Solving Quadratic Equations & Graphs | 二次方程与图像求解

📚 Solving Quadratic Equations & Graphs | 二次方程与图像求解

Quadratic equations are one of the most heavily tested topics in IGCSE Mathematics. Every year, examiners report that a large portion of lost marks comes from careless algebra in factorisation, misuse of the quadratic formula, and failure to interpret the discriminant correctly. This article covers every core skill you need: solving by factorisation, completing the square, using the quadratic formula, analysing the discriminant, sketching graphs, and applying these ideas to realistic problems. By the end, you will have a complete revision reference that follows the IGCSE syllabus precisely.

二次方程是 IGCSE 数学中考查频率最高的主题之一。每年考官报告都指出，大量失分源自因式分解中的粗心代数错误、二次公式的误用，以及对判别式的错误解读。这篇文章涵盖你需要的全部核心技能：因式分解法、配方法、二次公式法、判别式分析、图像绘制，以及将这些知识应用于实际问题。学完之后，你将获得一份严格对应 IGCSE 考纲的完整复习资料。

1. What Is a Quadratic Equation? | 什么是二次方程

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The reason the coefficient a cannot be zero is simple: if a = 0, the x² term disappears and the equation becomes linear, no longer quadratic.

二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为实数，且 a ≠ 0。为什么 a 不能为零？原因很简单：如果 a = 0，x² 项消失，方程就变成了线性方程，不再是二次方程。

The highest power of the variable is 2, so a quadratic equation has at most two distinct real roots. These roots are also called solutions or zeros. Geometrically, they are the x-coordinates where the parabola y = ax² + bx + c crosses the x-axis.

变量的最高次数是 2，因此二次方程最多有两个不同的实数根。这些根也称为解或零点。从几何上看，它们就是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。

Example: 2x² − 5x + 3 = 0 中 a = 2, b = −5, c = 3

Always rearrange the equation into the form ax² + bx + c = 0 before applying any solving method. A common mistake is trying to factorise an expression that still has nonzero terms on both sides of the equals sign.

在应用任何求解方法之前，务必先把方程整理成 ax² + bx + c = 0 的形式。一个常见错误是：等号两边仍有非零项时就急于因式分解。

2. Solving by Factorisation | 因式分解法求解

Factorisation is the fastest method, but it only works when the quadratic has rational roots. The principle is based on the zero product property: if p × q = 0, then p = 0 or q = 0. This means once you factor the expression into two brackets, you can set each bracket to zero and solve two linear equations.

因式分解是最快捷的方法，但只适用于具有有理数根的情况。其原理基于零积性质：若 p × q = 0，则 p = 0 或 q = 0。这意味着一旦你把表达式分解为两个括号相乘，就可以令每个括号为零，并解出两个一次方程。

For a quadratic with a = 1, such as x² + bx + c, you look for two numbers that multiply to give c and add to give b. For example, x² − 7x + 12 = 0: the numbers −3 and −4 multiply to 12 and add to −7, so (x − 3)(x − 4) = 0, giving x = 3 or x = 4.

对于 a = 1 的二次式，例如 x² + bx + c，你需要找到两个数，使它们的乘积为 c、和为 b。例如 x² − 7x + 12 = 0：数 −3 和 −4 相乘得 12、相加得 −7，所以 (x − 3)(x − 4) = 0，得 x = 3 或 x = 4。

When a ≠ 1, use the factorisation by grouping method or the “ac” method. For 2x² + 7x + 3 = 0, multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7, which are 6 and 1. Then rewrite: 2x² + 6x + x + 3 = 0, and group: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0, so x = −½ or x = −3.

当 a ≠ 1 时，使用分组因式分解法或”ac 法”。对于 2x² + 7x + 3 = 0，先算 a × c：2 × 3 = 6。找到两个数，乘积为 6、和为 7，即 6 和 1。然后改写：2x² + 6x + x + 3 = 0，再分组：2x(x + 3) + 1(x + 3) = 0，得 (2x + 1)(x + 3) = 0，所以 x = −½ 或 x = −3。

Always check your brackets by expanding them back to the original expression.
如果二次式是完全平方，如 x² − 6x + 9 = (x − 3)²，则方程只有一个重根 x = 3。
If the expression cannot be factorised neatly, move to another method instead of guessing.

Examiners award method marks even if your final answer is slightly wrong, so always show the factorised form clearly before stating the solutions.

