📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are a core topic in IGCSE Mathematics. They appear in algebra, graphs, geometry, and word problems, and mastering them is essential for achieving high grades. This article covers the standard form, three solving methods, the discriminant, inequalities, graphing, and common mistakes.
二次方程是 IGCSE 数学的核心考点，广泛出现在代数、图像、几何和应用题中，掌握它是取得高分的关键。本文将系统讲解标准形式、三种求解方法、判别式、不等式、图像以及常见错误。
1. What Is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2. Its general form is:
二次方程是最高次数为 2 的多项式方程，其一般形式为：
ax² + bx + c = 0, where a ≠ 0
ot be zero; if a = 0, the equation becomes linear (bx + c = 0) and is no longer quadratic.
其中 a、b、c 是常数，x 是未知数。系数 a 不能为零；若 a = 0，方程就变成一次方程 bx + c = 0，不再属于二次方程。
A quadratic equation always has at most two solutions, called roots. These roots may be real and distinct, real and equal, or not real at all, depending on the value of the discriminant.
二次方程至多有两个解，称为根。根据判别式的取值，根可以是两个不等实根、两个相等实根，或者没有实根。
2. Rearranging into Standard Form | 化为标准形式
0. This means moving every term to one side of the equals sign and collecting like terms.
在采用任何求解方法之前，必须先将方程整理成标准形式 ax² + bx + c = 0，即把所有项移到等号一侧并合并同类项。
rd form. Subtract 5x from both sides to get 2x² − 5x + 3 = 0. Now a = 2, b = −5 and c = 3.
例如，方程 2x² + 3 = 5x 不是标准形式。两边减去 5x 得到 2x² − 5x + 3 = 0。此时 a = 2，b = −5，c = 3。
- Always check that the x² term is written first, then the x term, then the constant. 注意按 x² 项、x 项、常数项的顺序书写。
- If both sides contain x² terms, subtract them first before factorising. 若等式两边都有 x² 项，先相减消去再进行因式分解。
A common mistake is to attempt factorisation before moving all terms to one side. This often hides the true relationship between the factors and the roots.
常见错误是在把所有项移到同一侧之前就尝试因式分解，这样往往会掩盖因式与根之间的真实关系。
3. Solving by Factorisation | 因式分解法
Factorisation is usually the fastest method when the quadratic has simple integer coefficients. The principle is that if the product of two expressions is zero, then at least one of them must be zero.
当二次方程具有简单的整数系数时，因式分解通常是最快的解法。其原理是：如果两个式子相乘等于零，则其中至少有一个式子必须为零。
If (px + q)(rx + s) = 0, then px + q = 0 or rx + s = 0.
5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5. These are −2 and −3.
例：解 x² − 5x + 6 = 0。寻找两个数，使其乘积为 6 且和为 −5，这两个数是 −2 和 −3。
(x − 2)(x − 3) = 0, so x = 2 or x = 3.
ind two numbers whose product is 6 and sum is 5: these are 2 and 3. Split the middle term:
当 x² 的系数不为 1 时，需要更加小心。例如，解 2x² + 5x + 3 = 0。将 a 与 c 相乘：2 × 3 = 6。找出两个数，使乘积为 6 且和为 5，即 2 和 3。然后拆开中间项：
2x² + 2x + 3x + 3 = 0 → 2x(x + 1) + 3(x + 1) = 0 → (2x + 3)(x + 1) = 0
Therefore x = −3/2 or x = −1. Always expand your answer to verify the factorisation.
因此 x = −3/2 或 x = −1。务必展开结果以检验因式分解是否正确。
4. The Quadratic Formula | 二次公式法
When factorisation is difficult, impossible, or too time-consuming, the quadratic formula provides a universal method. It works for every quadratic equation, including those with irrational roots.
当因式分解困难、不可行或耗时过多时，二次公式提供了一种通用解法。它适用于所有二次方程，包括带有无理根的方程。
x = [−b ± √b² − 4ac] / 2a
Example: Solve 2x² + 4x − 6 = 0. Here a = 2, b = 4, c = −6. Substitute into the formula:
例：解 2x² + 4x − 6 = 0。这里 a = 2，b = 4，c = −6。代入公式：
x = [−4 ± √(4² − 4 × 2 × (−6))] / (2 × 2) = (−4 ± √64) / 4 = (−4 ± 8) / 4
So x = (−4 + 8)/4 = 1, or x = (−4 − 8)/4 = −3. The solution set is x = 1 or x = −3.
