📚 Solving Quadratic Equations | 解二次方程
nts and a ≠ 0. This article will guide you through every major method of solving quadratics, show you how to analyse the number of solutions using the discriminant, and help you avoid the most common exam pitfalls.
二次方程是 IGCSE 数学中最重要的话题之一，它出现在代数、坐标几何、微积分乃至物理学中。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。本文将引导你掌握求解二次方程的全部主要方法，教你如何利用判别式分析解的个数，并帮助你避开最常见的考试陷阱。
1. What Is a Quadratic Equation? | 什么是二次方程？
nt of x, and c is the constant term. If a = 0, the equation becomes linear, not quadratic.
二次方程是次数为 2 的多项式方程，其中变量 x 的最高次数恰好是 2。其一般形式为 ax² + bx + c = 0。其中，a 称为首项系数，b 是 x 的系数，c 是常数项。如果 a = 0，方程就退化为一次方程，而非二次方程。
h fits the general form. If an equation contains x² and no higher powers, it is quadratic.
二次方程的例子包括：2x² + 3x - 5 = 0；x² - 9 = 0；4x² = 16。注意 x² = 16 可以改写为 x² - 16 = 0，符合一般形式。如果一个方程含有 x² 且没有更高次项，它就是二次方程。
General form: ax² + bx + c = 0 (a ≠ 0)
一般形式：ax² + bx + c = 0（a ≠ 0）
2. Solving by Factorisation | 因式分解法求解
(px + q)(rx + s) = 0, then by the zero-product property, either px + q = 0 or rx + s = 0.
因式分解法通常是二次方程具有简单整数根时最快的解法。其原理是将 ax² + bx + c 改写为两个一次因式的乘积。如果我们能写成 (px + q)(rx + s) = 0，那么根据零积性质，必有 px + q = 0 或 rx + s = 0。
ause (-2) × (-3) = 6 and (-2) + (-3) = -5. Thus (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
当 a = 1 时，我们寻找两个数，使它们的乘积等于 c、和等于 b。例如，解 x² - 5x + 6 = 0。我们需要两个乘积为 6、和为 -5 的数。这两个数是 -2 和 -3，因为 (-2) × (-3) = 6 且 (-2) + (-3) = -5。因此 (x - 2)(x - 3) = 0，解得 x = 2 或 x = 3。
. Now group: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. Hence x = -½ or x = -3.
当 a ≠ 1 时，例如 2x² + 7x + 3 = 0，我们可以使用分组法。先计算 a 与 c 的乘积：2 × 3 = 6。寻找两个乘积为 6、和为 7 的数，它们是 6 和 1。改写中间项：2x² + 6x + x + 3 = 0。然后分组：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0。因此 x = -½ 或 x = -3。
- k) = 0, giving x = ±k. For example, x² - 25 = 0 → (x + 5)(x - 5) = 0 → x = 5 or x = -5.
特殊情况：平方差公式。如果方程为 x² - k² = 0，则可分解为 (x + k)(x - k) = 0，解得 x = ±k。例如，x² - 25 = 0 → (x + 5)(x - 5) = 0 → x = 5 或 x = -5。
3. Solving by Completing the Square | 配方法求解
the equation x² + bx + c = 0, we take half of b, square it, and then adjust the constant.
配方法是一种强大的技巧，可以将二次式改写为 m(x + n)² + p 的形式。这种形式对于求极值点和推导求根公式非常有用。对于方程 x² + bx + c = 0，我们取 b 的一半、将其平方，然后调整常数项。
2 = 0, which simplifies to (x + 3)² = 7. Now take the square root of both sides: x + 3 = ±√7. Therefore x = -3 ± √7, giving two solutions: x ≈ 0.646 and x ≈ -6.646.
