📚 Solving Quadratic Equations | 解二次方程
nts and a ≠ 0. These equations appear frequently in algebra, geometry and real-life problem solving, so mastering the methods to solve them is essential for exam success.
二次方程是 IGCSE 数学中最重要的主题之一。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 为常数，且 a ≠ 0。这类方程在代数、几何和实际问题中频繁出现，因此掌握其求解方法对于考试成功至关重要。
1. The Standard Form of a Quadratic Equation | 二次方程的标准形式
tion, you must ensure it is written in the standard form ax² + bx + c = 0. The coefficient 'a' is called the leading coefficient, 'b' is the coefficient of x, and 'c' nstant term. For example, 2x² − 5x + 3 = 0 is in standard form, while 3x² = 2x − 1 is not.
在求解任何二次方程之前，必须确保方程已写成标准形式 ax² + bx + c = 0。系数 a 称为首项系数，b 是 x 的系数，c 是常数项。例如，2x² − 5x + 3 = 0 是标准形式，而 3x² = 2x − 1 不是。
ce rule. For 3x² = 2x − 1, subtract 2x and add 1 to both sides to obtain 3x² − 2x + 1 = 0.
要将非标准方程转换为标准形式，需要利用平衡规则将所有项移到一边。对于 3x² = 2x − 1，两边同时减去 2x 并加上 1，得到 3x² − 2x + 1 = 0。
2. Solving by Factorisation | 因式分解法
oduct of two brackets, then use the zero product property: if AB = 0, then A = 0 or B = 0.
因式分解法通常是当二次表达式具有简单整数因式时最快捷的方法。其原理是将 ax² + bx + c 重写为两个括号的乘积，然后利用零积性质：若 AB = 0，则 A = 0 或 B = 0。
mbers are −2 and −3. Hence the equation becomes (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
考虑 x² − 5x + 6 = 0。我们寻找两个数，它们相乘等于 6 且相加等于 −5。这两个数是 −2 和 −3。因此方程变为 (x − 2)(x − 3) = 0，解得 x = 2 或 x = 3。
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 或 x = 3
nd 6, allowing us to split the middle term: 2x² + x + 6x + 3 = 0, then factor by grouping.
对于 a ≠ 1 的方程，如 2x² + 7x + 3 = 0，因式分解需要更加小心。我们寻找两个数，它们相乘等于 a × c = 2 × 3 = 6，且相加等于 7。这两个数是 1 和 6，于是可以拆分中间项：2x² + x + 6x + 3 = 0，然后分组因式分解。
2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3) = 0 → x = −½ 或 x = −3
3. Solving by Completing the Square | 配方法
is method always works and is especially useful when the equation has irrational or complex roots. The key step involves halving the coefficient of x and squaring it.
配方法将二次方程改写为 a(x − h)² + k = 0 的形式。这种方法始终有效，尤其在方程具有无理数根或复数根时特别有用。关键步骤是将 x 的系数取半再平方。
3)² − 9 − 7 = 0, so (x + 3)² = 16. Taking square roots, x + 3 = ±4, hence x = 1 or x = −7.
以 x² + 6x − 7 = 0 为例。6 的一半是 3，3² = 9。我们写 x² + 6x = (x + 3)² − 9。代回原方程得 (x + 3)² − 9 − 7 = 0，即 (x + 3)² = 16。两边开平方，x + 3 = ±4，因此 x = 1 或 x = −7。
x² + 6x − 7 = 0 → (x + 3)² − 16 = 0 → (x + 3)² = 16 → x = 1 或 x = −7
2x² − 4x + 1 = 0 becomes 2(x² − 2x) + 1 = 0, then complete the square inside the bracket.
当 x² 的系数不为 1 时，先从前两项中提出该系数。例如，2x² − 4x + 1 = 0 变为 2(x² − 2x) + 1 = 0，然后在括号内配方。
2(x² − 2x) + 1 = 2[(x − 1)² − 1] + 1 = 2(x − 1)² − 1 = 0 → (x − 1)² = ½ → x = 1 ± √½
4. The Quadratic Formula | 求根公式
the square and is the most reliable method. You should memorise it and always check that your solutions satisfy the original equation.
