📚 Solving Quadratic Equations: IGCSE Core and Extended Guide | 解二次方程：IGCSE 核心与扩展指南
Quadratic equations are a central topic in the IGCSE Mathematics syllabus, appearing in both Core and Extended papers. This guide explains the standard form, the main solution methods, the discriminant, graphical interpretation, and common exam techniques.
二次方程是 IGCSE 数学课程大纲中的核心主题，在核心和扩展试卷中都会出现。本指南将解释标准形式、主要求解方法、判别式、图形解释以及常见考试技巧。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree two. The highest power of the variable x is x².
二次方程是一个二次多项式方程。变量 x 的最高次数是 x²。
ax² + bx + c = 0, where a ≠ 0
condition a ≠ 0 is essential because if a = 0 the equation becomes linear, not quadratic.
在这个标准形式中，a、b 和 c 是常数。a ≠ 0 这一条件非常重要，因为如果 a = 0，方程就变成一次方程，而不是二次方程。
- 2x² + 3x − 5 = 0 has a = 2, b = 3, c = −5.
- 2x² + 3x − 5 = 0 中 a = 2，b = 3，c = −5。
- x² − 4 = 0 has a = 1, b = 0, c = −4.
- x² − 4 = 0 中 a = 1，b = 0，c = −4。
- −x² + 6x = 0 has a = −1, b = 6, c = 0.
- −x² + 6x = 0 中 a = −1，b = 6，c = 0。
You must be able to rearrange an equation into this standard form before choosing a solution method.
在选择求解方法之前，你必须能够将方程整理成这种标准形式。
2. Solving by Factorising | 因式分解法求解
Factorising is usually the fastest method when the quadratic expression can be written as a product of two linear brackets.
当二次表达式可以写成两个一次括号的乘积时，因式分解通常是最快的方法。
or two numbers that multiply to give +6 and add to give +5. The numbers are +2 and +3, so:
求解 x² + 5x + 6 = 0 时，找出两个数，它们的乘积为 +6，和为 +5。这两个数是 +2 和 +3，因此：
(x + 2)(x + 3) = 0
sing the zero product property, either x + 2 = 0 or x + 3 = 0. Therefore x = −2 or x = −3.
根据零乘积性质，要么 x + 2 = 0，要么 x + 3 = 0。因此 x = −2 或 x = −3。
carefully. Splitting the middle term gives 2x² + 6x + x + 3 = 0, then factor by grouping:
当 x² 的系数不是 1 时，例如 2x² + 7x + 3 = 0，你需要更仔细地进行因式分解。拆项得到 2x² + 6x + x + 3 = 0，然后分组分解：
2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0
The solutions are x = −½ or x = −3.
解为 x = −½ 或 x = −3。
Always check your factorisation by expanding the brackets back to the original expression.
一定要通过展开括号回到原表达式来检查因式分解是否正确。
3. Completing the Square | 配方法求解
Completing the square is useful when a quadratic does not factorise easily, and it is essential for understanding the vertex form of a quadratic function.
当二次方程不容易因式分解时，配方法非常有用，而且它是理解二次函数顶点式的基础。
For x² + 6x + 2 = 0, first move the constant term: x² + 6x = −2.
对于 x² + 6x + 2 = 0，首先移项：x² + 6x = −2。
of the coefficient of x, which is 6 ÷ 2 = 3, then square it: 3² = 9. Add 9 to both sides:
取 x 系数的一半，即 6 ÷ 2 = 3，然后平方：3² = 9。两边同时加上 9：
x² + 6x + 9 = −2 + 9
(x + 3)² = 7
Take square roots of both sides, remembering the ± sign:
两边开平方，记得加上 ± 号：
x + 3 = ±√7
Therefore x = −3 ± √7. These are the exact solutions.
因此 x = −3 ± √7。这些是精确解。
y term by that coefficient first. For example, 2x² + 8x + 5 = 0 becomes x² + 4x + 5/2 = 0.
如果 x² 的系数不是 1，首先用该系数除以每一项。例如，2x² + 8x + 5 = 0 变为 x² + 4x + 5/2 = 0。
4. The Quadratic Formula | 二次公式法
uadratic equation ax² + bx + c = 0. It is especially useful when factorising is difficult.
