Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear in nearly every IGCSE Mathematics paper, often in both the non-calculator and calculator sections. Mastering them is essential because they also form the foundation for inequalities, functions and graph sketching tested at A Level. This revision guide covers the standard form, four reliable solution methods, the discriminant and the graphical meaning of roots, with worked examples in the style of recent exam questions.

一元二次方程几乎出现在每张 IGCSE 数学试卷中，既出现在非计算器部分，也出现在计算器部分。掌握这一专题至关重要，因为它同时也是 A Level 中不等式、函数与图象问题的基础。本复习指南将系统讲解标准形式、四种可靠的解法、判别式以及根的图象意义，并配有贴近近年真题风格的例题。

1. The Standard Form | 标准形式

Every quadratic equation can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The coefficient a is the number in front of x², b is the number in front of x, and c is the constant term. If a = 0, the equation becomes linear and many of the methods below no longer apply.

任何一个一元二次方程都可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。系数 a 是 x² 前的数，b 是 x 前的数，c 是常数项。若 a = 0，方程就会变成一次方程，下面很多方法都不再适用。

ax² + bx + c = 0 (a ≠ 0)

Consider the equation 2x² − 7x + 3 = 0. Here a = 2, b = −7 and c = 3. A common mistake is to read b as 7; the sign belongs to the coefficient, so b is negative. Always write the equation with the x² term, the x term and the constant in the same order before identifying coefficients.

以方程 2x² − 7x + 3 = 0 为例，此时 a = 2，b = −7，c = 3。一个常见错误是把 b 当作 7；正负号属于系数本身，所以 b 是负数。在识别系数前，务必先把方程按 x² 项、x 项和常数项的顺序排列整齐。

2. Solving by Factorisation | 因式分解法

When the quadratic expression factorises neatly, this is the fastest method. First rearrange the equation so that one side is zero, then factorise the expression and apply the zero product property: if two factors multiply to zero, at least one of them must be zero.

当二次多项式可以顺利分解时，因式分解法是最快的方法。先把方程整理成一边为零，再对表达式进行因式分解，并运用零积性质：若两个因式相乘为零，则至少有一个因式为零。

Solve x² − 5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5, namely −2 and −3.

解方程 x² − 5x + 6 = 0。我们需要找到两个数，它们相乘得 6、相加得 −5，这两个数就是 −2 和 −3。

(x − 2)(x − 3) = 0 → x = 2 or x = 3

When the coefficient of x² is not 1, factorisation is still possible but requires more care. For 2x² − 5x − 3 = 0, the factor pair is (2x + 1)(x − 3), giving x = −½ or x = 3. Always expand your brackets to check the original expression.

当 x² 的系数不为 1 时，因式分解仍然可行，但需要更加细心。对于 2x² − 5x − 3 = 0，其因式分解为 (2x + 1)(x − 3)，于是 x = −½ 或 x = 3。务必把括号展开检验是否还原为原表达式。

A special case is the difference of two squares: x² − 16 = 0 becomes (x − 4)(x + 4) = 0, so x = 4 or x = −4. This pattern also works with non-square coefficients, such as x² − 7 = (x − √7)(x + √7).

一个特殊情形是平方差公式：x² − 16 = 0 可化为 (x − 4)(x + 4) = 0，因此 x = 4 或 x = −4。这一模式对非平方数系数同样适用，例如 x² − 7 = (x − √7)(x + √7)。

3. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q, which makes solving straightforward and also reveals the turning point of the graph. For x² + px + q, the identity is (x + p/2)² − (p/2)² + q.

配方法把二次式改写为 (x + p)² + q 的形式，这不仅让求解变得直接，还能揭示图象的顶点。对于 x² + px + q，恒等式为 (x + p/2)² − (p/2)² + q。

Solve x² + 6x + 5 = 0 by completing the square. Take half of 6, which is 3, and write (x + 3)². Since (x + 3)² = x² + 6x + 9, we must subtract 9 to compensate.