考官即使在你最终答案略有错误时也会给方法分，所以一定要清晰地写出因式分解的形式，再给出解。

3. Solving by Completing the Square | 配方法求解

Completing the square is a powerful algebraic technique that transforms a quadratic expression into the form a(x + p)² + q. For equations with a = 1, take half of the coefficient of x, square it, and add and subtract this value. For x² + 6x − 7 = 0, half of 6 is 3, and 3² = 9, so the expression becomes x² + 6x + 9 − 9 − 7 = (x + 3)² − 16.

配方法是一种强大的代数技巧，可以把二次式转化为 a(x + p)² + q 的形式。对于 a = 1 的方程，取 x 系数的一半，将其平方，然后加一项、减一项。对于 x² + 6x − 7 = 0，6 的一半是 3，3² = 9，所以表达式变为 x² + 6x + 9 − 9 − 7 = (x + 3)² − 16。

To solve, set the completed-square form to zero: (x + 3)² − 16 = 0, so (x + 3)² = 16, then x + 3 = ±4, giving x = 1 or x = −7. Remember the ± sign: when you take the square root of both sides, you must consider both the positive and negative roots.

求解时，令配方后的形式为零：(x + 3)² − 16 = 0，所以 (x + 3)² = 16，然后 x + 3 = ±4，得 x = 1 或 x = −7。记住 ± 号：两边开平方时，必须同时考虑正根和负根。

When a ≠ 1, factor out a first. For 2x² − 8x + 5 = 0, write 2(x² − 4x) + 5 = 0, complete the square inside the bracket: 2[(x − 2)² − 4] + 5 = 0, then simplify to 2(x − 2)² − 8 + 5 = 0, giving 2(x − 2)² − 3 = 0.

当 a ≠ 1 时，先把 a 提取出来。对于 2x² − 8x + 5 = 0，写成 2(x² − 4x) + 5 = 0，在括号内配方：2[(x − 2)² − 4] + 5 = 0，再化简为 2(x − 2)² − 8 + 5 = 0，即 2(x − 2)² − 3 = 0。

从 (x + p)² = k 中直接读出方程的顶点坐标 (−p, k − q)。
Completed-square form also helps you find the minimum or maximum value of a quadratic easily.
配方形式还能帮助你轻松求出二次式的最小值或最大值。

Completing the square is not just a solving technique; it is the bridge to understanding the vertex form of a quadratic graph.

配方法不仅仅是一种求解技巧；它还是理解二次函数图像顶点形式的桥梁。

4. Solving by the Quadratic Formula | 二次公式法求解

The quadratic formula is the most reliable method because it works for every quadratic equation, regardless of whether the roots are rational, irrational or even complex (though complex roots are beyond IGCSE). Derived from completing the square, the formula is:

二次公式是最可靠的方法，因为它对每一个二次方程都适用，无论根是有理数、无理数（复数根超出 IGCSE 范围）。它由配方法推导而来，公式为：

x = (−b ± √(b² − 4ac)) / 2a

To use it, identify a, b and c from the equation ax² + bx + c = 0, substitute them into the formula, and simplify carefully. Take the equation 3x² − 4x − 2 = 0. Here a = 3, b = −4, c = −2. Substituting gives:

使用该公式时，先从方程 ax² + bx + c = 0 中确定 a、b、c，代入公式并仔细化简。以方程 3x² − 4x − 2 = 0 为例，这里 a = 3，b = −4，c = −2。代入得：

x = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6

Since √40 = 2√10, the exact solutions are x = (2 ± √10) / 3. Leave surds in exact form unless the question asks for a decimal approximation. If a decimal is required, give three significant figures.

因为 √40 = 2√10，所以精确解为 x = (2 ± √10) / 3。除非题目要求小数近似值，否则把根式保留为精确形式。如果要求小数，请保留三位有效数字。

小心 b 为负数时的符号：−b 会变成正数。
Always write the formula down first; examiners often award marks for the correct substitution even if arithmetic fails later.
先写下公式再代入；即使后续计算出错，考官也常会给正确的代入步骤分。
使用计算器检验：把根代回原方程，看等号两边是否接近相等。

5. The Discriminant Δ | 判别式 Δ

The expression inside the square root of the quadratic formula, b² − 4ac, is called the discriminant and is denoted by Δ (Delta). Its sign tells us exactly how many real roots a quadratic equation has, without solving it at all.

