Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in nearly every IGCSE Mathematics paper, either as direct solving questions or as part of graphs, sequences and word problems. A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. This revision guide covers the three core algebraic methods — factorisation, completing the square and the quadratic formula — together with the discriminant and common exam pitfalls.

二次方程几乎出现在每一份 IGCSE 数学试卷中，无论是直接求解，还是作为函数图像、数列和应用题的一部分。二次方程是指任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。本复习指南涵盖三种核心代数方法——因式分解、配方法和求根公式——以及判别式和常见考试陷阱。

1. The Standard Form | 标准形式

Before solving, always rewrite the equation as ax² + bx + c = 0, with all terms on one side and zero on the other. The coefficient a is the number in front of x², b is the number in front of x, and c is the constant term. For example, in 2x² − 7x + 3 = 0, we have a = 2, b = −7 and c = 3.

在求解之前，务必把方程写成 ax² + bx + c = 0 的形式，即将所有项移到一侧，另一侧为 0。系数 a 是 x² 前的数字，b 是 x 前的数字，c 是常数项。例如，在 2x² − 7x + 3 = 0 中，a = 2，b = −7，c = 3。

If the equation is given as 3x² + 5x = 2, you must first subtract 2 from both sides to obtain 3x² + 5x − 2 = 0. A quadratic equation always has degree 2, meaning the highest power of x is 2. If a = 0, the equation becomes linear, so the quadratic condition a ≠ 0 is essential.

如果方程给出为 3x² + 5x = 2，必须先将两边同时减去 2，得到 3x² + 5x − 2 = 0。二次方程的次数总是 2，即 x 的最高次幂为 2。如果 a = 0，方程就变成一次方程，因此 a ≠ 0 这一二次条件至关重要。

Ensure the right-hand side is zero before factorising.
Identify a, b and c carefully, including their signs.
If the equation has fractions, multiply through by the common denominator first.

因式分解前确保右侧为 0。
仔细辨认 a、b、c 及其正负号。
若方程含有分数，先乘以公分母化为整数。

2. Solving by Factorisation | 因式分解法

Factorisation is usually the quickest method when the quadratic has simple integer factors. For a monic quadratic x² + bx + c, look for two numbers that multiply to give c and add to give b. These two numbers become the constants in the factorised form (x + p)(x + q) = 0.

当二次方程具有简单整数因子时，因式分解通常是最快的方法。对于首项系数为 1 的二次式 x² + bx + c，找出两个数，使其乘积为 c、和为 b。这两个数成为分解形式 (x + p)(x + q) = 0 中的常数项。

Once factorised, set each bracket equal to zero and solve the two linear equations. This works because of the zero product property: if A × B = 0, then either A = 0 or B = 0. For example, x² − 5x + 6 = 0 factorises as (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

分解后，令每个括号等于 0，并解两个一次方程。这基于零积性质：若 A × B = 0，则 A = 0 或 B = 0。例如，x² − 5x + 6 = 0 分解为 (x − 2)(x − 3) = 0，解得 x = 2 或 x = 3。

For a non-monic quadratic such as 2x² − 7x + 3 = 0, you may use the product of a and c: find two numbers that multiply to give a × c = 6 and add to give b = −7. These numbers are −1 and −6. Split the middle term and factor by grouping to obtain (2x − 1)(x − 3) = 0, so x = ½ or x = 3.

对于非首项系数为 1 的二次式，如 2x² − 7x + 3 = 0，可使用 a 与 c 的乘积：找出两个数，使其乘积为 a × c = 6、和为 b = −7。这两个数是 −1 和 −6。拆开中项并分组分解，得到 (2x − 1)(x − 3) = 0，故 x = ½ 或 x = 3。

3. Difference of Two Squares | 平方差公式

A special factorisation pattern is the difference of two squares: x² − a² = (x − a)(x + a). This applies when there is no x term and the constant is a perfect square. For instance, x² − 49 = 0 becomes (x − 7)(x + 7) = 0, giving x = ±7.