用配方法解 x² + 6x + 5 = 0。取 6 的一半得到 3，写成 (x + 3)²。因为 (x + 3)² = x² + 6x + 9，所以要减 9 以保持等价。

(x + 3)² − 9 + 5 = 0 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5

Notice that completing the square naturally produces the ± sign. This method is especially useful when the equation has irrational roots, and it is the basis of the quadratic formula in Section 4. If the coefficient of x² is not 1, factor a out of the first two terms before completing the square.

注意配方法会自然产生 ± 号。当方程出现无理根时，这一方法尤其有用，它也是第 4 节求根公式的基础。若 x² 的系数不为 1，先把 a 从前两项中提出，再进行配方。

4. The Quadratic Formula | 求根公式

The quadratic formula solves every quadratic equation, including those that cannot be factorised. In the IGCSE exam, the formula is given in the formula booklet, but you must know how to substitute a, b and c correctly and simplify the result.

求根公式可以求解所有一元二次方程，包括那些无法因式分解的方程。在 IGCSE 考试中，公式会印在公式册里，但你必须知道如何正确代入 a、b、c 并化简结果。

x = (−b ± √(b² − 4ac)) ÷ (2a)

Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4 and c = −3. Substitute into the formula.

解 2x² − 4x − 3 = 0。此时 a = 2，b = −4，c = −3。代入公式：

x = (4 ± √(16 + 24)) ÷ 4 = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = 1 ± (√10)/2

Using √10 ≈ 3.162, we obtain x ≈ 2.58 or x ≈ −0.58 to two decimal places. In a non-calculator paper, leaving the answer as 1 ± (√10)/2 is preferred; in a calculator paper, give both decimal values.

取 √10 ≈ 3.162，可得 x ≈ 2.58 或 x ≈ −0.58（精确到两位小数）。在非计算器试卷中，保留答案 1 ± (√10)/2 更为妥当；在计算器试卷中，则应给出两个小数形式的解。

5. The Discriminant | 判别式

The discriminant is the expression Δ = b² − 4ac under the square root in the formula. Its sign tells us how many real roots the equation has, without solving it fully.

判别式是求根公式中被开平方的部分 Δ = b² − 4ac。它的符号可以判断方程有多少个实数根，而不必完整解方程。

Discriminant Δ	Nature of roots	Graph meaning
Δ > 0	Two distinct real roots	Crosses the x-axis twice
Δ = 0	One repeated real root	Touches the x-axis once
Δ < 0	No real roots	Does not meet the x-axis

For x² + 2x + 5 = 0, the discriminant is Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16. Since Δ < 0, this equation has no real roots. The expression √(−16) is not a real number, so the quadratic formula also confirms the same conclusion.

对于 x² + 2x + 5 = 0，判别式为 Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16。因为 Δ < 0，该方程没有实数根。√(−16) 不是实数，因此用求根公式也能得到同样的结论。

6. Quadratic Graphs and Roots | 二次函数图象与根

The roots of ax² + bx + c = 0 are exactly the x-coordinates where the graph y = ax² + bx + c intersects the x-axis. The graph of a quadratic is a parabola: U-shaped when a > 0 and n-shaped when a < 0.

方程 ax² + bx + c = 0 的根正是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。二次函数图象是抛物线：当 a > 0 时开口向上，呈 U 形；当 a < 0 时开口向下，呈 n 形。

The axis of symmetry is the vertical line x = −b/(2a), and the turning point lies on this line. Substituting this x-value into the equation gives the minimum or maximum value of the quadratic. Completing the square also reveals the vertex directly: y = (x − h)² + k has vertex (h, k).

对称轴是竖直直线 x = −b/(2a)，顶点就在这条直线上。将这个 x 值代入方程，即可得到二次函数的最小值或最大值。配方后也能直接看出顶点：y = (x − h)² + k 的顶点为 (h, k)。

For example, the graph y = x² − 4x + 3 has axis x = 2 and roots x = 1 and x = 3, because (x − 1)(x − 3) = 0. Substituting x = 2 gives y = −1, so the turning point is (2, −1). Sketch this parabola to check that it opens upwards and crosses the x-axis at 1 and 3.