二次公式根号内的表达式 b² − 4ac 被称为判别式，记作 Δ（Delta）。它的符号无需解方程就能直接告诉我们二次方程有多少个实数根。

Δ = b² − 4ac

Δ 的值	根的情况	图像含义
Δ > 0	两个不同实数根	与 x 轴有两个交点
Δ = 0	一个重根（两个相等实根）	与 x 轴相切于一点
Δ < 0	无实数根	与 x 轴无交点

For example, the equation x² − 5x + 6 = 0 has Δ = 25 − 24 = 1 > 0, so it has two distinct roots. The equation x² − 2x + 5 = 0 has Δ = 4 − 20 = −16 < 0, so it has no real roots — the parabola never crosses the x-axis.

例如，方程 x² − 5x + 6 = 0 的 Δ = 25 − 24 = 1 > 0，所以它有两个不同的实数根。方程 x² − 2x + 5 = 0 的 Δ = 4 − 20 = −16 < 0，所以它没有实数根——抛物线永远不会与 x 轴相交。

If a problem states that a quadratic equation has a repeated root or touches the x-axis, immediately set Δ = 0 and solve the resulting equation. This is a very common exam question pattern.

如果题目说一个二次方程有重根或图像与 x 轴相切，立即令 Δ = 0 并解出所得方程。这是非常常见的考题模式。

6. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, there are two elegant relationships you can memorise. These let you write a quadratic equation when you know its roots, without expanding brackets every time.

对于根为 α 和 β 的二次方程 ax² + bx + c = 0，有两个非常简洁的关系式。它们让你在已知根的情况下直接写出二次方程，而不必每次都展开括号。

α + β = −b/a, so αβ = c/a

For the equation 2x² − 8x + 3 = 0, the sum of the roots is −(−8)/2 = 4, and the product is 3/2. You can verify this: the roots are x = (8 ± √40)/4, and indeed their sum is 4 and product is 1.5.

对于方程 2x² − 8x + 3 = 0，根的和为 −(−8)/2 = 4，根的积为 3/2。你可以验证：根为 x = (8 ± √40)/4，它们之和确实为 4，之积确为 1.5。

A quadratic equation with roots α and β can be written as x² − (α + β)x + αβ = 0. This is especially useful for forming equations from given roots in problems, such as “find a quadratic whose roots are twice the roots of x² − 3x + 2 = 0”.

以 α 和 β 为根的二次方程可以写成 x² − (α + β)x + αβ = 0。这种方法在”求一个二次方程，使其根是 x² − 3x + 2 = 0 的根的两倍”这类问题中特别有用。

7. Sketching Quadratic Graphs | 二次函数图像绘制

The graph of y = ax² + bx + c is always a parabola. When a > 0, the parabola opens upwards (a “smile” shape) and the quadratic has a minimum point. When a < 0, it opens downwards (a "frown" shape) and has a maximum point. Understanding this from the sign of a is the first step of every sketch.

y = ax² + bx + c 的图像永远是抛物线。当 a > 0 时，抛物线开口向上（”微笑”形状），二次函数有最小值点。当 a < 0 时，抛物线开口向下（"悲伤"形状），有最大值点。从 a 的符号判断这一点，是所有绘图的第一个步骤。

To sketch a parabola accurately, find three key features: the y-intercept (set x = 0), the x-intercepts or roots (set y = 0 and solve), and the vertex. The x-coordinate of the vertex is given by x = −b/(2a); substitute this back into the equation to find the y-coordinate.

要准确绘制抛物线，需要找三个关键特征：y 轴截距（令 x = 0）、x 轴截距即根（令 y = 0 并求解）、以及顶点。顶点的横坐标由 x = −b/(2a) 给出；把它代回原方程即可求出纵坐标。

顶点坐标: ( −b/(2a) , c − b²/(4a) )

Alternatively, if the quadratic is written in completed-square form y = a(x + p)² + q, the vertex is simply (−p, q). This is why completing the square is often required in graph-sketching questions.

另一种方法：如果二次式写成配方法形式 y = a(x + p)² + q，顶点坐标直接就是 (−p, q)。这就是为什么作图题中常常要求配方法。

先标出与 y 轴的交点 (0, c)，因为这是最容易找的。
If Δ > 0, mark the two roots clearly with their exact values; if Δ = 0, the vertex touches the x-axis; if Δ < 0, the parabola floats entirely above or below the x-axis.
若 Δ < 0，抛物线整体位于 x 轴上方或下方，此时只需标出顶点和 y 截距。

8. Quadratic Inequalities | 二次不等式

Quadratic inequalities ask for the range of x values that make the quadratic expression positive or negative. The method is always the same: first solve the corresponding quadratic equation to find critical values, then test intervals on a number line.

