📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. They appear in both Core and Extended papers, across short-answer questions and multi-step problems. Mastering the three standard solution methods — factorisation, completing the square and the quadratic formula — is essential for exam success.
二次方程是 IGCSE 数学中最常考的考点之一，在 Core 和 Extended 两份试卷中都会出现，既包括基础小题，也包括多步综合题。掌握因式分解法、配方法和二次求根公式这三大解法，是取得高分的关键。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. Its general form is:
ax² + bx + c = 0, where a ≠ 0
nd a must never be zero. If a = 0, the x² term disappears and the equation becomes linear.
其中 a、b、c 是实数，且 a 不能为 0。如果 a = 0，x² 项消失，方程就退化为一次方程。
2 = 0 is not. The highest power of x is 2, which is why it is called a quadratic equation.
例如，x² + 5x + 6 = 0 是二次方程，而 3x + 2 = 0 不是。方程中 x 的最高次数为 2，因此称为二次方程。
2. Standard Form and Key Terms | 标准形式与关键术语
into the standard form ax² + bx + c = 0. Only then can you identify a, b and c correctly.
解任何二次方程之前，都必须先将其整理成标准形式 ax² + bx + c = 0。只有这样，才能正确确定 a、b、c 的值。
| Equation | a | b | c |
| x² + 5x + 6 = 0 | 1 | 5 | 6 |
| 2x² − 3x + 1 = 0 | 2 | −3 | 1 |
| x² − 7 = 0 | 1 | 0 | −7 |
The values of x that satisfy the equation are called roots or solutions. A quadratic equation always has two roots, but they may be equal or non-real.
满足方程的 x 值称为根（roots）或解（solutions）。二次方程一定有两个根，但它们可能相等，也可能不是实数。
example, 2x² = 3x + 5 must be rewritten as 2x² − 3x − 5 = 0 before any method is applied.
例如，2x² = 3x + 5 必须改写成 2x² − 3x − 5 = 0 后才能继续求解。
3. Solving by Factorisation | 因式分解法
Factorisation is the quickest method when the quadratic expression can be written as the product of two linear factors.
当二次式能写成两个一次因式的乘积时，因式分解法是最快的方法。
Follow these steps:
Write the equation in standard form ax² + bx + c = 0. 先将方程写成标准形式 ax² + bx + c = 0。
Factorise the left-hand side into two brackets. 将左边因式分解为两个括号。
Use the Zero Product Property: if AB = 0, then A = 0 or B = 0. 利用零积性质：若 AB = 0，则 A = 0 或 B = 0。
Solve the two resulting linear equations. 分别解两个一次方程。
Worked example: solve x² + 5x + 6 = 0.
例题：解 x² + 5x + 6 = 0。
x² + 5x + 6 = (x + 2)(x + 3) = 0
So x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
因此 x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。
example, 2x² + 5x + 2 = 0 can be written as (2x + 1)(x + 2) = 0, giving x = −½ or x = −2.
当 x² 的系数不为 1 时，因式分解稍难一些。例如，2x² + 5x + 2 = 0 可分解为 (2x + 1)(x + 2) = 0，得到 x = −½ 或 x = −2。
Always expand your brackets to check the factorisation before solving.
在求解前，务必展开括号验证因式分解是否正确。
4. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form (x + p)² + q. This method is especially useful when the equation cannot be factorised.
配方法将二次式改写为 (x + p)² + q 的形式。当方程无法因式分解时，这种方法特别有用。
x² + bx = (x + b/2)² − (b/2)²
Worked example: solve x² + 6x + 2 = 0 by completing the square.
例题：用配方法解 x² + 6x + 2 = 0。
(x + 3)² − 9 + 2 = 0
(x + 3)² = 7
Take the square root of both sides, remembering the ± sign:
两边开平方，注意加上 ± 号：
x + 3 = ±√7
x = −3 ± √7
So the two roots are x = −3 + √7 and x = −3 − √7.
因此两根为 x = −3 + √7 和 x = −3 − √7。
If the coefficient of x² is not 1, divide the whole equation by that coefficient first before completing the square.