例题：用配方法解 x² + 6x + 2 = 0。6 的一半是 3；3² = 9。写成 (x + 3)² - 9 + 2 = 0，化简为 (x + 3)² = 7。两边开平方：x + 3 = ±√7。因此 x = -3 ± √7，得到两个解：x ≈ 0.646 和 x ≈ -6.646。
. Half of -4 is -2; (-2)² = 4. Then (x - 2)² - 4 + 2.5 = 0, so (x - 2)² = 1.5, and x = 2 ± √1.5.
如果 x² 的系数不是 1，先将整个方程除以 a。例如，2x² - 8x + 5 = 0 变为 x² - 4x + 2.5 = 0。-4 的一半是 -2；(-2)² = 4。于是 (x - 2)² - 4 + 2.5 = 0，所以 (x - 2)² = 1.5，即 x = 2 ± √1.5。
Key formula: x² + bx = (x + b/2)² - (b/2)²
关键公式：x² + bx = (x + b/2)² - (b/2)²
4. The Quadratic Formula | 二次求根公式
ven by the formula below. You are expected to memorise this formula for the IGCSE examination.
二次求根公式适用于所有二次方程，包括那些不易因式分解的方程。对于 ax² + bx + c = 0，解由以下公式给出。IGCSE 考试要求你牢记这个公式。
x = (-b ± √b² - 4ac) / 2a
5, c = -2. Substitute into the formula: x = (-5 ± √(5² - 4 × 3 × (-2))) / (2 × 3) = (-5 ± √25 + 24) / 6 = (-5 ± √49) / 6 = (-5 ± 7) / 6. Hence x = 2/6 = 1/3 or x = -12/6 = -2.
我们将此公式应用于 3x² + 5x - 2 = 0。这里 a = 3，b = 5，c = -2。代入公式：x = (-5 ± √(5² - 4 × 3 × (-2))) / (2 × 3) = (-5 ± √25 + 24) / 6 = (-5 ± √49) / 6 = (-5 ± 7) / 6。因此 x = 2/6 = 1/3 或 x = -12/6 = -2。
mple above, c = -2, so -4ac becomes +24. Many students lose marks here due to sign errors.
使用公式时，要特别注意 c 为负值的情况。计算 b² - 4ac 时，必须正确计算 4 × a × c，注意 c 的符号。在上面的例子中，c = -2，所以 -4ac 变为 +24。许多学生在这里因为符号错误而失分。
Always check your answers by substituting them back into the original equation. This is a quick and reliable way to catch arithmetic mistakes before the final answer.
务必通过将答案代入原方程来检验。这是一个快速可靠的检查方法，可以帮助你在提交最终答案前发现算术错误。
5. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression b² - 4ac inside the square root is called the discriminant, often denoted by Δ (the Greek letter delta). It tells us how many real solutions a quadratic equation has, without actually solving the equation.
根号内的表达式 b² - 4ac 称为判别式，通常用希腊字母 Δ 表示。它告诉我们二次方程有多少个实数解，而无需真正求解方程。
| Discriminant value | Number of real roots | Graph interpretation |
| Δ > 0 | 2 distinct real roots | Curve crosses the x-axis twice |
| Δ = 0 | 1 repeated real root | Curve touches the x-axis at one point |
| Δ < 0 | No real roots | Curve does not meet the x-axis |
| 判别式取值 | 实数根的个数 | 图像含义 |
| Δ > 0 | 两个不同实数根 | 曲线与 x 轴相交两次 |
| Δ = 0 | 一个重根 | 曲线与 x 轴相切于一点 |
| Δ < 0 | 无实数根 | 曲线不与 x 轴相交 |
exactly one repeated root. Indeed, x² - 4x + 4 = (x - 2)², so x = 2 is the only solution.
例如：判断 x² - 4x + 4 = 0 的根的性质。这里 Δ = (-4)² - 4 × 1 × 4 = 16 - 16 = 0。因为 Δ = 0，方程只有一个重根。事实上，x² - 4x + 4 = (x - 2)²，所以 x = 2 是唯一解。
If a question states that a quadratic equation has two equal roots, or touches the x-axis, you can set the discriminant equal to zero and solve for an unknown parameter. This is a common exam question type.
