📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
In IGCSE Mathematics, quadratic equations appear in algebra, graphs, and real-life problems. A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. This article explains three main solving methods: factorising, completing the square, and the quadratic formula, together with the discriminant and graph features.
在 IGCSE 数学中，二次方程出现在代数、图像和实际应用题中。二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。本文讲解三种主要求解方法：因式分解法、配方法和求根公式，并介绍判别式与图像特征。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. The highest power of the unknown x is 2. It can have two real solutions, one repeated solution, or no real solutions, depending on the discriminant.
二次方程是次数为 2 的多项式方程。未知数 x 的最高次数为 2。根据判别式的不同，它可能有两个实数解、一个重根，或没有实数解。
ax² + bx + c = 0, a ≠ 0
The condition a ≠ 0 is essential. If a were zero, the x² term would disappear and the equation would become linear rather than quadratic. In IGCSE questions, you are often asked to recognise a quadratic equation from its standard form.
a ≠ 0 这一条件至关重要。如果 a 为零，x² 项就会消失，方程就变成一次方程而不是二次方程。在 IGCSE 题目中，经常要求你根据标准形式识别二次方程。
2. Standard Form and Key Features | 标准形式与关键特征
The standard form is ax² + bx + c = 0. The coefficient a cannot be zero. The constants b and c may be zero, giving special cases such as ax² + c = 0 or ax² + bx = 0. The coefficient a determines the shape and direction of the parabola.
标准形式为 ax² + bx + c = 0。系数 a 不能为零。常数 b 和 c 可以为零，从而产生如 ax² + c = 0 或 ax² + bx = 0 的特殊情况。系数 a 决定抛物线的形状和开口方向。
- a > 0: the parabola opens upwards. | a > 0：抛物线开口向上。
- a < 0: the parabola opens downwards. | a < 0：抛物线开口向下。
- c: the y-intercept is (0, c). | c：y 轴截距为 (0, c)。
When a quadratic is given in a different order, such as c + bx + ax² = 0, always rearrange it into descending powers of x before applying any method.
当二次方程以不同顺序给出时，例如 c + bx + ax² = 0，在应用任何方法之前，一定要将其重新排列为 x 的降幂形式。
3. Solving by Factorising | 因式分解法
Factorising works when the quadratic expression can be written as a product of two linear brackets. If (px + q)(rx + s) = 0, then at least one bracket must equal zero. This gives x = -q/p or x = -s/r. The method is quick when p, q, r and s are simple integers.
当二次式能写成两个一次括号的乘积时，可使用因式分解法。若 (px + q)(rx + s) = 0，则至少一个括号必须等于零。这给出 x = -q/p 或 x = -s/r。当 p、q、r、s 为简单整数时，这种方法非常快捷。
x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = -2 or x = -3
To factorise x² + bx + c, look for two numbers that multiply to give c and add to give b. For x² + 5x + 6, the numbers 2 and 3 multiply to 6 and add to 5, so the factors are (x + 2) and (x + 3).
要对 x² + bx + c 进行因式分解，需要找到两个数，它们的乘积等于 c，和等于 b。对于 x² + 5x + 6，数字 2 和 3 的乘积为 6，和为 5，因此因式为 (x + 2) 和 (x + 3)。
Always check that the equation equals zero before factorising. If it does not, move all terms to one side first. For example, x² + 5x = -6 must be rearranged to x² + 5x + 6 = 0.
因式分解前一定要确认方程右边等于零。如果不等，先把所有项移到一边。例如，x² + 5x = -6 必须重新整理为 x² + 5x + 6 = 0。
4. Completing the Square: The Idea | 配方法：核心思想
Completing the square rewrites the quadratic expression x² + bx + c in the form (x + p)² + q. Since (x + p)² is never negative, the minimum or maximum value of the quadratic becomes visible. This method is especially useful for finding the vertex of a parabola and deriving the quadratic formula.
配方法将二次式 x² + bx + c 改写为 (x + p)² + q 的形式。由于 (x + p)² 始终非负，二次式的最小值或最大值就变得清晰。这种方法特别适用于求抛物线的顶点，以及推导求根公式。
The expression (x + p)² + q is called the completed square form. It is also known as the vertex form. IGCSE exam papers often ask you to express a quadratic in this form and then state the minimum or maximum value.
表达式 (x + p)² + q 称为完全平方形式，也称为顶点式。IGCSE 试卷经常要求你将二次式表示为这种形式，然后写出最小值或最大值。
5. Completing the Square: Worked Steps | 配方法：步骤详解
To complete the square for x² + bx + c, take half of b, square it, and adjust the constant. For example, x² + 6x + 5 becomes (x + 3)² – 4. The steps are: halve the coefficient of x, square it, add and subtract that square, then factorise the perfect square.
对 x² + bx + c 进行配方时，取 b 的一半，平方，并调整常数项。例如，x² + 6x + 5 变为 (x + 3)² – 4。步骤为：将 x 的系数减半，平方，加上并减去该平方，然后对完全平方部分进行因式分解。
x² + 6x + 5 = (x² + 6x + 9) + 5 – 9 = (x + 3)² – 4
Solving (x + 3)² – 4 = 0 gives (x + 3)² = 4, so x + 3 = ±2, hence x = -1 or x = -5. This produces the same solutions as factorising but can always be used, even when the quadratic cannot be factorised neatly.