二次不等式要求求出使二次表达式为正或为负的 x 的取值范围。方法始终一致：先解对应的二次方程求出临界值，然后在数轴上测试区间。

Take x² − x − 6 > 0. Factorise to (x − 3)(x + 2) > 0. The critical values are x = 3 and x = −2. These split the number line into three intervals: x < −2, −2 < x < 3, and x > 3. Test a value from each interval in the original inequality.

以 x² − x − 6 > 0 为例。因式分解得 (x − 3)(x + 2) > 0。临界值为 x = 3 和 x = −2。它们把数轴分为三个区间：x < −2、−2 < x < 3、x > 3。在每个区间取一个值代入原不等式测试。

Testing x = −3 gives (−6)(−1) = 6 > 0, so the first interval works. Testing x = 0 gives (−3)(2) = −6 < 0, so the middle interval fails. Testing x = 4 gives (1)(6) = 6 > 0, so the third interval works. Therefore the solution is x < −2 or x > 3.

测试 x = −3 得 (−6)(−1) = 6 > 0，所以第一个区间可行。测试 x = 0 得 (−3)(2) = −6 < 0，所以中间区间不可行。测试 x = 4 得 (1)(6) = 6 > 0，所以第三个区间可行。因此解为 x < −2 或 x > 3。

For a “less than” inequality such as x² − x − 6 < 0, the solution is the middle interval only: −2 < x < 3. A helpful memory rule: for a positive leading coefficient, ax² + bx + c > 0 is satisfied outside the roots, while ax² + bx + c < 0 is satisfied between the roots.

对于”小于”不等式，如 x² − x − 6 < 0，解只有中间区间：−2 < x < 3。一个有用的记忆规则：当首项系数为正时，ax² + bx + c > 0 的解在两根之外，而 ax² + bx + c < 0 的解在两根之间。

9. Word Problems with Quadratics | 二次方程应用题

Examiners love presenting quadratics in real-life contexts. Common scenarios include area problems, projectile motion, and number puzzles. The key is to translate the words into an equation carefully, and then state which solution is physically valid.

考官喜欢把二次方程放在实际生活情境中。常见场景包括面积问题、抛体运动和数字谜题。关键是把文字仔细地翻译成方程，然后说明哪一个解在物理上有效。

Example: A rectangular field has length 5 metres more than its width. Its area is 84 m². Find the dimensions. Let the width be x, so the length is x + 5. The area equation is x(x + 5) = 84, or x² + 5x − 84 = 0. Factorise to (x + 12)(x − 7) = 0, giving x = −12 or x = 7. Since a width cannot be negative, x = 7, and the length is 12 metres.

示例：一块长方形田地，长比宽多 5 米，面积为 84 平方米。求长和宽。设宽为 x，则长为 x + 5。面积方程为 x(x + 5) = 84，即 x² + 5x − 84 = 0。因式分解得 (x + 12)(x − 7) = 0，可得 x = −12 或 x = 7。由于宽度不能为负，x = 7，长为 12 米。

Define variables clearly at the start of your working.
开始时清晰地定义变量。
Always reject negative roots when the context involves length, time or quantity.
当情境涉及长度、时间或数量时，务必舍去负根。
Check your answer by substituting back into the original worded condition.

For projectile motion, the height h of a ball at time t often follows h = ut − ½gt². Setting h = 0 gives the time when the ball lands, and completing the square reveals the maximum height and the time at which it occurs.

对于抛体运动，球在时间 t 时的高度通常满足 h = ut − ½gt²。令 h = 0 可求出球落地的时间，而配方则能揭示最大高度及其发生时刻。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

After years of marking IGCSE papers, examiners consistently report the same errors. Knowing them in advance is your advantage. The first and most frequent error is attempting to factorise without first moving all terms to one side of the equation.

经过多年的 IGCSE 阅卷，考官们一致反馈同样的错误。提前了解它们就是你的优势。第一个也是最常见的错误是：在把所有项移到等号一侧之前就尝试因式分解。

常见错误	正确做法
x² = 9 直接写成 x = 3，漏掉负数根	x = ±3，始终写 ±
在二次公式中用错 −b 的符号	先圈出 a、b、c 再代入
配方时忘记在括号外乘回 a	2(x − 2)² − 3，展开检查
画抛物线时忽略对称轴	顶点 x = −b/(2a) 就是对折线

Time management is critical. In the IGCSE exam, a typical quadratic equation is worth 3 to 4 marks: one for rearrangement, one for the correct factorisation or substitution, and one or two for the final solutions. Never skip steps that earn method marks.

