📚 Quadratic Equations: Solving and Sketching | 二次方程：求解与图像
Quadratic equations sit at the centre of IGCSE and A-Level algebra, appearing in pure algebra questions, coordinate geometry, graph sketching and real-life modelling. This revision guide explains what a quadratic equation is, how to solve one by factorising, by completing the square and by using the quadratic formula, how the discriminant tells you how many roots to expect, and how to sketch the parabola accurately with all its key features labelled.
二次方程是 IGCSE 与 A-Level 代数的核心内容，广泛出现在纯代数题、坐标几何、图像绘制以及实际建模中。本复习指南讲解什么是二次方程，如何通过因式分解法、配方法和二次公式法求解，判别式如何预示根的个数，以及如何准确绘制抛物线并标注全部关键特征。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation in which the highest power of the unknown is 2. The name comes from the Latin word quadratus, meaning square, because the variable is squared. A quadratic equation always involves a term in x², may involve a term in x and a constant term, and its graph is a curve called a parabola.
二次方程是指未知数最高次数为 2 的方程。其名称源自拉丁语 quadratus，意为平方，因为变量被平方。二次方程一定含有 x² 项，可能含有 x 项和常数项，其图像是一条称为抛物线的曲线。
ese look very different, but every one of them can be rearranged into the same standard form, and each has at most two real solutions. Those solutions are called the roots of the equation.
典型例子包括 x² + 3x − 4 = 0、2x² − 8 = 0、(x − 1)² = 9 以及 x(x + 5) = 0。它们看起来差别很大，但都可以整理成同一个标准形式，而且每个方程最多有两个实数解。这些解称为方程的根。
- An equation contains an equals sign and can be solved; an expression such as x² + 3x − 4 cannot be solved, only factorised or simplified.
方程含有等号、可以求解；而像 x² + 3x − 4 这样的表达式不能求解，只能因式分解或化简。 - A quadratic has at most two real roots, but it may have one repeated root or no real roots at all.
二次方程最多有两个实根，但也可能只有一个重根，或者完全没有实根。 - The word "root" means a value of x that makes the equation true, and it is also the x-coordinate where the parabola crosses the x-axis.
"根"是指使方程成立的 x 值，同时也是抛物线与 x 轴交点的横坐标。
The key skill in exam questions is choosing the right method for the form you are given: factorising for simple integer roots, completing the square when you need the vertex or an exact surd answer, and the formula when factorising is awkward or impossible.
考试中的核心技能是根据所给形式选择正确的方法：根为简单整数时用因式分解，需要顶点或精确根式答案时用配方法，因式分解困难或不可行时用公式法。
2. Standard Form and Coefficients | 标准形式与系数
The standard form of a quadratic equation is shown below, where a, b and c are real numbers and a ≠ 0. If a were 0, the x² term would vanish and the equation would become linear.
二次方程的标准形式如下，其中 a、b、c 为实数，且 a ≠ 0。若 a 为 0，x² 项将消失，方程就退化为一次方程。
ax² + bx + c = 0, a ≠ 0
Here a is called the quadratic coefficient, b the linear coefficient and c the constant term. Always rearrange an equation into this form before doing anything else, because the sign of each coefficient matters enormously in the formula and in the discriminant.
其中 a 称为二次项系数，b 称为一次项系数，c 称为常数项。在动笔之前，务必先把方程整理成这一形式，因为每个系数的正负号在公式法和判别式中都极其重要。
| Equation | a | b | c |
| 3x² − 5x + 2 = 0 | 3 | −5 | 2 |
| x² + 4x = 0 | 1 | 4 | 0 |
| 2x² − 9 = 0 | 2 | 0 | −9 |
| 4x² = 9 − 3x | 4 | 3 | −9 |
4x² + 3x − 9 = 0. Missing out this step is one of the most frequent sources of lost marks.
注意最后一行：必须先整理方程，两边同时加 3x 再减 9，得到 4x² + 3x − 9 = 0。漏掉这一步是失分最常见的原因之一。
3. Solving by Factorising | 因式分解法求解
Factorising is the fastest method when the roots are simple integers or fractions. It relies on the zero product rule: if the product of two numbers is zero, then at least one of them must be zero.