解 (x + 3)² – 4 = 0 得到 (x + 3)² = 4，因此 x + 3 = ±2，所以 x = -1 或 x = -5。这与因式分解法得到相同的解，但即使二次式不能很好地因式分解，配方法也总是可以使用。
If the coefficient of x² is not 1, first divide every term by a before completing the square. For example, 2x² + 8x + 6 = 0 becomes x² + 4x + 3 = 0, then complete the square.
如果 x² 的系数不是 1，在配方前应先将每一项除以 a。例如，2x² + 8x + 6 = 0 变为 x² + 4x + 3 = 0，然后再配方。
6. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation ax² + bx + c = 0. It is obtained by completing the square on the general form. The formula is:
求根公式可以解任意二次方程 ax² + bx + c = 0。它是对一般形式进行配方法后得到的。公式为：
x = (-b ± √(b² – 4ac)) ÷ 2a
You must substitute a, b and c carefully, including negative signs. The expression under the square root, b² – 4ac, is called the discriminant. It determines the number and type of solutions.
代入 a、b、c 时必须小心，包括负号。平方根下的表达式 b² – 4ac 称为判别式。它决定了解的个数和类型。
For example, to solve 2x² – 4x – 6 = 0, identify a = 2, b = -4 and c = -6. Then x = (4 ± √(16 + 48)) ÷ 4 = (4 ± √64) ÷ 4, giving x = 3 or x = -1.
例如，解 2x² – 4x – 6 = 0，确定 a = 2，b = -4，c = -6。然后 x = (4 ± √(16 + 48)) ÷ 4 = (4 ± √64) ÷ 4，得到 x = 3 或 x = -1。
7. The Discriminant | 判别式
The discriminant D = b² – 4ac tells us how many real solutions a quadratic equation has. If D > 0, there are two distinct real solutions. If D = 0, there is exactly one repeated real solution. If D < 0, there are no real solutions, only complex solutions.
判别式 D = b² – 4ac 告诉我们二次方程有多少个实数解。如果 D > 0，有两个不同的实数解；如果 D = 0，恰好有一个重根；如果 D < 0，没有实数解，只有复数解。
D = b² – 4ac
In IGCSE, you usually do not need to calculate complex solutions, but you may be asked to explain why an equation has no real roots. For instance, x² + x + 1 = 0 has D = 1 – 4 = -3, so it has no real solutions.
在 IGCSE 中，通常不需要计算复数解，但可能会要求你解释为什么方程没有实数根。例如，x² + x + 1 = 0 的判别式 D = 1 – 4 = -3，所以它没有实数解。
8. Graphs of Quadratic Functions | 二次函数图像
The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point. The y-intercept is (0, c).
y = ax² + bx + c 的图像是抛物线。如果 a > 0，抛物线开口向上，具有最小值点；如果 a < 0，抛物线开口向下，具有最大值点。y 轴截距为 (0, c)。
Axis of symmetry: x = -b ÷ (2a)
The axis of symmetry passes through the vertex and divides the parabola into two mirror-image halves. The x-intercepts, or roots, are the solutions of ax² + bx + c = 0. They may be read from the graph where the curve crosses the x-axis.
对称轴经过顶点，将抛物线分成两个镜像对称的部分。x 轴截距，也就是根，是 ax² + bx + c = 0 的解。它们可以从图像中曲线与 x 轴相交的位置读出。
9. Turning Point and Axis of Symmetry | 顶点与对称轴
The turning point or vertex can be found by completing the square or using x = -b/(2a). If the quadratic is written as (x + p)² + q, the vertex is (-p, q). The x-coordinate is -p, and the y-coordinate is q.
顶点（转折点）可以通过配方法或使用 x = -b/(2a) 求出。如果二次式写成 (x + p)² + q，则顶点为 (-p, q)。x 坐标为 -p，y 坐标为 q。
y = (x + 3)² – 4 → vertex = (-3, -4)
For the general form, first find x = -b/(2a), then substitute this x into the original equation to find y. This gives the coordinates of the maximum or minimum point. Knowing the vertex helps sketch the graph and solve optimisation problems.
对于一般形式，先求 x = -b/(2a)，再将 x 代入原方程求 y。这样就得到最大值点或最小值点的坐标。掌握顶点有助于绘制草图并解决最优化问题。
10. Applications and Word Problems | 实际应用与文字题
Quadratic equations model area, projectile motion, profit, and speed problems. For example, a rectangle with length x + 3 and width x may have area 40 m². The equation is x(x + 3) = 40, which expands to x² + 3x – 40 = 0. Factorising gives (x + 8)(x – 5) = 0, so x = 5, ignoring the negative length.
二次方程可用于面积、抛射运动、利润和速度等问题。例如，一个长为 x + 3、宽为 x 的矩形，面积可能为 40 m²。方程为 x(x + 3) = 40，展开为 x² + 3x – 40 = 0。因式分解得到 (x + 8)(x – 5) = 0，所以 x = 5，舍去负的长度。
In projection problems, the height h of an object at time t may be modelled by h = -5t² + 20t + 15. Setting h = 0 gives a quadratic equation that tells when the object hits the ground. Negative time solutions are ignored because time cannot be negative.
在抛射问题中，物体在时间 t 时的高度 h 可以表示为 h = -5t² + 20t + 15。令 h = 0 得到二次方程，从而得知物体何时落地。负的时间解应舍去，因为时间不可能为负数。
11. Common Mistakes | 常见
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