如果题目说明某个二次方程有两个相等的根，或与 x 轴相切，你可以令判别式等于零，然后求解未知参数。这是常见的考试题型。
6. Solving Quadratic Equations by Graphs | 利用图像解二次方程
x² + bx + c = 0 are the x-coordinates of the points where the parabola crosses the x-axis.
图解法有时用于不需要精确解、或方程难以用代数方法求解的场合。y = ax² + bx + c 的图像是一条抛物线。ax² + bx + c = 0 的解就是抛物线与 x 轴交点的横坐标。
= c. Then find the x-intercepts by solving the equation, or estimate them from the graph.
要绘制草图，首先判断抛物线的开口方向：若 a > 0，开口向上；若 a < 0，开口向下。然后令 x = 0 求 y 截距，即 y = c。再通过求解方程确定 x 截距，或者从图像上估算。
midpoint of the two x-intercepts, or x = -b / (2a) when the equation is in standard form.
抛物线的顶点（极值点）可以通过配方法求出。对于 y = a(x + n)² + p，顶点在 (-n, p)。在二次曲线上，顶点的横坐标是两个 x 截距的中点，标准形式下也就是 x = -b / (2a)。
as x² - 2x - 3 = 4 by drawing a horizontal line y = 4 and reading the intersection points.
如果给你一张图像，要求解诸如 x² - 2x - 3 = 0 的方程，只需读出曲线 y = x² - 2x - 3 与 x 轴交点的 x 坐标。在此例中，交点为 x = 3 和 x = -1，它们就是方程的解。你也可能被要求通过绘制水平线 y = 4 并读取交点，来解相关方程 x² - 2x - 3 = 4。
7. Forming a Quadratic Equation from Given Roots | 由已知根构造二次方程
f the roots is -b/a and the product of the roots is c/a for the equation ax² + bx + c = 0.
有时考试会要求你根据给定的根来求二次方程。如果根为 p 和 q，则方程可以写成 (x - p)(x - q) = 0。展开得 x² - (p + q)x + pq = 0。这表明，对于方程 ax² + bx + c = 0，根之和为 -b/a，根之积为 c/a。
fy: 2² + 3 × 2 - 10 = 4 + 6 - 10 = 0; (-5)² + 3 × (-5) - 10 = 25 - 15 - 10 = 0. Both work.
例如：求一个以 2 和 -5 为根的二次方程。利用上述公式：x² - (2 + (-5))x + (2 × (-5)) = 0 → x² - (-3)x - 10 = 0 → x² + 3x - 10 = 0。将每个根代入验证：2² + 3 × 2 - 10 = 4 + 6 - 10 = 0；(-5)² + 3 × (-5) - 10 = 25 - 15 - 10 = 0。两个都成立。
This technique is especially useful when the roots are surds, such as x = 3 ± √2 - 6x + 7 = 0, which is harder to solve by factorisation but easy to construct from roots.
这个技巧在根是无理数时尤其有用，例如 x = 3 ± √2。所得方程为 x² - 6x + 7 = 0，这个方程用因式分解法较难解，但从根构造却很容易。
8. Quadratic Equations in Real-World Problems | 二次方程在实际问题中的应用
IGCSE exam papers frequently present word problems that lead to quadratic equations. These include area problems, projectile motion, number puzzles, and geometry questions. The key steps are: read the problem carefully, define the unknown variable, form the equation, solve it, and then check that the answers make sense in context.
IGCSE 试卷中经常出现引导出二次方程的文字题，包括面积问题、抛体运动、数字谜题和几何问题。关键步骤是：仔细阅读题目，设出未知变量，建立方程，求解，然后检查答案在题目情境中是否合理。
e x = -7 or x = 4. Since a width cannot be negative, x = 4 cm. The length is 4 + 3 = 7 cm.
