📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are a central topic in the IGCSE Mathematics syllabus. You are expected to move fluently between factorised form, expanded form and the graph of a quadratic function. This revision guide covers the main algebraic methods: factorising, completing the square, using the quadratic formula, and interpreting the discriminant. Each section pairs English explanations with Chinese translations so you can check understanding quickly.
二次方程是 IGCSE 数学课程的核心内容。你需要能够在因式分解形式、展开形式和二次函数图像之间灵活转换。本复习指南涵盖主要代数方法：因式分解法、配方法、使用求根公式，以及判别式的解释。每一部分都提供英文解释与中文翻译，方便你快速核对理解。
1. What is a Quadratic Equation? | 什么是二次方程？
a, b and c are constants and a ≠ 0. The highest power of x is 2, which is why it is called quadratic. If a were zero, the equation would become linear, not quadratic.
一个一元二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数且 a ≠ 0。x 的最高次数是 2，因此称为二次方程。如果 a 为零，方程就变成一次方程，而不是二次方程。
here is no x term: here b = 0. The coefficient a must not be zero, but b or c may be zero.
例如，2x² + 5x - 3 = 0 是二次方程，因为 x 的最高次数是 2。方程 x² - 4 = 0 也是二次方程，尽管没有 x 项：此时 b = 0。系数 a 不能为零，但 b 或 c 可以为零。
2. Expanding Binomial Products | 展开二项式乘积
Many quadratic expressions come from multiplying two linear brackets. To expand (x + p)(x + q), use the distributive law: multiply each term in the first bracket by each term in the second bracket, then collect like terms.
许多二次表达式来自两个一次因式的乘积。要展开 (x + p)(x + q)，使用分配律：将第一个括号中的每一项分别乘以第二个括号中的每一项，然后合并同类项。
neral, (x + p)(x + q) = x² + (p + q)x + pq. This pattern is useful when factorising later.
例如，(x + 3)(x - 2) 展开为 x² - 2x + 3x - 6，化简为 x² + x - 6。一般地，(x + p)(x + q) = x² + (p + q)x + pq。这个模式在后面的因式分解中很有用。
xam questions often test these special forms because they speed up algebraic manipulation.
你还应能够展开完全平方式，如 (x + 5)² = x² + 10x + 25，以及平方差式，如 (x + 4)(x - 4) = x² - 16。考试题目经常考查这些特殊形式，因为它们能加快代数运算速度。
3. Factorising Quadratic Expressions | 因式分解二次表达式
ct is c and whose sum is b. If the numbers are p and q, then x² + bx + c = (x + p)(x + q).
因式分解是展开的逆运算。对于 x² + bx + c 形式的二次式，寻找两个数，使它们的乘积为 c，和为 b。如果这两个数是 p 和 q，则 x² + bx + c = (x + p)(x + q)。
5. Therefore x² + 7x + 10 = (x + 2)(x + 5). Always check by expanding the brackets again.
例如，要对 x² + 7x + 10 进行因式分解，需要找到两个数相乘为 10、相加为 7：2 和 5。因此 x² + 7x + 10 = (x + 2)(x + 5)。务必通过重新展开括号来检验。
term, then factor by grouping: 2x² + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3).
当 a ≠ 1 时，过程稍难一些。对于 2x² + 5x - 3，找到两个数，使它们相乘为 2 × (-3) = -6，相加为 5：这两个数是 6 和 -1。拆分中间项，然后分组分解：2x² + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)。
4. Solving by Factorisation: The Zero Product Principle | 因式分解法：零乘积原理
. This works because the only way a product can be zero is if at least one factor is zero.
一旦二次式被因式分解，你就可以利用零乘积原理来解方程：如果 AB = 0，那么 A = 0 或 B = 0。这个原理成立，因为乘积为零的唯一可能是其中至少一个因式为零。
ro: x + 3 = 0 gives x = -3, and x - 2 = 0 gives x = 2. The solutions are x = -3 and x = 2.
例如，解 x² + x - 6 = 0。因式分解得到 (x + 3)(x - 2) = 0。令每个括号等于零：x + 3 = 0 得到 x = -3，x - 2 = 0 得到 x = 2。解为 x = -3 和 x = 2。
o x = 0 or x = 2. Never divide both sides by x, because you might lose the solution x = 0.
如果方程右边不为零，需要先移项。对于 3x² = 6x，两边减去 6x 得到 3x² - 6x = 0。因式分解为 3x(x - 2) = 0，所以 x = 0 或 x = 2。千万不要两边同除以 x，因为你可能会丢掉 x = 0 这个解。
5. Completing the Square | 配方法
Completing the square transforms a quadratic into the form (x + p)² + q. This method is especially useful when the quadratic cannot be factorised easily, and it also helps find the vertex of a graph.
配方法将二次式转化为 (x + p)² + q 的形式。当二次式不容易因式分解时，这种方法特别有用，它也有助于求图像顶点。
act 9: x² + 6x + 9 - 9 + 2 = (x + 3)² - 7. The expression is now in completed square form.
对于 x² + 6x + 2，取 x 的系数一半，即 3，平方得到 9。加 9 再减 9：x² + 6x + 9 - 9 + 2 = (x + 3)² - 7。该表达式现在已写成完全平方形式。
+ 2 = 0 by completing the square, write (x + 3)² - 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7. Therefore x = -3 ± √7. Leave your answer in surd form unless a decimal is requested.
用配方法解 x² + 6x + 2 = 0 时，写成 (x + 3)² - 7 = 0。然后 (x + 3)² = 7，所以 x + 3 = ±√7。因此 x = -3 ± √7。除非题目要求小数，否则保留根式形式。
If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 5, write 2(x² + 4x) + 5, complete the square inside the bracket, and then distribute the 2 carefully.