因此 x = (−4 + 8)/4 = 1，或 x = (−4 − 8)/4 = −3。解集为 x = 1 或 x = −3。
- Be very careful with negative values of c: the term −4ac becomes positive when c is negative. 特别注意 c 为负值的情况：当 c 为负数时，−4ac 会变成正数。
- Show every substitution step in the exam to avoid arithmetic errors. 考试时写出完整的代入步骤，以免运算出错。
- If the question says "give your answer correct to 2 decimal places", round only at the final step. 若题目要求“答案保留两位小数”，只在最后一步进行四舍五入。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This form is extremely useful for finding turning points and solving equations without using the full formula.
配方法将二次表达式改写成 a(x + p)² + q 的形式。这种形式在求顶点坐标和解方程时非常有用。
For a quadratic x² + bx, add (b/2)² and subtract it. For example, x² + 6x becomes (x + 3)² − 9.
对于 x² + bx，加上 (b/2)² 再减去它。例如，x² + 6x 可写成 (x + 3)² − 9。
Example: Solve x² + 6x + 5 = 0 by completing the square.
例：用配方法解 x² + 6x + 5 = 0。
x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4
+ 3)² = 4. Taking the square root of both sides gives x + 3 = ±2. Hence x = −1 or x = −5.
令该表达式为零：(x + 3)² − 4 = 0，即 (x + 3)² = 4。两边开平方得 x + 3 = ±2，因此 x = −1 或 x = −5。
t first. For example, 2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5.
当 a 不等于 1 时，先把 a 提出来。例如：2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5。
6. The Discriminant | 判别式
: Δ = b² − 4ac. It tells us the nature of the roots without actually solving the equation.
判别式是二次公式根号内的表达式：Δ = b² − 4ac。它无需真正解方程就可以判断根的性质。
| Value of Δ | Nature of Roots | Graph Interpretation |
| Δ > 0 | Two distinct real roots 两个不等实根 | Graph crosses the x-axis twice 图像与 x 轴有两个交点 |
| Δ = 0 | One repeated real root 两个相等实根（重根） | Graph touches the x-axis once 图像与 x 轴相切 |
| Δ < 0 | No real roots 无实根 | Graph never meets the x-axis 图像与 x 轴无交点 |
Example: Find the value of k for which the equation x² + kx + 9 = 0 has equal roots.
例：求使方程 x² + kx + 9 = 0 有等根时 k 的值。
Δ = k² − 4 × 1 × 9 = k² − 36. Set Δ = 0: k² = 36, so k = ±6.
If the question states "the equation has real roots", the condition is Δ ≥ 0. If it states "two distinct real roots", the condition is Δ > 0.
若题目说“方程有实根”，则条件为 Δ ≥ 0；若说“有两个不等实根”，则条件为 Δ > 0。
7. Quadratic Inequalities | 二次不等式
Quadratic inequalities are solved by first finding the roots of the corresponding equation, then testing intervals on a number line.
解二次不等式需先求出对应方程的根，然后在数轴上测试各区间的符号。
and x = 2. The parabola opens upward, so the graph is below the x-axis between the roots.
例：解 x² − 4 < 0。方程 x² − 4 = 0 的根为 x = −2 和 x = 2。抛物线开口向上，因此图像在两个根之间位于 x 轴下方。
Solution: −2 < x < 2
For x² − 4 > 0, the solution is x < −2 or x > 2. Note that the inequality sign tells you whether the solution is "between the roots" or "outside the roots".
对于 x² − 4 > 0，解为 x < −2 或 x > 2。注意不等号的方向决定解是“两根之间”还是“两根之外”。
- Use open circles ( ) on the number line for strict inequalities < and >. 严格不等式 < 和 > 在数轴上用空心圆圈表示。
- Use closed circles [ ] for ≤ and ≥. 非严格不等式 ≤ 和 ≥ 用实心圆圈表示。
8. Sketching Quadratic Graphs | 二次函数图像
ion y = ax² + bx + c is a parabola. The sign of a determines the direction of the opening.
