📚 Solving Quadratic Equations: A Complete IGCSE Guide | IGCSE 数学：二次方程求解完全指南
Quadratic equations run through the whole of IGCSE Mathematics, from basic algebra and graph sketching to geometry and problem solving. This guide covers every method you need: factorisation, completing the square, the quadratic formula and the discriminant, with worked examples, common mistakes and exam tips along the way.
二次方程贯穿整个 IGCSE 数学，从基础代数、图像绘制到几何与问题解决都会出现。本指南涵盖你需要的全部方法：因式分解、配方法、二次公式和判别式，并配有例题、常见错误分析和考试技巧。
1. What is a Quadratic Equation? | 什么是二次方程？
nts and a ≠ 0. The highest power of the unknown is 2, and that is what makes the equation quadratic rather than linear.
二次方程是指任何可以整理成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数且 a ≠ 0。未知数的最高次数是 2，这正是它被称为二次方程而不是一次方程的原因。
ax² + bx + c = 0, a ≠ 0
ion would collapse into a linear equation such as bx + c = 0, which has only one solution.
条件 a ≠ 0 至关重要。如果 a 等于 0，x² 项就会消失，方程退化成 bx + c = 0 这样的一次方程，而一次方程只有一个解。
ions once they are rearranged. The solutions of a quadratic equation are also called its roots, zeros, or the x-intercepts of its graph.
例如，x² + 3x + 2 = 0、4x² - 9 = 0 和 2x² = 5x - 7 在整理之后都是二次方程。二次方程的解也叫做它的根、零点，或者其图像的 x 轴截距。
2. Standard Form and Coefficients | 标准形式与系数
Before using any method, rearrange the equation so that one side is zero and the terms are written in descending powers of x. This standard form makes the coefficients easy to read: a multiplies x², b multiplies x, and c is the constant term.
在使用任何方法之前，先把方程整理成一边为零、各项按 x 的降幂排列。这个标准形式让系数一目了然：a 乘 x²，b 乘 x，c 是常数项。
example, x² - 4 = 0 has a = 1, b = 0 and c = -4, and 5x² = 3x has a = 5, b = -3 and c = 0.
缺失的项系数记为零。例如，x² - 4 = 0 中 a = 1、b = 0、c = -4；而 5x² = 3x 中 a = 5、b = -3、c = 0。
als sign, because signs change when you move them. Worked example: rearrange 3x + 5 = 2x².
当等号两边都有项时要格外小心，因为移项时符号会改变。例题：把 3x + 5 = 2x² 整理成标准形式。
-2x² + 3x + 5 = 0 is also valid, but the positive leading coefficient is more convenient.
两边同时减去 3x 和 5：得到 2x² - 3x - 5 = 0，所以 a = 2、b = -3、c = -5。把同一个方程写成 -2x² + 3x + 5 = 0 也是对的，但首项为正数更便于后续计算。
| Original | 原方程 | Standard form | 标准形式 | a, b, c |
| x² + 3x + 2 = 0 | x² + 3x + 2 = 0 | 1, 3, 2 |
| 4x² - 9 = 0 | 4x² + 0x - 9 = 0 | 4, 0, -9 |
| 2x² = 5x - 7 | 2x² - 5x + 7 = 0 | 2, -5, 7 |
3. Solving by Factorisation | 用因式分解求解
Factorisation is the fastest method when the quadratic factorises nicely. It relies on the zero product rule: if two numbers multiply to give zero, then at least one of them must be zero.
当二次式可以整齐地分解时，因式分解是最快的方法。它依据的是零乘积法则：如果两个数相乘等于零，那么其中至少有一个数为零。
If (x - p)(x - q) = 0, then x = p or x = q
0 gives (x + 1)(x + 2) = 0, so x = -1 or x = -2. Check mentally: 1 × 2 = 2 and 1 + 2 = 3.
对于 a = 1 的简单二次式，找两个数：它们相乘等于 c，相加等于 b。分解 x² + 3x + 2 = 0 得到 (x + 1)(x + 2) = 0，所以 x = -1 或 x = -2。心算验证：1 × 2 = 2，1 + 2 = 3。
ecomes (2x - 3)(2x + 3) = 0, so x = 3 ÷ 2 = 1.5 or x = -1.5, often written as x = ± 3 ÷ 2.
平方差是很有用的特例：A² - B² = (A + B)(A - B)。例如 4x² - 9 = 0 变为 (2x - 3)(2x + 3) = 0，所以 x = 3 ÷ 2 = 1.5 或 x = -1.5，通常写作 x = ± 3 ÷ 2。
+ 3 = 0, giving 2x(x + 3) + 1(x + 3) = 0 and finally (2x + 1)(x + 3) = 0, so x = -1 ÷ 2 or x = -3.