例如，抛物线 y = x² − 4x + 3 的对称轴为 x = 2，根为 x = 1 和 x = 3，因为 (x − 1)(x − 3) = 0。代入 x = 2 得 y = −1，所以顶点为 (2, −1)。画出这条抛物线即可验证它开口向上，并与 x 轴相交于 1 和 3。

7. Word Problems Involving Quadratics | 二次方程应用题

Many IGCSE questions present a practical situation that leads to a quadratic equation. The method is the same as always: define the unknown, form the equation, solve it, and then check which answers make sense in the context.

许多 IGCSE 题目给出一个实际问题，最终会归结为一元二次方程。解法步骤始终相同：设未知数、列方程、解方程，再检查哪些答案在具体情境中合理。

Example: A rectangle has length (x + 4) cm and width x cm. Its area is 21 cm². Find the dimensions of the rectangle.

例题：一个矩形的长为 (x + 4) cm，宽为 x cm，面积为 21 cm²。求该矩形的尺寸。

x(x + 4) = 21 → x² + 4x − 21 = 0 → (x + 7)(x − 3) = 0

This gives x = −7 or x = 3. A length cannot be negative, so we reject x = −7. Therefore the width is 3 cm and the length is 7 cm. Always write a clear final sentence that answers the original question.

于是 x = −7 或 x = 3。长度不可能为负数，所以舍去 x = −7。因此宽为 3 cm，长为 7 cm。务必写出明确的答句来回应原题的问题。

8. Common Exam Pitfalls | 常见考试陷阱

Factorising before moving all terms to one side: always solve ax² + bx + c = 0, not ax² + bx = −c.
Losing the negative sign when b or c is negative; write coefficients in brackets during substitution.
Forgetting the ± sign when taking square roots, so only half of the solutions are reported.
Rejecting a negative root without justification when the context uses length, time or distance.
Applying the formula with a = 0; this is not a quadratic equation and the denominator 2a becomes zero.
Rounding too early in calculator questions; keep exact values until the final step.

中文提示：因式分解前没有把方程移项成 ax² + bx + c = 0；当 b 或 c 为负数时丢失负号；开平方时忘记写 ± 号，只写了一半解；在涉及长度、时间或距离的问题中，无故舍去负根；a = 0 时仍套用求根公式导致分母为零；计算器题中过早四舍五入，应保留精确值到最后一步。

9. Methods Compared | 方法对比与选择

How do you know which method to use? Factorisation is quickest when the roots are integers or simple fractions. Completing the square is helpful when asked to find the turning point or when a is small. The quadratic formula always works and is the safest choice when factorisation is not obvious.

如何选择方法？当根是整数或简单分数时，因式分解最快。当题目要求找顶点或 a 较小时，配方法很有帮助。求根公式永远有效，当因式分解不明显时是最稳妥的选择。

Method	Best used when	Example
Factorisation	Roots are integers or simple fractions	x² − 5x + 6 = 0
Completing the square	Need the vertex or irrational roots	x² + 6x + 5 = 0
Quadratic formula	Any quadratic; use when factorisation is hard	2x² − 4x − 3 = 0

10. Revision Checklist | 复习清单

I can rearrange any quadratic equation into standard form ax² + bx + c = 0 and identify a, b and c with their signs.
I can solve by factorisation, including the difference of two squares and cases with a > 1.
I can complete the square for x² + px + q and solve equations of the form (x + p)² = k.
I can substitute into the quadratic formula and simplify answers with surds.
I can use the discriminant to state the number and nature of the roots.
I can relate roots to the x-intercepts of the parabola and find the axis of symmetry.
I can set up a quadratic equation from a word problem, solve it and reject invalid solutions.

中文清单：能把方程化为标准形式并识别带符号的 a、b、c；能使用因式分解法，包括平方差公式和 a > 1 的情形；能对 x² + px + q 配方并解 (x + p)² = k 形式的方程；能代入求根公式并用根式化简答案；能利用判别式判断根的个数与类型；能建立根与抛物线 x 轴交点之间的关系并求对称轴；能根据应用题列方程、求解并舍去不合理的解。

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Solving Quadratic Equations | 解一元二次方程