例题：一个长方形的长比宽长 3 cm，面积为 28 cm²。设宽为 x cm，则长为 (x + 3) cm。面积方程为 x(x + 3) = 28，即 x² + 3x - 28 = 0。因式分解：(x + 7)(x - 4) = 0。因此 x = -7 或 x = 4。由于宽度不可能为负，所以 x = 4 cm。长为 4 + 3 = 7 cm。
ets you find the time when the object hits the ground. Only the positive root is physically meaningful.
在抛体运动中，物体向上抛出的高度 h 可以用 h = -4.9t² + v₀t + h₀ 建模，其中 t 为时间，v₀ 为初速度，h₀ 为初始高度。令 h = 0 可以求出物体落地的时间。只有正的根才具有物理意义。
Always read the question to see whether you need exact answers or decimal approximations. Many word problems require you to round to a sensible degree of accuracy, usually 1 or 2 decimal places, unless the question states otherwise.
务必阅读题目要求，判断需要精确答案还是近似值。许多文字题要求你保留合理的精度，通常精确到 1 位或 2 位小数，除非题目另有说明。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
sides to get x² - 3x - 4 = 0. Only then can you factorise or apply the formula correctly.
一个常见错误是忘记在求解前将方程整理为标准形式 ax² + bx + c = 0。例如，在方程 x² = 3x + 4 中，你必须先在两边减去 3x + 4，得到 x² - 3x - 4 = 0。只有这样，你才能正确地进行因式分解或套用公式。
ct approach is to bring all terms to one side: 2x² - 5x = 0, then factorise x(2x - 5) = 0.
另一个常见错误是在 x 不明确非零的情况下两边同除以 x。如果 x = 0 是可能解，这一步会丢失一个根。例如，在 2x² = 5x 中，两边同除以 x 得到 2x = 5，从而 x = 2.5，但丢失了 x = 0 这个解。正确做法是移项：2x² - 5x = 0，然后因式分解 x(2x - 5) = 0。
= 1, b = -4 and c = -5, we have Δ = (-4)² - 4 × 1 × (-5) = 16 + 20 = 36, not 16 - 20 = -4.
在二次求根公式中出现符号错误也非常普遍。当把负数 c 代入 b² - 4ac 时，要小心处理。例如，当 a = 1，b = -4，c = -5 时，Δ = (-4)² - 4 × 1 × (-5) = 16 + 20 = 36，而不是 16 - 20 = -4。
Exam tips: (1) Always show your working clearly; method marks are awarded even if the final answer is wrong. (2) Check whether the question specifies the form of the answer, such as "give your answer correct to 2 decimal places" or "leave your answer in surd form". (3) In multiple-choice questions, you can often use substitution to test which solutions satisfy the equation.
考试技巧：(1) 始终清晰地写出你的求解过程；即使最终答案出错，也能获得步骤分。(2) 注意题目是否指定了答案的形式，例如"将答案精确到 2 位小数"或"以根式形式保留答案"。(3) 在选择题中，你通常可以通过代入法检验哪些选项满足方程。
10. Practice Questions | 练习题
To master quadratic equations, you must practise solving a variety of problems. Below are five practice questions that cover different skills. Attempt each one before checking the solution method.
要掌握二次方程，你必须练习各种类型的问题。下面有五道练习题，覆盖了不同的技能。请先尝试解答每一题，再查看解题方法。
| No. | Question | Suggested method |
| 1 | Solve x² - 7x + 12 = 0 | Factorisation → (x - 3)(x - 4) = 0 |
| 2 | Solve 2x² + 5x - 3 = 0 | Quadratic formula or grouping |
| 3 | Find the value of k if x² + kx + 9 = 0 has one repeated root | Set Δ = k² - 36 = 0 |
| 4 | Solve x² - 6x + 2 = 0, giving answers in surd form | Completing the square |
| 5 | A square has an area of x² + 4x + 4 = 25. Find x. | Recognise perfect square → (x + 2)² = 25Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
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