当根为简单整数或分数时，因式分解是最快的方法。它依赖零乘积法则：若两个数的乘积为零，则其中至少有一个数为零。
If (x − p)(x − q) = 0, then x = p or x = q
For a monic quadratic x² + bx + c, you look for two numbers that multiply to give c and add to give b. For a non-monic quadratic ax² + bx + c, look for two numbers that multiply to give ac and add to give b, then split the middle term.
对于首项系数为 1 的二次式 x² + bx + c，要找两个数，它们的乘积为 c、和为 b。对于首项系数不为 1 的二次式 ax² + bx + c，要找两个数，乘积为 ac、和为 b，然后拆分中间项。
- 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3. Here −2 and −3 multiply to +6 and add to −5.
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 或 x = 3。此处 −2 与 −3 相乘得 +6、相加得 −5。 - → (x + 5)(x − 3) = 0 → x = −5 or x = 3. Different signs are needed because c is negative.
x² + 2x − 15 = 0 → (x + 5)(x − 3) = 0 → x = −5 或 x = 3。因为 c 为负，所以两数符号相反。 - (2x + 1)(x + 3) = 0 → x = −½ or x = −3. The product ac = 6 and the sum b = 7 give 6 and 1.
2x² + 7x + 3 = 0 → 2x² + 6x + x + 3 = 0 → (2x + 1)(x + 3) = 0 → x = −½ 或 x = −3。乘积 ac = 6、和 b = 7，对应 6 与 1。 - x² − 49 = 0 → (x + 7)(x − 7) = 0 → x = ±7. This is the difference of two squares.
x² − 49 = 0 → (x + 7)(x − 7) = 0 → x = ±7。这是平方差公式。
− 5) = 0 and conclude x = 0 or x = 5. Dividing by x would lose the root x = 0 completely.
考官特别喜欢设一个陷阱：千万不要在两边同时除以 x。对于 x² = 5x，应整理为 x² − 5x = 0，因式分解为 x(x − 5) = 0，得出 x = 0 或 x = 5。若除以 x，就会完全丢失 x = 0 这个根。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic as a perfect square plus a constant. It works for every quadratic, gives exact answers in surd form, and is the standard way to find the vertex of a parabola.
配方法把二次式改写为一个完全平方加上一个常数。它适用于所有二次式，能给出根式形式的精确答案，也是求抛物线顶点的标准方法。
x² + bx + c = (x + b/2)² − (b/2)² + c
For a monic quadratic, take half of the coefficient of x, square it, and subtract it again so the value of the expression is unchanged. Then solve by isolating the bracket.
对于首项系数为 1 的二次式，取 x 的系数的一半，平方后再减去它，使表达式的值保持不变。然后通过分离括号内部分求解。
(x + 3)² = 16. Taking the square root of both sides gives x + 3 = ±4, so x = 1 or x = −7.
例题：x² + 6x − 7 = 0。6 的一半是 3，于是 (x + 3)² − 9 − 7 = 0，得到 (x + 3)² = 16。两边开平方得 x + 3 = ±4，所以 x = 1 或 x = −7。
2)² − 4] + 3 = 2(x − 2)² − 5. Setting this equal to zero gives (x − 2)² = 5/2, so x = 2 ± √5/2 = 2 ± √10/2 ≈ 2 ± 1.581.
当 a ≠ 1 时，先把系数 a 提到括号外：2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5。令其等于零，得 (x − 2)² = 5/2，所以 x = 2 ± √5/2 = 2 ± √10/2 ≈ 2 ± 1.581。
5. The Quadratic Formula | 二次公式法
ts, without guessing or factorising. It is worth memorising exactly, including the ± sign.
二次公式可直接由系数求解任意二次方程，无需猜测或因式分解。它值得逐字背诵，包括 ± 符号。
x = (−b ± √b² − 4ac) / (2a)
s works. In exams you must substitute carefully, using brackets around negative numbers, and then simplify the surd if possible.
该公式来自对一般形式 ax² + bx + c = 0 进行配方，因此它总是成立。考试中必须小心代入，负数要加括号，最后尽可能化简根式。
2x² − 4x − 3 = 0, so a = 2, b = −4 and c = −3. Then b² − 4ac = 16 + 24 = 40, and x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2. Therefore x ≈ 2.581 or x ≈ −0.581, both correct to three decimal places.