一个特殊的因式分解模式是平方差公式：x² − a² = (x − a)(x + a)。当没有 x 项且常数为完全平方数时适用。例如，x² − 49 = 0 可化为 (x − 7)(x + 7) = 0，解得 x = ±7。

This pattern also works with coefficients: 4x² − 25 = 0 can be written as (2x)² − 5², so (2x − 5)(2x + 5) = 0, giving x = ±5/2. If the right-hand side is not zero, for example x² = 18, then rearranging gives x² − 18 = 0, which factorises as (x − 3√2)(x + 3√2) = 0.

这一模式同样适用于带系数的情形：4x² − 25 = 0 可写成 (2x)² − 5²，因此 (2x − 5)(2x + 5) = 0，解得 x = ±5/2。若右侧不为 0，例如 x² = 18，则移项得 x² − 18 = 0，分解为 (x − 3√2)(x + 3√2) = 0。

x² − a² = (x − a)(x + a)

Candidates often forget that x² + a² cannot be factorised using real numbers. Only a difference of two squares, never a sum, has this simple factorisation.

考生常忘记 x² + a² 在实数范围内不能分解。只有两个平方数的差，而不是和，才具有这种简单分解形式。

4. Completing the Square | 配方法

Completing the square rewrites x² + bx as (x + b/2)² − (b/2)². This method is especially useful when factorisation does not work, and it directly reveals the turning point of a quadratic graph. For x² + 6x, half of 6 is 3, so x² + 6x = (x + 3)² − 9.

配方法将 x² + bx 改写为 (x + b/2)² − (b/2)²。该方法在因式分解行不通时尤为实用，并能直接揭示二次函数图像的顶点。对于 x² + 6x，6 的一半是 3，所以 x² + 6x = (x + 3)² − 9。

To solve x² + 6x + 5 = 0 by completing the square, first rewrite as (x + 3)² − 9 + 5 = 0, so (x + 3)² − 4 = 0. Then (x + 3)² = 4, giving x + 3 = ±2, so x = −1 or x = −5.

用配方法解 x² + 6x + 5 = 0，先改写为 (x + 3)² − 9 + 5 = 0，即 (x + 3)² − 4 = 0。于是 (x + 3)² = 4，得 x + 3 = ±2，所以 x = −1 或 x = −5。

For a non-monic quadratic, factor out the coefficient of x² first: 2x² − 8x + 3 = 0 becomes 2(x² − 4x) + 3 = 0. Complete the square inside the bracket: 2[(x − 2)² − 4] + 3 = 0, so 2(x − 2)² − 8 + 3 = 0, hence 2(x − 2)² = 5.

对于首项系数不为 1 的二次式，先提出 x² 的系数：2x² − 8x + 3 = 0 变为 2(x² − 4x) + 3 = 0。在括号内配方：2[(x − 2)² − 4] + 3 = 0，即 2(x − 2)² − 8 + 3 = 0，因此 2(x − 2)² = 5。

The completed square form y = a(x − h)² + k gives the vertex (h, k) directly. In the example above, the vertex is (2, −5/2), and the axis of symmetry is x = 2.

配平方形式 y = a(x − h)² + k 直接给出顶点坐标 (h, k)。在上述例子中，顶点为 (2, −5/2)，对称轴为 x = 2。

5. The Quadratic Formula | 求根公式

When factorisation is difficult or impossible, use the quadratic formula. It works for any quadratic equation ax² + bx + c = 0 and gives the exact solutions:

当因式分解困难或无法分解时，使用求根公式。它适用于任意二次方程 ax² + bx + c = 0，并给出精确解：

x = (−b ± √(b² − 4ac)) / 2a

For 2x² − 7x + 3 = 0, substitute a = 2, b = −7 and c = 3. The discriminant part is b² − 4ac = (−7)² − 4 × 2 × 3 = 49 − 24 = 25. Then x = (7 ± 5) / 4, so x = 3 or x = ½. Notice that the ± sign produces two solutions.