如果 x² 的系数不是 1，需要先将其提出来。对于 2x² + 8x + 5，写成 2(x² + 4x) + 5，在括号内完成平方，然后小心地分配系数 2。
6. The Quadratic Formula | 二次方程求根公式
The quadratic formula gives the roots of ax² + bx + c = 0 directly. It is:
二次方程求根公式可以直接给出 ax² + bx + c = 0 的根。公式为：
x = (-b ± √b² - 4ac) ÷ (2a)
This formula works for every quadratic equation, including those that cannot be factorised. Memorise it exactly, and make sure you identify a, b and c from the standard form before substituting.
该公式适用于所有二次方程，包括无法因式分解的方程。必须准确记住公式，并在代入前确保从标准形式中识别出 a、b、c。
and c = -3. Substitute into the formula: x = (-5 ± √(5² - 4 × 2 × (-3))) ÷ (2 × 2) = (-5 ± √25 + 24) ÷ 4 = (-5 ± √49) ÷ 4 = (-5 ± 7) ÷ 4. This gives x = 2/4 = 1/2 or x = -12/4 = -3.
对于 2x² + 5x - 3 = 0，a = 2，b = 5，c = -3。代入公式：x = (-5 ± √(5² - 4 × 2 × (-3))) ÷ (2 × 2) = (-5 ± √25 + 24) ÷ 4 = (-5 ± √49) ÷ 4 = (-5 ± 7) ÷ 4。得到 x = 2/4 = 1/2 或 x = -12/4 = -3。
When using the formula, be careful with negative coefficients. If c is negative, the expression inside the square root becomes b² + 4|a||c|, which often reduces errors. Show your substitution clearly in the exam.
使用公式时要特别注意负数系数。如果 c 为负，平方根内的表达式变为 b² + 4|a||c|，这样常常能减少错误。考试中要清楚地写出代入过程。
7. The Discriminant and Number of Roots | 判别式与根的个数
The discriminant is the expression b² - 4ac inside the square root. It tells you how many real solutions a quadratic equation has without solving it.
判别式是平方根内的表达式 b² - 4ac。它可以在不解方程的情况下判断二次方程有多少个实数解。
is exactly one repeated real root. If b² - 4ac < 0, there are no real roots, because you cannot take the square root of a negative number in the real number system.
如果 b² - 4ac > 0，方程有两个不同的实数根。如果 b² - 4ac = 0，方程有一个重复实数根。如果 b² - 4ac < 0，方程没有实数根，因为在实数系中不能对负数开平方。
it has one repeated root. Indeed x² - 6x + 9 = (x - 3)², giving the single solution x = 3.
例如，x² - 6x + 9 = 0 的判别式为 (-6)² - 4×1×9 = 36 - 36 = 0，因此它有一个重复根。事实上 x² - 6x + 9 = (x - 3)²，得到唯一解 x = 3。
hich a quadratic has two equal roots. Set b² - 4ac = 0 and solve for the unknown constant.
IGCSE 题目可能要求你说明方程有多少个根，或者求出使二次方程有两个相等根的 k 值。此时令 b² - 4ac = 0 并解出未知常数。
8. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point. The graph is symmetric about a vertical line through the vertex.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上且有最小值点；如果 a < 0，抛物线开口向下且有最大值点。图像关于经过顶点的竖直线对称。
are the x-intercepts of the parabola. If the discriminant is positive, the graph crosses the x-axis twice. If it is zero, the graph touches the x-axis at one point. If it is negative, the graph does not meet the x-axis.
方程 ax² + bx + c = 0 的根是抛物线与 x 轴交点的横坐标。如果判别式为正，图像与 x 轴相交两次；如果为零，图像与 x 轴相切于一点；如果为负，图像与 x 轴不相交。
q). For y = x² + 6x + 2, completed square form is (x + 3)² - 7, so the vertex is (-3, -7).
顶点可以通过配方法找到。在 y = (x + p)² + q 的形式中，顶点位于 (-p, q)。对于 y = x² + 6x + 2，完全平方形式为 (x + 3)² - 7，因此顶点为 (-3, -7)。
9. Applied Problems and IGCSE Exam Tips | 应用题与 IGCSE 考试技巧
Quadratic equations often appear in real-life contexts such as area, projectile motion, or product puzzles. When solving word problems, define your variable clearly, form the equation carefully, then solve and check whether all solutions make sense in the original context.
二次方程经常出现在面积、抛体运动或乘积谜题等实际情境中。解应用题时，要明确定义变量，仔细列出方程，然后求解并检验所有解在原情境中是否合理。
) = 0, so x = -8 or x = 5. A width cannot be negative, so x = 5 cm and the length is 8 cm.
例如，一个矩形的长比宽多 3 cm，面积为 40 cm²。设宽为 x cm，则长为 x + 3 cm，面积关系给出 x(x + 3) = 40。展开并移项得 x² + 3x - 40 = 0，因式分解为 (x + 8)(x - 5) = 0，所以 x = -8 或 x = 5。宽不能为负，因此 x = 5 cm，长为 8 cm。
In the IGCSE exam, always write down your method, even if the final answer is wrong. Show substitution into the formula, factorised brackets, or completed square steps. This earns method marks. Simplify surds fully, and give answers to three significant figures if the question specifies that degree of accuracy.
在 IGCSE 考试中，即使最终答案错误，也要写出解题过程。写出代入公式、因式分解括号或配方法步骤，这样可以获得方法分。要完全化简根式，如果题目要求精确到三位有效数字，就按要求给出答案。
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