二次公式可以给出任何二次方程 ax² + bx + c = 0 的解。当因式分解困难时，它尤其有用。
x = (−b ± √b² − 4ac) ÷ 2a
To solve 3x² − 4x − 2 = 0, identify a = 3, b = −4, c = −2. Substitute into the formula:
求解 3x² − 4x − 2 = 0 时，确定 a = 3，b = −4，c = −2。代入公式：
x = (−(−4) ± √((−4)² − 4(3)(−2))) ÷ (2 × 3)
x = (4 ± √16 + 24) ÷ 6
x = (4 ± √40) ÷ 6
Simplify √40 = √4 × 10 = 2√10, so x = (4 ± 2√10) ÷ 6 = (2 ± √10) ÷ 3.
化简 √40 = √4 × 10 = 2√10，因此 x = (4 ± 2√10) ÷ 6 = (2 ± √10) ÷ 3。
You may leave exact answers in surd form unless the question asks for decimals.
除非题目要求用小数表示，否则你可以将精确答案保留为根式形式。
5. The Discriminant and Nature of Roots | 判别式与根的性质
scriminant is the expression under the square root in the quadratic formula: D = b² − 4ac.
判别式是二次公式中平方根下的表达式：D = b² − 4ac。
It tells you how many real solutions a quadratic equation has without actually solving it.
它可以在不实际求解方程的情况下，告诉你二次方程有多少个实数解。
- If b² − 4ac > 0, there are two distinct real roots.
- 如果 b² − 4ac > 0，则有两个不同的实数根。
- If b² − 4ac = 0, there is exactly one repeated real root.
- 如果 b² − 4ac = 0，则恰好有一个重根。
- If b² − 4ac < 0, there are no real roots, only two complex roots.
- 如果 b² − 4ac < 0，则没有实数根，只有两个复数根。
+ 9 = 0 has discriminant (−6)² − 4(1)(9) = 36 − 36 = 0, so it has one repeated root x = 3.
例如，x² − 6x + 9 = 0 的判别式为 (−6)² − 4(1)(9) = 36 − 36 = 0，因此它有一个重根 x = 3。
n equation has real roots. You should set b² − 4ac ≥ 0 and solve the resulting inequality.
IGCSE 题目可能会要求你找出使方程有实数根的 k 的取值范围。你应设 b² − 4ac ≥ 0，然后求解得到的不等式。
6. Graphical Interpretation of Quadratic Equations | 二次方程的图形解释
he solutions of ax² + bx + c = 0 are the x-coordinates where the graph crosses the x-axis.
y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的解是图像与 x 轴交点的 x 坐标。
If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
如果 a > 0，抛物线开口向上并有最小值点。如果 a < 0，抛物线开口向下并有最大值点。
The number of x-intercepts corresponds exactly to the discriminant: two intercepts, one intercept, or no intercepts.
x 轴截距的个数与判别式完全对应：两个截距、一个截距或没有截距。
he x-intercepts. Alternatively, you may be given the graph and asked to solve the equation.
要用图形方法求解二次方程，画出 y = ax² + bx + c 的图像并读取 x 轴截距。或者，题目可能给出图像并要求你求解方程。
l for identifying the turning point (h, k). Completing the square converts a quadratic into this form.
顶点式 y = a(x − h)² + k 有助于识别转折点 (h, k)。配方法可以将二次函数化为这种形式。
7. Solving Quadratic Inequalities | 求解二次不等式
A quadratic inequality such as x² − 4x + 3 < 0 can be solved by first finding the roots of the corresponding quadratic equation.
像 x² − 4x + 3 < 0 这样的二次不等式，可以通过先求出对应二次方程的根来求解。
) < 0. The roots are x = 1 and x = 3; these split the number line into three intervals.
因式分解得到 (x − 1)(x − 3) < 0。根为 x = 1 和 x = 3；它们将数轴分为三个区间。
1)(2 − 3) = −1, which is < 0. For x = 4, it is (4 − 1)(4 − 3) = 3, which is not < 0.
在每个区间中选一个值代入原不等式检验。x = 0 时，表达式为 (0 − 1)(0 − 3) = 3，不小于 0。x = 2 时，为 (2 − 1)(2 − 3) = −1，小于 0。x = 4 时，为 (4 − 1)(4 − 3) = 3，不小于 0。
Therefore the solution is 1 < x < 3.
因此解为 1 < x < 3。
the endpoints in the solution set. Always sketch a graph or number line to support your answer.
