the graph of y = ax² + bx + c crosses the x-axis, that is, where y = 0.
ax² + bx + c = 0 的实数根就是函数 y = ax² + bx + c 的图像与 x 轴交点的横坐标，即 y = 0 的位置。
= 0, it touches the x-axis at exactly one point; when Δ < 0, it never crosses the x-axis.
当 Δ > 0 时，图像与 x 轴相交于两点；当 Δ = 0 时，图像与 x 轴只有一个切点；当 Δ < 0 时，图像与 x 轴没有交点。
x-axis at (−1, 0) and (3, 0), then the solutions of x² − 2x − 3 = 0 are x = −1 and x = 3.
如果给出图像，你可以直接读出近似根。例如，抛物线 y = x² − 2x − 3 与 x 轴交于 (−1, 0) 和 (3, 0)，那么方程 x² − 2x − 3 = 0 的解就是 x = −1 和 x = 3。
7. The Graph of y = ax² + bx + c | 二次函数 y = ax² + bx + c 的图像
The graph of a quadratic function is always a parabola. If a > 0, the parabola opens upward and has a minimum point; if a < 0, it opens downward and has a maximum point.
二次函数的图像始终是一条抛物线。如果 a > 0，抛物线开口向上，存在最小值点；如果 a < 0，抛物线开口向下，存在最大值点。
- The y-intercept is at (0, c). / y 轴截距为 (0, c)。
- The x-intercepts (if any) are the real roots of ax² + bx + c = 0. / x 轴截距（如果存在）就是 ax² + bx + c = 0 的实数根。
- The parabola is symmetric about a vertical line called the axis of symmetry. / 抛物线关于一条垂直线对称，这条线称为对称轴。
- The vertex (turning point) lies on the axis of symmetry. / 顶点（转向点）位于对称轴上。
. The y-intercept is 3, and factorising gives (x − 1)(x − 3), so the x-intercepts are 1 and 3.
对于 y = x² − 4x + 3，有 a = 1，b = −4，c = 3。y 轴截距为 3，因式分解得 (x − 1)(x − 3)，因此 x 轴截距为 1 和 3。
8. Finding the Vertex and Axis of Symmetry | 求顶点与对称轴
me x-coordinate, so its y-coordinate is found by substituting x = −b/2a into the function.
对于二次函数 y = ax² + bx + c，对称轴是垂直线 x = −b/2a。顶点的 x 坐标与此相同，将其代入函数即可求出顶点的 y 坐标。
x = −b/2a, y = f(−b/2a)
2 = 3. Then y = 9 − 18 + 5 = −4. The vertex is (3, −4), and the axis of symmetry is x = 3.
求 y = x² − 6x + 5 的顶点。此处 a = 1，b = −6，所以 x = 6/2 = 3。代入得 y = 9 − 18 + 5 = −4。顶点为 (3, −4)，对称轴为 x = 3。
= (x − 3)² − 4, which directly reveals the vertex (3, −4). Both methods are equally valid.
另外，配方可得 y = (x − 3)² − 4，直接显示出顶点 (3, −4)。两种方法同样有效。
9. Applications: Word Problems | 应用题
Quadratic equations frequently appear in geometry and number problems. A common type involves area relationships.
二次方程经常出现在几何和数字问题中，其中常见的一类是涉及面积关系的问题。
A rectangle is 3 cm longer than it is wide, and its area is 40 cm². Find its width.
一个矩形的长比宽长 3 cm，面积为 40 cm²。求它的宽。
- Let the width be x cm, so the length is (x + 3) cm. / 设宽为 x cm，则长为 (x + 3) cm。
- Area equation: x(x + 3) = 40. / 面积方程：x(x + 3) = 40。
- Expand and rearrange: x² + 3x − 40 = 0. / 展开并整理：x² + 3x − 40 = 0。
- Factorise: (x + 8)(x − 5) = 0, so x = −8 or x = 5. / 因式分解：(x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。
- Since width cannot be negative, the width is 5 cm. / 由于宽度不能为负，宽度为 5 cm。
Always reject solutions that do not fit the context of the problem, and state your final answer with the correct units.
务必舍弃不符合题目实际情况的解，并在最终答案中注明正确的单位。
10. Sum and Product of Roots | 根的和与积
+ bx + c = 0 with roots α and β, the sum and product of the roots follow simple formulas.
对于根为 α 和 β 的二次方程 ax² + bx + c = 0，根的和与积遵循两个简单的公式。
α + β = −b/a, αβ = c/a
product is 6/2 = 3. You can verify this: the roots are 1 and 3, and 1 + 3 = 4, 1 × 3 = 3.
对于方程 2x² − 8x + 6 = 0，根的和为 −(−8)/2 = 4，根的积为 6/2 = 3。你可以验证：两个根是 1 和 3，且 1 + 3 = 4，1 × 3 = 3。
This relationship is useful for checking answers and for constructing a quadratic equation when the roots are known.
这个关系可以用来检验答案，也可以在已知根的情况下构造一个二次方程。
11. Quadratic Inequalities | 一元二次不等式
nequality such as x² − 4x + 3 < 0, first solve the corresponding equation x² − 4x + 3 = 0.
要解二次不等式，例如 x² − 4x + 3 < 0，先解对应的方程 x² − 4x + 3 = 0。
- Factorise: (x − 1)(x − 3) = 0, so x = 1 or x = 3. / 因式分解：(x − 1)(x − 3) = 0，所以 x = 1 或 x = 3。
- etch the parabola y = (x − 1)(x − 3). It opens upward and crosses the x-axis at 1 and 3. / 画出抛物线 y = (x − 1)(x − 3)。开口向上，与 x 轴交于 1 和 3。
- The inequality is < 0, so we need the region below the x-axis, which lies between the two roots. / 不等式为 < 0，因此我们需要 x 轴下方的区域，即位于两个根之间。
- Answer: 1 < x < 3. / 答案：1 < x < 3。
If the inequality were (x − 1)(x − 3) > 0, the solution would be x < 1 or x > 3, because the graph is above the x-axis outside the interval between the roots.
如果不等式是 (x − 1)(x − 3) > 0，解则为 x < 1 或 x > 3，因为在两根区间之外图像位于 x 轴上方。
12. Common Mistakes and Exam Tips | 常见错误与考试技巧
Many students lose marks on quadratic questions due to avoidable errors. Keep the following points in mind.
许多学生在二次方程题目上失分，
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图