对于 2x² − 7x + 3 = 0，代入 a = 2、b = −7、c = 3。判别式部分为 b² − 4ac = (−7)² − 4 × 2 × 3 = 49 − 24 = 25。于是 x = (7 ± 5) / 4，所以 x = 3 或 x = ½。注意 ± 号产生两个解。

Always write the formula in your working before substituting. A common error is using −b incorrectly when b is already negative. If b = −7, then −b = 7. Also ensure you divide the whole numerator by 2a, not just one term.

代入之前务必先写出公式。常见错误是当 b 本身为负数时误用 −b。若 b = −7，则 −b = 7。同时，要用整个分子除以 2a，而非只除其中一项。

6. The Discriminant | 判别式

The expression b² − 4ac is called the discriminant, often denoted by Δ. It determines the nature of the roots without solving the full equation. This is a favourite topic in IGCSE multiple-choice and short-answer questions.

表达式 b² − 4ac 称为判别式，通常记作 Δ。它无需解整道方程即可判断根的性质。这是 IGCSE 选择题和简答题中的热门考点。

If Δ > 0, there are two distinct real roots.
If Δ = 0, there is one repeated real root (a perfect square quadratic).
If Δ < 0, there are no real roots (the graph never crosses the x-axis).

若 Δ > 0，有两个不同的实数根。
若 Δ = 0，有一个重根（二次式为完全平方式）。
若 Δ < 0，没有实数根（图像不与 x 轴相交）。

For example, x² − 4x + 4 = 0 has Δ = 16 − 16 = 0, so it has exactly one root x = 2. Meanwhile, x² + 2x + 5 = 0 has Δ = 4 − 20 = −16, so it has no real roots. If a question asks for equal roots, set b² − 4ac = 0 and solve for the unknown coefficient.

例如，x² − 4x + 4 = 0 的 Δ = 16 − 16 = 0，因此只有一个根 x = 2。而 x² + 2x + 5 = 0 的 Δ = 4 − 20 = −16，因此没有实数根。若题目要求等根，令 b² − 4ac = 0 并解出未知系数即可。

7. Interpreting Quadratic Graphs | 二次函数图像

A quadratic function y = ax² + bx + c has a graph called a parabola. The sign of a tells the shape: if a > 0, the parabola opens upwards (a cup shape); if a < 0, it opens downwards (a cap shape). The roots found by solving ax² + bx + c = 0 are exactly the x-intercepts of the graph.

二次函数 y = ax² + bx + c 的图像称为抛物线。a 的符号决定开口方向：若 a > 0，抛物线开口向上（杯形）；若 a < 0，开口向下（帽形）。解 ax² + bx + c = 0 所求得的根正是图像与 x 轴的交点。

The y-intercept is simply c, found by setting x = 0. The axis of symmetry lies exactly halfway between the two roots, or is given by the formula x = −b / 2a. The vertex, or turning point, lies on this axis; its y-coordinate is found by substituting the x-coordinate back into the equation.

y 截距就是 c，令 x = 0 即得。对称轴恰好位于两个根的正中间，也可由公式 x = −b / 2a 给出。顶点（即转向点）位于该对称轴上；将其 x 坐标代回方程即可求得 y 坐标。

If the discriminant is negative, the parabola does not cross the x-axis at all. The vertex then lies entirely above or below the axis, telling you whether the function is always positive or always negative. Sketching the graph after solving the equation is an excellent way to check your answers.

若判别式为负，抛物线完全不与 x 轴相交。此时顶点完全位于 x 轴上方或下方，由此可判断函数恒为正还是恒为负。解完方程后画出草图是检验答案的绝佳方法。

8. Word Problems and Applications | 应用题

Quadratic equations often arise in geometry and number problems. A classic IGCSE example: a rectangle has length (x + 3) cm and width x cm, with an area of 40 cm². The area equation is x(x + 3) = 40, which expands to x² + 3x − 40 = 0.