当 a ≠ 1 时，找两个数：它们相乘等于 ac，相加等于 b，然后拆开中间项。对于 2x² + 7x + 3 = 0，ac = 6，而 6 与 1 相加得 7，于是 2x² + 6x + x + 3 = 0，得 2x(x + 3) + 1(x + 3) = 0，最终 (2x + 1)(x + 3) = 0，所以 x = -1 ÷ 2 或 x = -3。
Not every quadratic factorises with integers. If you cannot find a pair after a reasonable effort, switch to the quadratic formula rather than losing time in the exam.
并不是每个二次式都能用整数分解。如果尝试一段时间后仍找不到那一对数，就改用二次公式，不要在考试中继续浪费时间。
4. Completing the Square | 配方法
Completing the square rewrites the quadratic in the form (x + p)² + q, which is extremely useful for solving equations that do not factorise and for finding the vertex of a parabola.
配方法把二次式改写成 (x + p)² + q 的形式，这对求解无法分解的方程以及寻找抛物线的顶点都非常有用。
x² + bx + c = (x + b ÷ 2)² - (b ÷ 2)² + c
3, write (x + 3)² and subtract 3² = 9, giving (x + 3)² - 9 + 2 = 0, that is (x + 3)² = 7.
例题：解 x² + 6x + 2 = 0。把 x 的系数除以 2 得 3，写成 (x + 3)²，再减去 3² = 9，得到 (x + 3)² - 9 + 2 = 0，即 (x + 3)² = 7。
Taking square roots of both sides gives x + 3 = ± √7, so x = -3 + √7 ≈ -0.354 or x = -3 - √7 ≈ -5.646. Remember that every square root produces both a positive and a negative value.
两边开平方得 x + 3 = ± √7，所以 x = -3 + √7 ≈ -0.354 或 x = -3 - √7 ≈ -5.646。记住每个平方根都有正负两个取值。
7 = 2(x - 3)² - 18 + 7 = 2(x - 3)² - 11, so the minimum point of the curve is at (3, -11).
当 a ≠ 1 时，先提取 a。对于 2x² - 12x + 7，写成 2(x² - 6x) + 7 = 2[(x - 3)² - 9] + 7 = 2(x - 3)² - 18 + 7 = 2(x - 3)² - 11，所以曲线的最低点是 (3, -11)。
5. The Quadratic Formula | 二次公式
t factorises. It is derived by completing the square on the general form ax² + bx + c = 0.
二次公式可以求解任何二次方程，无论它能否分解。它正是对一般形式 ax² + bx + c = 0 配方推导出来的。
x = (-b ± √b² - 4ac) ÷ 2a
example: solve 3x² + 4x - 2 = 0. Here a = 3, b = 4 and c = -2, so b² - 4ac = 16 + 24 = 40.
例题：解 3x² + 4x - 2 = 0。这里 a = 3、b = 4、c = -2，所以 b² - 4ac = 16 + 24 = 40。
x = (-4 ± √40) ÷ 6 = (-2 ± √10) ÷ 3
three decimal places. Note how the whole of the numerator is divided by 2a, not just the square-root part - a very common slip.
数值上，x ≈ 0.387 或 x ≈ -1.721（保留三位小数）。注意是整个分子除以 2a，而不是只除根号那一部分——这是非常常见的失误。
ce the square root of a negative number is not real, this equation has no real solutions, which the graph shows as a parabola that never crosses the x-axis.
再看第 2 节中的例子：2x² - 5x + 7 = 0 给出 b² - 4ac = 25 - 56 = -31。由于负数没有实数平方根，这个方程没有实数解，图像上表现为一条始终不与 x 轴相交的抛物线。
6. The Discriminant | 判别式
The expression b² - 4ac is called the discriminant and is usually written as Δ. It tells you how many real roots a quadratic equation has before you do any further work.
表达式 b² - 4ac 叫做判别式，通常记作 Δ。它能在你进一步计算之前告诉你这个二次方程有几个实数根。
Δ = b² - 4ac
Δ > 0: two distinct real roots, and the curve cuts the x-axis twice.
Δ > 0：有两个不同的实数根，曲线与 x 轴相交两次。
Δ = 0: one repeated real root, and the curve touches the x-axis at its vertex.
Δ = 0：有一个重根，曲线在顶点处与 x 轴相切。
Δ < 0: no real roots, and the curve never meets the x-axis.
Δ < 0：没有实数根，曲线与 x 轴永不相交。
root at x = 2. Example: x² + x + 1 = 0 has Δ = 1 - 4 = -3 < 0, so it has no real roots.
例：x² - 4x + 4 = 0 的 Δ = 16 - 16 = 0，所以在 x = 2 处有一个重根。例：x² + x + 1 = 0 的 Δ = 1 - 4 = -3 < 0，因此没有实数根。
Always read whether the question wants two distinct roots, equal roots, or no real roots.
考试常要求你求未知常数。如果 x² + kx + 9 = 0 有相等的根，那么 Δ = k² - 36 = 0，所以 k = 6 或 k = -6。一定要看清题目要求的是两个不同实根、重根，还是没有实数根。
7. Roots, Intercepts and Graphs | 根、截距与图像
the x-intercepts of that parabola, which is why the two ideas are always taught together.
y = ax² + bx + c 的图像是一条抛物线。相关方程 ax² + bx + c = 0 的根正好就是这条抛物线与 x 轴交点的横坐标，这正是这两个概念总是放在一起讲的原因。
a, and the vertex lies on that line; completing the square gives its coordinates directly.