时间管理至关重要。在 IGCSE 考试中，一道典型的二次方程题值 3 到 4 分：一分给移项整理，一分给正确的因式分解或代入，一分或两分给最终解。永远不要跳过能拿方法分的步骤。

Finally, always leave answers in the form the question requests: exact surds, three significant figures, or two decimal places. Writing 1.414 when the question asks for √2 can cost you the final answer mark.

最后，务必将答案保留在题目要求的格式中：精确根式、三位有效数字或两位小数。题目要求 √2 时写成 1.414 可能会丢失最终答案分。

11. Worked Exam-Style Example | 考试风格例题精解

Let us work through a full exam-style question step by step to demonstrate every skill in one go. The question states: “The equation 2x² − 4x − 6 = 0 has roots p and q. Without solving, find the value of p² + q².”

让我们完整地解一道考试风格的综合题，一次性展示所有技能。题目为：”已知方程 2x² − 4x − 6 = 0 的根为 p 和 q。不解方程，求 p² + q² 的值。”

From the sum and product rules, p + q = −(−4)/2 = 2 and pq = −6/2 = −3. Now use the identity p² + q² = (p + q)² − 2pq. Substituting gives p² + q² = (2)² − 2(−3) = 4 + 6 = 10.

由根的和与积公式，p + q = −(−4)/2 = 2，pq = −6/2 = −3。利用恒等式 p² + q² = (p + q)² − 2pq。代入得 p² + q² = (2)² − 2(−3) = 4 + 6 = 10。

This problem cleverly tests algebra rather than arithmetic: you never have to solve for p or q at all. It is exactly the kind of “no solving required” question that distinguishes high-achieving students.

这道题巧妙地考查代数而不是算术：你根本不需要解出 p 和 q。这正是那种”无需解方程”的问题，能区分出高分段学生。

For practice, try similar combinations: p² − q² can be found by first computing (p − q)² = (p + q)² − 4pq, then using p² − q² = (p + q)(p − q). You should confirm that p² − q² = 4√5 for this equation.

练习时，尝试类似组合：p² − q² 可以先计算 (p − q)² = (p + q)² − 4pq，再利用 p² − q² = (p + q)(p − q)。你可以验证此方程中 p² − q² = 4√5。

12. Final Summary and Revision Checklist | 最终总结与复习清单

Quadratic equations reward practice. Every solving method connects to every other: the quadratic formula comes from completing the square, the discriminant emerges from the formula, and the sum and product of roots come from factorisation. Master the connections and you master the entire topic.

二次方程回报勤奋的练习。每种求解方法都相互关联：二次公式源自配方，判别式来自公式本身，根的和与积来自因式分解。掌握了这些联系，你就掌握了整个主题。

能快速将任意二次方程整理为标准形式 ax² + bx + c = 0。
能判断何时使用因式分解、配方或二次公式。
能够用 Δ 判断根的数量，并解决”重根”和”无实根”问题。
能画出抛物线草图：开口方向、y 截距、根、顶点。
能解二次不等式并在数轴上正确书写解集。
能将应用题转化为二次方程，并舍去不合情理的根。
能以要求的形式（精确值或有效数字）给出最终答案。

Before entering the exam hall, work through at least ten past-paper quadratic questions under timed conditions. Mark them strictly, study every mistake, and re-attempt any question where you lost marks. This active recall method is far more effective than simply re-reading notes.

进入考场前，请在计时条件下至少完成十道历年二次方程真题。严格批改，研究每一个错误，并重新尝试任何失分的题目。这种主动回忆法远比简单重读笔记更有效。

If the quadratic formula works for every equation, why practise factorisation at all? Because factorisation is faster, it gives you exact values without surds, and the IGCSE exam rewards efficiency. Choose the shortest reliable method for each question, and you will finish the paper with time to check your work.

如果二次公式对所有方程都适用，那为什么还要练习因式分解呢？因为因式分解更快，能在没有根式的情况下给出精确值，而且 IGCSE 考试奖励高效。为每道题选择最短的可靠方法，你就能留出时间检查整张试卷。

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Solving Quadratic Equations & Graphs | 二次方程与图像求解