二次函数 y = ax² + bx + c 的图像是一条抛物线。a 的符号决定开口方向。
- If a > 0, the parabola opens upward (U-shape), and has a minimum point. 若 a > 0，抛物线开口向上（U 形），有最小值点。
- If a < 0, the parabola opens downward (n-shape), and has a maximum point. 若 a < 0，抛物线开口向下（n 形），有最大值点。
. The axis of symmetry is the vertical line x = −b/(2a), and the vertex lies on this line.
与 y 轴的交点始终为 (0, c)。与 x 轴的交点是方程 ax² + bx + c = 0 的根（若存在实根）。对称轴是竖直线 x = −b/(2a)，顶点在这条线上。
nd −1. The y-intercept is −3. The axis of symmetry is x = 1, and the vertex is at (1, −4).
例：画出 y = x² − 2x − 3 的草图。因式分解得 (x − 3)(x + 1)，所以 x 轴截距为 3 和 −1。y 轴截距为 −3。对称轴为 x = 1，顶点为 (1, −4)。
9. Word Problems Involving Quadratics | 二次方程应用题
Quadratic equations frequently arise in geometry, physics, and number problems. The key is to translate the given information into a quadratic equation, solve it, and then interpret the solutions in the context of the problem.
二次方程常见于几何、物理和数字问题中。关键在于将题目信息转化为二次方程，求解后再结合实际情境解释答案。
Example: A rectangle has length 3 cm greater than its width. Its area is 40 cm². Find the width.
例：一个长方形的长比宽多 3 cm，面积为 40 cm²。求宽。
Let width = w, then length = w + 3. The area is w(w + 3) = 40, so w² + 3w − 40 = 0.
设宽为 w，则长为 w + 3。面积为 w(w + 3) = 40，即 w² + 3w − 40 = 0。
w² + 3w − 40 = (w + 8)(w − 5) = 0, so w = −8 or w = 5.
= 5 cm and the length is 8 cm. Always reject solutions that are not physically meaningful.
宽度不能为负，所以 w = 5 cm，长为 8 cm。务必舍去不符合实际的解。
scribe height h at time t. Solving h = 0 gives the times when the object is on the ground.
抛体运动问题常用 h = ut − 5t² 描述物体在时刻 t 的高度 h。令 h = 0 可求出物体落地的时间。
10. Common Mistakes and How to Avoid Them | 常见错误与规避方法
Even strong students lose marks on quadratic equations due to small but repeated errors. Here are the most common pitfalls and strategies to avoid them.
即使是成绩优秀的学生也常因细小但反复出现的错误在二次方程上失分。以下是最常见的陷阱及应对策略。
| Mistake 错误 | Correction 纠正方法 |
| Forgetting to rearrange into standard form 忘记化为标准形式 | Always write ax² + bx + c = 0 first 先写出标准形式再求解 |
| Forgetting the ± sign when taking square roots 开平方时忘记 ± 号 | Write ± explicitly at the square-root step 在开平方步骤明确写出 ± |
| Incorrect expansion of (x + p)² 错误展开 (x + p)² | Remember (x + p)² = x² + 2px + p² 牢记完全平方公式 |
| Substituting wrong signs into the quadratic formula 代入公式时符号错误 | Write a = , b = , c = before substituting 代入前先列出 a、b、c 的值 |
| Keeping negative roots in word problems 应用题中保留负根 | Check every answer against the context 结合实际情境检验每个答案 |
11. Practice Questions | 练习与自测
Try the following questions on your own, then check the answers below.
请独立完成以下练习，再对照文末答案。
- Solve by factorisation: x² − 7x + 12 = 0. 用因式分解法解：x² − 7x + 12 = 0。
- Solve using the quadratic formula: 3x² + 5x − 2 = 0. 用二次公式法解：3x² + 5x − 2 = 0。
- Complete the square for x² + 8x + 15 and hence solve the equation. 用配方法解：x² + 8x + 15 = 0。
- Find the value of k such that kx² + 6x + 1 = 0 has two distinct real roots. 求使 kx² + 6x + 1 = 0 有两个不等实根时 k 的取值范围。
- A rectangle of area 28 m² has length 3 m longer than its width. Find the dimensions. 一个面积为 28 m² 的矩形，长比宽多 3 m，求长和宽。
Answers 答案:
1. x = 3 or x = 4. 2. x = 1/
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