求根公式给出了任何二次方程 ax² + bx + c = 0 的解。它由配方法推导而来，是最可靠的方法。你应该牢记该公式，并始终检查解是否满足原方程。
x = (−b ± √b² − 4ac) / 2a
ese into the formula, simplify carefully, and you will obtain two distinct rational roots.
使用公式时，从标准形式中代入 a、b、c 的值。考虑 3x² − 5x − 2 = 0。这里 a = 3，b = −5，c = −2。将这些值代入公式，仔细化简，将得到两个不同的有理数根。
x = (5 ± √25 + 24) / 6 = (5 ± 7) / 6 → x = 2 或 x = −⅓
Always simplify the square root fully. If the discriminant b² − 4ac is a perfect square, the roots are rational; otherwise they are irrational and must be left in simplified surd form.
始终将平方根完全化简。如果判别式 b² − 4ac 是一个完全平方数，则根为有理数；否则根为无理数，必须保留为化简后的根式形式。
5. The Discriminant and the Nature of Roots | 判别式与根的性质
ac, determines the nature of the roots without actually solving the equation. This concept is frequently tested in IGCSE exams, especially multiple-choice questions.
判别式，记作 Δ = b² − 4ac，无需实际解方程即可确定根的性质。这一概念在 IGCSE 考试中经常考查，尤其是选择题。
xactly one repeated real root. If Δ < 0, the equation has no real roots (two complex roots).
若 Δ > 0，方程有两个不同的实数根。若 Δ = 0，方程恰好有一个重根（两个相等的实数根）。若 Δ < 0，方程没有实数根（有两个复数根）。
| 判别式 Δ | 根的性质 |
| Δ > 0 | 两个不同的实数根 |
| Δ = 0 | 一个重复实数根 |
| Δ < 0 | 无实数根 |
y one repeated root x = 2. For x² + 1 = 0, Δ = 0 − 4 = −4 < 0, so there are no real roots.
例如，对于 x² − 4x + 4 = 0，Δ = 16 − 16 = 0，因此恰好有一个重根 x = 2。对于 x² + 1 = 0，Δ = 0 − 4 = −4 < 0，因此没有实数根。
6. Solving Quadratic Equations by Graphing | 图象法解二次方程
² + bx + c intersects the x-axis. Each x-intercept represents a real root of the equation.
从图象上看，ax² + bx + c = 0 的解对应于抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。每个 x 轴交点代表方程的一个实数根。
If the parabola touches the x-axis at exactly one point, the equation has one repeated root. If the parabola never crosses the x-axis, the equation has no real roots. This visual interpretation reinforces the discriminant results.
如果抛物线恰好与 x 轴相切于一点，则方程有一个重根。如果抛物线从不与 x 轴相交，则方程没有实数根。这种直观解释强化了判别式的结论。
When asked to solve a quadratic graphically, plot at least five points, sketch the parabola carefully, and read off the intersections to one decimal place if necessary. You should also learn to use a graphing calculator or graphical display calculator efficiently.
当被要求利用图象解二次方程时，至少取五个点，仔细绘制抛物线，如有必要则精确到一位小数读取交点。你还应学会高效使用图形计算器或图形显示计算器。
7. Quadratic Equations from Word Problems | 实际应用题中的二次方程
Many IGCSE problems require you to translate a real-life situation into a quadratic equation. Common contexts include area problems, number problems and projectile motion. The key is to define your variable clearly and form the equation from the given conditions.
许多 IGCSE 题目要求你将实际情况转化为二次方程。常见情境包括面积问题、数字问题和抛体运动。关键在于清晰地定义变量，并根据给定条件建立方程。
(x + 3) cm. The area condition gives x(x + 3) = 40, which simplifies to x² + 3x − 40 = 0.
例：一个长方形的长比宽长 3 cm，面积为 40 cm²。设宽为 x cm，则长为 (x + 3) cm。面积条件给出 x(x + 3) = 40，化简得 x² + 3x − 40 = 0。
x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0 → x = 5 或 x = −8
ngth is 8 cm. Always interpret your answers in the context of the problem and reject unrealistic negative solutions.
由于长度不能为负，舍去 x = −8。宽为 5 cm，长为 8 cm。始终结合题目背景解释答案，并舍弃不合实际的负数解。
8. Common Mistakes and Exam Tips | 常见错误与考试技巧
ll lose the solution x = 0. Always factorise instead of dividing by a variable expression.
