📚 Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程求解
A quadratic equation is one of the central algebra skills tested in IGCSE Mathematics. It appears as a standalone solving question, as part of a larger problem on area, motion or sequences, and as the tool behind graph sketching and the discriminant. Mastering the three main techniques - factorising, completing the square, and using the quadratic formula - together with the discriminant, gives you a reliable toolkit for any question style.
二次方程是 IGCSE 数学代数部分的核心技能之一。它经常作为独立的求解题出现，也会结合面积、运动或数列等问题一起考查，同时也是图像绘制与判别式问题的底层工具。掌握因式分解法、配方法和求根公式这三种主要方法，再加上判别式，你就能应对任何题型。
1. What Makes an Equation Quadratic? | 什么使方程成为二次方程
A quadratic equation is an equation of degree 2, which means the highest power of the variable is 2. The most common variable is x, but any letter can be used. The equation must include an x² term; it may also include an x term and a constant term. If the x² term has coefficient zero, the equation collapses into a linear equation and is no longer quadratic.
二次方程是次数为 2 的方程，也就是说未知数的最高次数是 2。最常用的未知数是 x，但也可以使用其他字母。方程中必须含有 x² 项，也可以同时含有 x 项和常数项。如果 x² 项的系数为 0，方程就退化为一次方程，不再是二次方程。
ax² + bx + c = 0, where a ≠ 0
n in one unknown has at most two solutions, called roots, because a curve of degree 2 can cross the x-axis at most twice.
条件 a ≠ 0 非常重要。例如 3x² − 2x + 1 = 0 是二次方程，而 0x² + 4x − 7 = 0 只是线性方程 4x − 7 = 0。含一个未知数的二次方程最多有两个解，称为根，因为二次曲线最多与 x 轴相交两次。
2. Standard Form and Identifying Coefficients | 标准形式与系数的确定
can you read off a, b and c correctly. For example, 5x² − 3x + 2 = 0 gives a = 5, b = −3 and c = 2.
求解之前，应先把方程写成标准形式 ax² + bx + c = 0，即各项按 x 的降幂排列，并且把全部项移到等号一边。只有这样，才能正确读出 a、b、c。例如 5x² − 3x + 2 = 0 中 a = 5，b = −3，c = 2。
terms first. For example, 2x + x² = 6 becomes x² + 2x − 6 = 0, so a = 1, b = 2 and c = −6.
如果各项分布在等号两边，需要先移项、合并同类项。例如 2x + x² = 6 可化为 x² + 2x − 6 = 0，于是 a = 1，b = 2，c = −6。
- A term like −x has coefficient −1, because −x means −1 × x. / 像 −x 这样的项系数为 −1，因为 −x 就是 −1 × x。
- If there is no x term, then b = 0, as in x² − 16 = 0. / 如果没有 x 项，那么 b = 0，例如 x² − 16 = 0。
- If there is no constant term, then c = 0, as in x² − 5x = 0. / 如果没有常数项，那么 c = 0，例如 x² − 5x = 0。
- The sign in front of a term belongs to that term, so it must be carried into the formula. / 项前面的负号属于该项本身，代入公式时必须一并保留。
3. Solving by Factorising when a = 1 | 当 a = 1 时用因式分解求解
wo factors is zero, then at least one factor must be zero. So if (x − p)(x − q) = 0, then x = p or x = q.
当二次式可以分解为整数因式时，因式分解是最快的方法。核心思想是零积法则：如果两个因式的乘积为 0，那么至少有一个因式为 0。因此若 (x − p)(x − q) = 0，则 x = p 或 x = q。
ply to give c and add to give b. Write them as p and q, then x² + bx + c = (x + p)(x + q).
对于 a = 1 的 x² + bx + c，找出两个数：它们的乘积等于 c，它们的和等于 b。记这两个数为 p 和 q，则 x² + bx + c = (x + p)(x + q)。
ply to 12 and add to −7 are −3 and −4, so (x − 3)(x − 4) = 0 and therefore x = 3 or x = 4.
例：解 x² − 7x + 12 = 0。乘积为 12、和为 −7 的两个数是 −3 和 −4，所以 (x − 3)(x − 4) = 0，因此 x = 3 或 x = 4。
, dividing by x loses the root x = 0. Factorise instead: x(x + 5) = 0, so x = 0 or x = −5.
千万不要在两边同时除以 x。对于 x² + 5x = 0，若除以 x 就会丢掉 x = 0 这个根。应当因式分解：x(x + 5) = 0，所以 x = 0 或 x = −5。
4. Solving by Factorising when a ≠ 1 | 当 a ≠ 1 时用因式分解求解
When the coefficient of x² is not 1, the same zero product rule applies, but finding the brackets needs a little more work. One reliable technique is to multiply a by c, find two numbers with that product and a sum equal to b, then split the middle term.
当 x² 的系数不是 1 时，零积法则依然适用，但找出括号需要多一步。一个可靠的方法是先把 a 与 c 相乘，找出乘积为 ac、和为 b 的两个数，再把中间项拆开。
6, and the two numbers that multiply to 6 and add to 7 are 1 and 6. Split the middle term and factorise in pairs:
例：解 2x² + 7x + 3 = 0。此处 ac = 6，乘积为 6、和为 7 的两个数是 1 和 6。拆开中间项，再分组提取公因式：
2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)
x = −3. Always expand your brackets mentally to check you recover the original expression.