例题：2x² − 4x − 3 = 0，所以 a = 2、b = −4、c = −3。则 b² − 4ac = 16 + 24 = 40，x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2。因此 x ≈ 2.581 或 x ≈ −0.581，均精确到三位小数。
- Write the formula first, then substitute, then simplify. This earns method marks even if the arithmetic slips.
先写出公式，再代入，最后化简。即使运算出错，这样也能拿到方法分。 - If the question asks for answers to a given number of decimal places or significant figures, keep full accuracy until the final rounding step.
若题目要求保留若干位小数或有效数字，请在最后一步才四舍五入，过程中保持完整精度。 - When the discriminant is a perfect square, the answers will be rational, and factorising would also have worked.
当判别式为完全平方数时，答案为有理数，此时因式分解法同样可行。
6. The Discriminant | 判别式
The discriminant is the part of the quadratic formula under the square root sign, and it tells you how many real roots the equation has without solving it.
判别式是二次公式中根号内的部分，它能在不解方程的情况下告诉你方程有几个实根。
Δ = b² − 4ac
| Discriminant | Nature of roots | Graph of y = ax² + bx + c |
| Δ > 0 | Two distinct real roots | Crosses the x-axis at two points |
| Δ = 0 | One repeated real root | Touches the x-axis at the vertex |
| Δ < 0 | No real roots | Does not meet the x-axis at all |
qual roots mean Δ = 0, so k² − 36 = 0, giving k = 6 or k = −6. Both values must be stated.
常见考题要求求未知常数。例如：已知 x² + kx + 9 = 0 有相等的根，求 k。相等根意味着 Δ = 0，所以 k² − 36 = 0，得 k = 6 或 k = −6。两个值都必须写出。
For an inequality question such as "find the values of m for which x² + 4x + m = 0 has two distinct real roots", set Δ > 0: 16 − 4m > 0, so m < 4.
对于不等式类问题，例如"求使 x² + 4x + m = 0 有两个不同实根的 m 的取值范围"，令 Δ > 0：16 − 4m > 0，所以 m < 4。
7. Sketching a Parabola: The Key Features | 抛物线的关键特征
A sketch is not an accurate plot; it is a clear diagram showing the shape, the orientation and the important coordinates. Before drawing, gather five pieces of information.
草图不是精确绘图，而是清楚展示形状、开口方向和重要坐标的示意图。动笔之前，先收集五项信息。
- Orientation: if a > 0 the parabola opens upwards (a minimum); if a < 0 it opens downwards (a maximum).
开口方向：若 a > 0，抛物线开口向上（有最小值）；若 a < 0，开口向下（有最大值）。 - y-intercept: substitute x = 0 to get y = c, so the curve passes through (0, c).
y 轴截距：令 x = 0 得 y = c，因此曲线经过点 (0, c)。 - x-intercepts: solve the equation y = 0 to find the roots.
x 轴截距：解 y = 0 求出根。 - Axis of symmetry: the vertical line x = −b/(2a) through the vertex.
对称轴：经过顶点的竖直直线 x = −b/(2a)。 - Vertex: substitute the axis value back into the equation to get the turning point.
顶点：把对称轴的横坐标代回方程，求出转折点坐标。
e x = 3 and x = −1, giving x-intercepts (−1, 0) and (3, 0). The axis of symmetry is x = −(−2)/(2 × 1) = 1, and substituting x = 1 gives y = 1 − 2 − 3 = −4, so the vertex is (1, −4).
例题：绘制 y = x² − 2x − 3 的草图。此处 a = 1 > 0，开口向上。y 轴截距为 (0, −3)。因式分解得 (x − 3)(x + 1) = 0，所以根为 x = 3 和 x = −1，x 轴截距为 (−1, 0) 与 (3, 0)。对称轴为 x = −(−2)/(2 × 1) = 1，代入 x = 1 得 y = 1 − 2 − 3 = −4，故顶点为 (1, −4)。
8. Vertex Form and the Axis of Symmetry | 顶点式与对称轴
When a quadratic is written in vertex (completed-square) form, the turning point can be read off immediately.
当二次式写成顶点式（配方式）时，转折点可以立刻读出。
y = a(x − h)² + k has vertex (h, k) and axis x = h
n reversal inside the bracket: (x − 3)² means h = +3, while (x + 3)² means h = −3. This is
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