若 x² 的系数不为 1，则先在整个方程两边除以该系数，再进行配方。
5. The Quadratic Formula | 二次求根公式
The quadratic formula works for every quadratic equation, including those that cannot be factorised or completed easily.
二次求根公式适用于所有二次方程，包括那些难以因式分解或配方的情况。
x = (−b ± √b² − 4ac) / 2a
Worked example: solve 2x² − 5x + 1 = 0 using the formula.
例题：用公式法解 2x² − 5x + 1 = 0。
Here a = 2, b = −5, c = 1. Substitute these values into the formula:
此处 a = 2，b = −5，c = 1。将这三个值代入公式：
x = (5 ± √((−5)² − 4 × 2 × 1)) / (2 × 2)
x = (5 ± √25 − 8) / 4
x = (5 ± √17) / 4
Notice that the minus sign of b changes when substituting, so −(−5) becomes +5. This is a very common source of errors.
注意代入时 b 的负号会发生变化，−(−5) 变为 +5。这是最常见的失分点之一。
The formula gives exact answers, so always leave your answer in surd form unless a decimal is requested.
公式给出精确答案，除非题目要求小数，否则应保留根式形式。
6. The Discriminant | 判别式 Δ
The expression b² − 4ac inside the square root is called the discriminant, written as Δ. It determines the nature of the roots without solving the equation.
根号内的表达式 b² − 4ac 称为判别式，记作 Δ。无需解方程，即可判断根的性质。
| Discriminant Δ | Nature of roots | 根的性质 |
| Δ > 0 | Two distinct real roots | 两个不相等的实数根 |
| Δ = 0 | One repeated real root | 一个二重实根 |
| Δ < 0 | No real roots | 无实数根 |
− 4x + 3 = 0. Here Δ = 16 − 12 = 4 > 0, so there are two distinct roots, x = 1 and x = 3.
验证第一个例子：x² − 4x + 3 = 0，此处 Δ = 16 − 12 = 4 > 0，所以有两个不等实根，x = 1 和 x = 3。
For x² − 4x + 4 = 0, Δ = 16 − 16 = 0, so there is exactly one repeated root, x = 2.
对于 x² − 4x + 4 = 0，Δ = 16 − 16 = 0，因此只有一个重根 x = 2。
For x² + 2x + 5 = 0, Δ = 4 − 20 = −16 < 0, so the equation has no real roots.
对于 x² + 2x + 5 = 0，Δ = 4 − 20 = −16 < 0，因此方程无实数根。
In IGCSE, the discriminant is often tested together with sketching the graph of a quadratic function.
在 IGCSE 考试中，判别式常与二次函数图像的性质结合考查。
7. Solving by Graphing | 图像法求解
re the x-coordinates of the points where the parabola y = ax² + bx + c crosses the x-axis.
ax² + bx + c = 0 的解，就是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。
If the graph crosses the x-axis at two points, there are two distinct roots. If it touches the axis at the vertex, there is one repeated root. If it never meets the axis, there are no real roots.
若图像与 x 轴有两个交点，则有两个不等实根；若在顶点处与 x 轴相切，则有一个重根；若与 x 轴无交点，则无实根。
Graphing is a helpful way to check your algebraic answers. It is also used in solving problems where an approximate answer is acceptable.
图像法可用于检验代数解是否正确，也常用于只需要近似解的实际问题。
For a quadratic with a > 0, the parabola opens upwards; for a < 0, it opens downwards. This affects the shape and the position of the vertex.
当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下，这会改变图像的形状和顶点位置。
8. Word Problems with Quadratics | 二次方程应用题
Many IGCSE exam questions require you to form a quadratic equation from a real-life situation and then solve it. The most common contexts are area, number and motion problems.
许多 IGCSE 考题要求从实际情境中建立二次方程并求解。最常见的背景有面积、数字和运动问题。
Worked example: the length of a rectangle is 3 cm more than its width, and its area is 28 cm². Find the dimensions of the rectangle.
例题：一个长方形的长比宽多 3 cm，面积为 28 cm²。求长方形的尺寸。
Let the width be x cm. Then the length is (x + 3) cm. The area gives:
设宽为 x cm，则长为 (x + 3) cm。由面积得：