因此 (2x + 1)(x + 3) = 0，得到 x = −½ 或 x = −3。养成心算展开括号的习惯，检查是否还原为原式。
- Check example: 6x² − 11x + 4 = (3x − 4)(2x − 1), so x = 4/3 or x = ½. / 检查例：6x² − 11x + 4 = (3x − 4)(2x − 1)，所以 x = 4/3 或 x = ½。
- If no integer pair works, the expression does not factorise with integers - switch to the formula. / 如果找不到整数对，说明该式不能整数分解，应改用求根公式。
5. The Difference of Two Squares | 平方差公式
A + B). When c is negative and the x term is missing, this pattern usually applies and gives two opposite roots.
一个非常常见的特例是平方差：A² − B² = (A − B)(A + B)。当常数项为负且缺少 x 项时，通常就是这种模式，会得到两个互为相反数的根。
a² − b² = (a − b)(a + b)
ncisely as x = ±3. Similarly, 4x² − 25 = 0 gives (2x − 5)(2x + 5) = 0, so x = ±5/2 = ±2.5.
例：x² − 9 = 0 化为 (x − 3)(x + 3) = 0，所以 x = 3 或 x = −3，可简写为 x = ±3。类似地，4x² − 25 = 0 化为 (2x − 5)(2x + 5) = 0，所以 x = ±5/2 = ±2.5。
1 + 4) = 0, so x = 3 or x = −5. Recognising this pattern saves a lot of time in the exam.
当一个平方项是括号时同样适用：(x + 1)² − 16 = 0 化为 (x + 1 − 4)(x + 1 + 4) = 0，所以 x = 3 或 x = −5。在考试中认出这个模式可以节省大量时间。
6. Completing the Square | 配方法
s the required method when a question asks for an exact answer in surd form, and it also reveals the turning point of the graph, which is (−p, q).
配方法把二次式改写为 (x + p)² + q = 0 的形式。当题目要求用根式给出精确答案时，这是指定方法；同时它还能直接给出图像的顶点坐标 (−p, q)。
x² + bx + c = 0 → (x + b/2)² = (b/2)² − c
)² − 9 + 1 = 0, that is (x + 3)² = 8. Taking the square root of both sides gives x + 3 = ±√8, so x = −3 ± √8 = −3 ± 2√2, which is approximately x = −0.172 or x = −5.83.
例：解 x² + 6x + 1 = 0。6 的一半是 3，于是 (x + 3)² − 9 + 1 = 0，即 (x + 3)² = 8。两边开平方得 x + 3 = ±√8，所以 x = −3 ± √8 = −3 ± 2√2，约为 x = −0.172 或 x = −5.83。
2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3, so 2x² + 8x + 5 = 0 gives (x + 2)² = 3/2 and x = −2 ± √3/2.
当 a ≠ 1 时，先把系数 a 只提取到 x² 项和 x 项前面。例如 2x² + 8x + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3，因此 2x² + 8x + 5 = 0 可得 (x + 2)² = 3/2，x = −2 ± √3/2。
7. Solving with the Quadratic Formula | 用求根公式求解
The quadratic formula solves any quadratic equation, whether or not it factorises, and whether or not the roots are rational. It is the safest fallback method, but you must substitute carefully because the negative signs are easy to lose.
求根公式可以解决任何二次方程，无论它能否分解，也无论根是否为有理数。它是最稳妥的兜底方法，但代入时必须小心，因为负号很容易出错。
x = (−b ± √b² − 4ac) / (2a)
a = 2, b = −4 and c = −3, so b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40. Then x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2, giving x ≈ 2.58 or x ≈ −0.581 to three significant figures.
例：解 2x² − 4x − 3 = 0。此处 a = 2，b = −4，c = −3，所以 b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40。于是 x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2，保留三位有效数字得 x ≈ 2.58 或 x ≈ −0.581。
- Write the formula first, then substitute inside brackets: (−4)² is 16, not −16. / 先抄写公式，再带括号代入：(−4)² 等于 16，而不是 −16。
- Keep full calculator accuracy and round only at the end. / 计算过程中保留完整精度，只在最后一步四舍五入。
- If the question says "give your answer in surd form", simplify √40 = 2√10 rather than using decimals. / 若题目要求"用根式形式作答"，应把 √40 化简为 2√10，而不写小数。
- If b² − 4ac is negative, stop and state that there are no real roots. / 如果 b² − 4ac 为负，立即停手并说明该方程无实根。
8. The Discriminant and the Number of Roots | 判别式与根的个数
The expression under the square root, b² − 4ac, is called the discriminant and is usually written as Δ. It tells you how many real roots the equation has before you do any solving, and it is a frequent one-mark or two-mark exam question.
根号下的表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在求解之前就告诉你方程有几个实根，是考试中经常出现的一两分小问。
| Discriminant 判别式 | Number of real roots 实根个数 | Graph 图像 |
| Δ > 0 | Two distinct real roots 两个不同实根 | Crosses the x-axis twice 与 x 轴交于两点 |
| Δ = 0 | One repeated root 一个重根 | Touches the x-axis 与 x 轴相切 |
| Δ < 0 |