如果不等式使用 ≤ 或 ≥，则在解集中包含端点。始终画出图像或数轴来支持你的答案。
8. Word Problems and Modelling | 文字题与建模
Many IGCSE questions ask you to form a quadratic equation from a real-life situation, then solve it and interpret the answer in context.
许多 IGCSE 题目要求你根据实际情况建立二次方程，然后求解并在情境中解释答案。
A typical example: 'The product of two consecutive positive integers is 42. Find the integers.' Let the smaller integer be n, so the next is n + 1.
一个典型例子：“两个连续正整数的乘积是 42。求这两个整数。”设较小的整数为 n，则下一个为 n + 1。
n(n + 1) = 42
are n = −7 or n = 6. Since the integers are positive, n = 6, so the integers are 6 and 7.
展开并整理：n² + n − 42 = 0。因式分解：(n + 7)(n − 6) = 0。解为 n = −7 或 n = 6。由于整数是正数，n = 6，因此这两个整数是 6 和 7。
In geometry problems, the Pythagorean theorem often produces a quadratic equation. For example, if the sides of a right-angled triangle are x, x + 1 and x + 2, then:
在几何问题中，勾股定理常常会产生二次方程。例如，如果直角三角形的三边为 x、x + 1 和 x + 2，则：
x² + (x + 1)² = (x + 2)²
Solve this by expanding, simplifying and factorising. Always reject solutions that are not valid in the given context, such as negative lengths.
通过展开、化简和因式分解来求解。始终舍弃在给定情境中无效的解，例如负长度。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
3x = 4 cannot be factorised as x(x + 3) = 4; you must first rewrite it as x² + 3x − 4 = 0.
一个常见错误是在因式分解前忘记将方程设为等于零。例如，x² + 3x = 4 不能因式分解为 x(x + 3) = 4；你必须先将其改写为 x² + 3x − 4 = 0。
because x = 0 is a valid solution. Always factorise: x(x − 5) = 0, giving x = 0 or x = 5.
另一个错误是在求解像 x² − 5x = 0 这样的方程时两边同时除以 x，因为 x = 0 是一个有效解。应始终进行因式分解：x(x − 5) = 0，得到 x = 0 或 x = 5。
Be careful with signs when substituting into the quadratic formula. If b is negative, −b becomes positive.
代入二次公式时要注意符号。如果 b 是负数，−b 就变成正数。
When completing the square, remember to add the same value to both sides, and if a ≠ 1, divide first.
配方法时，记得在等式两边同时加上相同的值，如果 a ≠ 1，则先除以 a。
Finally, always give exact answers unless the question asks for rounding. Surds such as √7 should be left in exact form in IGCSE non-calculator papers.
最后，除非题目要求四舍五入，否则始终给出精确答案。在 IGCSE 非计算器试卷中，像 √7 这样的根式应保留为精确形式。
10. Worked Examination Example | 考试真题示例
the equation 2x² − 3x − 5 = 0 by factorising, and then verify using the quadratic formula.
让我们通过因式分解法求解方程 2x² − 3x − 5 = 0，然后用二次公式进行验证。
wo numbers that multiply to −10 and add to −3. These are −5 and +2. Split the middle term:
将 a 乘以 c：2 × (−5) = −10。我们需要两个数，乘积为 −10，和为 −3。这两个数是 −5 和 +2。拆分中间项：
2x² − 5x + 2x − 5 = 0
Factor by grouping: x(2x − 5) + 1(2x − 5) = 0, so (2x − 5)(x + 1) = 0.
分组分解：x(2x − 5) + 1(2x − 5) = 0，因此 (2x − 5)(x + 1) = 0。
The solutions are x = 5/2 or x = −1.
解为 x = 5/2 或 x = −1。
Using the quadratic formula with a = 2, b = −3, c = −5 gives x = (3 ± √9 + 40) ÷ 4 = (3 ± √49) ÷ 4 = (3 ± 7) ÷ 4, which also gives x = 10/4 = 5/2 or x = −4/4 = −1.
使用二次公式，a = 2，b = −3，c = −5，得到 x = (3 ± √9 + 40) ÷ 4 = (3 ± √49) ÷ 4 = (3 ± 7) ÷ 4，同样得到 x = 10/4 = 5/2 或 x = −4/4 = −1。
The two methods agree. Use the method that is most efficient for the given coefficients.
两种方法结果一致。对给定系数使用最高效的方法。
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