一个常见错误是方程两边同时除以 x。这是不允许的，因为 x 可能为零，这样做会丢失 x = 0 这个解。应始终采用因式分解而不是除以含变量的表达式。
Another frequent error is forgetting to rearrange the equation into standard form before applying the quadratic formula. If you substitute a, b, c from a non-standard form, your answers will be incorrect.
另一个常见错误是在应用求根公式之前忘记将方程整理为标准形式。如果从非标准形式直接代入 a、b、c，答案将会出错。
ives x + 2 = ±3, so x = 1 or x = −5. Missing the negative root is a very common lost mark.
此外，两边同时开平方时，务必加上 ± 号。例如，(x + 2)² = 9 给出 x + 2 = ±3，所以 x = 1 或 x = −5。遗漏负根是常见的失分点。
Always check your solutions by substitution into the original equation.
始终通过代入原方程来检验解。
When the question says "give your answer correct to 2 decimal places", use the quadratic formula and your calculator, not factorisation.
当题目要求"答案精确到两位小数"时，应使用求根公式和计算器，而不是因式分解。
Read whether the answer should be exact or approximated; surd answers should be simplified fully.
注意题目要求精确值还是近似值；根式答案需要完全化简。
9. Practice Problems | 练习题目
Work through these problems to consolidate your understanding. Solutions are given below in factorised form or via the discriminant, but try each one independently first.
通过以下练习来巩固理解。下方给出了因式分解形式或判别式提示的解答，但请先独立思考解答每一题。
Question 1: Solve x² − 9 = 0.
问题 1：解 x² − 9 = 0。
Solution: This is a difference of squares: (x − 3)(x + 3) = 0, so x = 3 or x = −3.
解答：这是平方差公式：(x − 3)(x + 3) = 0，所以 x = 3 或 x = −3。
Question 2: Solve 2x² − 8x + 6 = 0.
问题 2：解 2x² − 8x + 6 = 0。
Solution: vide by 2: x² − 4x + 3 = 0, which factorises to (x − 1)(x − 3) = 0, giving x = 1 or x = 3.
解答：先除以 2：x² − 4x + 3 = 0，因式分解为 (x − 1)(x − 3) = 0，得 x = 1 或 x = 3。
Question 3: Find the value of the discriminant for x² + 2x + 5 = 0 and state the nature of the roots.
问题 3：求 x² + 2x + 5 = 0 的判别式值，并说明根的性质。
Solution: Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16 < 0, so there are no real roots.
解答：Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16 < 0，因此没有实数根。
Question 4: A right-angled triangle has hypotenuse 13 cm and one leg 5 cm longer than the other. Find the lengths of the two legs.
问题 4：一个直角三角形，斜边为 13 cm，一条直角边比另一条长 5 cm。求两条直角边的长度。
Solution: ide by 2: x² + 5x − 72 = 0 = (x + 9)(x − 8). Hence x = 8, and the legs are 8 cm and 13 cm.
解答：设较短的直角边为 x cm。则 x² + (x + 5)² = 169，即 2x² + 10x + 25 = 169，所以 2x² + 10x − 144 = 0。除以 2：x² + 5x − 72 = 0 = (x + 9)(x − 8)。因此 x = 8，两条直角边分别为 8 cm 和 13 cm。
10. Summary of Methods | 方法总结
Choose the most efficient method by examining the equation. Factorisation is fastest for simple integer roots; completing the square is useful for deriving the vertex form or solving equations with surd roots; the quadratic formula always works and is a safe fallback.
通过观察方程来选择最有效的方法。对于简单整数根，因式分解最快；配方法适于推导顶点形式或求解含根式的方程；求根公式始终有效，是稳妥的备选方案。
A quick decision guide: if the question tests factorisation, look for two brackets directly. If a calculator is allowed and decimals are requested, use the formula. If the coefficient of x is even, completing the square is often neater.
快速决策指南：如果题目考查因式分解，直接寻找两个括号。如果允许使用计算器且要求小数答案，使用求根公式。如果 x 的系数为偶数，配方法通常更简洁。
The graph of a quadratic equation gives a visual check: each x-intercept is a real root. Combined with the discriminant, you can determine the number and type of roots before you even begin solving.
二次方程的图象提供了直观验证：每个 x 轴交点都是一个实数根。结合判别式，你甚至可以在开始求解之前确定根的个数和类型。
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