令 x = 0 即可求出 y 轴截距，即 y = c。对称轴是竖直线 x = -b ÷ 2a，顶点就在这条直线上；配方可以直接给出顶点的坐标。
4). If a > 0 the parabola opens upwards and the vertex is a minimum; if a < 0 it opens downwards and the vertex is a maximum.
对于 y = x² - 6x + 5：根为 x = 1 和 x = 5，y 轴截距为 5，对称轴为 x = 3，顶点为 (3, -4)。若 a > 0，抛物线开口向上，顶点是最小值点；若 a < 0，开口向下，顶点是最大值点。
A quick sketch showing the intercepts, the vertex and the direction of opening often earns method marks even if the final algebra goes wrong.
画出标注截距、顶点和开口方向的草图，即使最后代数出错，也常常能拿到方法分。
8. Forming Equations from Word Problems | 由应用题建立方程
Many IGCSE questions hide a quadratic inside a word problem. The skill is to define a variable clearly, translate the words into an equation, rearrange to standard form and then solve.
许多 IGCSE 题目把二次方程藏在应用题里。关键技巧是清楚定义变量、把文字翻译成方程、整理成标准形式，然后再求解。
cm². Let the width be w, so the length is w + 3, and w(w + 3) = 40 gives w² + 3w - 40 = 0.
例：一个矩形的长比宽多 3 cm，面积为 40 cm²。设宽为 w，则长为 w + 3，由 w(w + 3) = 40 得 w² + 3w - 40 = 0。
th is 8 cm. Always reject the solution that does not fit the physical context and say why.
分解得 (w + 8)(w - 5) = 0，所以 w = -8 或 w = 5。由于长度不能为负，取 w = 5，长为 8 cm。一定要舍去不符合实际情境的解，并说明理由。
1 gives n² + n - 56 = 0, so (n + 8)(n - 7) = 0 and n = 7, with the integers being 7 and 8.
连续整数问题同理：若两个连续正整数的乘积为 56，设它们为 n 和 n + 1，得 n² + n - 56 = 0，所以 (n + 8)(n - 7) = 0，n = 7，这两个整数就是 7 和 8。
9. Quadratic and Linear Equations Together | 二次方程与一次方程联立
When a quadratic equation is given together with a linear equation, solve them simultaneously by substitution. Rearrange the linear equation to make one variable the subject, substitute into the quadratic, and solve the resulting equation.
当二次方程与一次方程同时出现时，用代入法联立求解。先把一次方程整理成某个变量作主元，代入二次方程，再解得到的方程。
ng gives x + 1 = x² - 1, so x² - x - 2 = 0 and (x - 2)(x + 1) = 0, giving x = 2 or x = -1.
例题：同时解 y = x + 1 与 y = x² - 1。代入得 x + 1 = x² - 1，所以 x² - x - 2 = 0，(x - 2)(x + 1) = 0，得 x = 2 或 x = -1。
ne meets the curve at (2, 3) and (-1, 0), which corresponds to two intersections on the graph.
务必要代回较简单的一次方程：当 x = 2 时 y = 3；当 x = -1 时 y = 0。因此直线与曲线交于 (2, 3) 和 (-1, 0)，对应图像上的两个交点。
not meet. If Δ = 0, the line is a tangent to the curve, touching it at exactly one point.
如果得到的二次方程 Δ < 0，则直线与曲线不相交。如果 Δ = 0，则直线与曲线相切，恰好只有一个公共点。
10. Common Mistakes to Avoid | 需要避免的常见错误
Most lost marks in quadratic questions come from a small number of repeated slips rather than from not knowing the method. Knowing them in advance is the cheapest way to raise your grade.
二次方程题目中失分大多来自少数几个重复出现的小失误，而不是不懂方法。提前了解这些失误，是提分最省力的方式。
Never divide both sides by x. Doing so destroys the root x = 0; factorise instead.
千万不要两边同时除以 x。这会丢掉 x = 0 这个根；应该改为因式分解。
Watch the signs when you rearrange, especially with negative leading coefficients.
移项时注意符号，尤其是首项为负时。
Do not round early. Keep full accuracy until the final answer, then round as the question asks.
不要过早取近似值。计算过程保持完整精度，最后再按题目要求保留位数。
Remember the ± in the formula and when taking square roots.
记住公式中和开平方时的 ± 符号。
For a quadratic with only an x² term and a constant, isolate x² and take the square root of both sides.
对于只含 x² 项和常数项的方程，把 x² 单独放到一边，再两边开平方。
Check your answers by substituting them back into the original equation.
把答案代回原方程进行检验。
11. Worked Exam-Style Examples | 考试风格例题精讲
Example A:
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