二次方程常出现在几何和数字问题中。一个经典 IGCSE 例题：长方形的长为 (x + 3) cm，宽为 x cm，面积为 40 cm²。面积方程为 x(x + 3) = 40，展开得 x² + 3x − 40 = 0。

Factorising gives (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since a length cannot be negative, reject x = −8 and conclude x = 5. The rectangle is therefore 5 cm by 8 cm. Always interpret solutions in the context of the problem and discard impossible ones.

因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于长度不能为负，舍去 x = −8，得出 x = 5。因此长方形为 5 cm × 8 cm。务必结合题目背景解释解的意义，并舍弃不合理的解。

When building a quadratic equation from a word problem, define the unknown clearly, translate each sentence into algebra step by step, and check both solutions against the original conditions. This systematic approach earns full method marks even if a small arithmetic slip occurs.

从应用题建立二次方程时，要明确设定未知数，逐步将每个句子转化为代数式，并用原题条件检验两个解。这种系统化方法即使出现小的运算失误，也能获得完整的方法分。

9. Common Mistakes | 常见错误

Many marks are lost to avoidable errors. The most frequent mistake is factorising before moving all terms to one side, such as trying to factorise x² − 5x = −6 directly instead of first writing x² − 5x + 6 = 0. The zero product property only works when one side is exactly zero.

许多分数都因可避免的错误而丢失。最常见的错误是在把所有项移到一侧之前就进行因式分解，例如直接分解 x² − 5x = −6，而非先写成 x² − 5x + 6 = 0。零积性质只在一边恰好为 0 时才能使用。

Forgetting the ± sign when taking square roots: x² = 9 gives x = ±3.
Sign errors when substituting negatives into the quadratic formula.
Dividing by x when x could be zero, losing a root in equations like x² = 3x.
Writing only one solution when two exist.
Confusing the formula for the axis of symmetry x = −b / 2a with the quadratic formula.

开平方时忘记 ± 号：x² = 9 应得 x = ±3。
将负数代入求根公式时出现符号错误。
两边同除以 x（x 可能为 0），在 x² = 3x 这类方程中丢失一个根。
存在两个解时只写出一个。
混淆对称轴公式 x = −b / 2a 与求根公式。

If x² = 3x, never divide both sides by x. Instead rearrange to x² − 3x = 0, factorise as x(x − 3) = 0, and obtain both roots x = 0 and x = 3. Dividing by x loses the root x = 0 completely.

若遇到 x² = 3x，切勿两边同除以 x。应移项得 x² − 3x = 0，分解为 x(x − 3) = 0，从而得到两个根 x = 0 和 x = 3。若除以 x，就会完全丢失根 x = 0。

10. Exam Tips | 应考技巧

In the examination, decide which method to use quickly. Try factorisation first, especially when the coefficient of x² is 1 or the question suggests it. If factorisation is not obvious within a few seconds, switch to the quadratic formula, which always works. Completing the square is best when the question asks for the turning point or exact surd forms.

考试时请快速决定使用哪种方法。先尝试因式分解，尤其是 x² 系数为 1 或题目有所提示时。若几秒内看不出分解方式，就改用求根公式，它总是有效。当题目要求顶点坐标或精确根式形式时，配方法最为合适。

Show all working: method marks are awarded even for incorrect final answers.
Check your roots by substituting them back into the original equation.
For surd answers, leave results like (3 ± √5) / 2 in exact form unless a decimal is requested.
Read whether the question asks for roots, intercepts, the vertex, or a sketch.
Use the discriminant to quickly verify whether a stated equation has real roots.

写出全部过程：即使最终答案有误，也会获得方法分。
将根代回原方程检验。
根式答案应保留 (3 ± √5) / 2 之类的精确形式，除非题目要求小数。
看清题目要求的是根、截距、顶点还是草图。
用判别式快速验证给定方程是否有实数根。

Finally, remember that solving a quadratic equation is only one part of a larger toolkit. Knowing when each method is most efficient, and how the roots connect to the graph, will save time and boost your confidence across the entire paper.

最后请记住，求解二次方程只是整个数学工具箱中的一部分。了解每种方法最适用的场景，以及根与图像之间的联系，将为你节省时间，并在整张试卷中提升自信。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Equations | 解二次方程