Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from pure algebra to graph sketching and word problems. Mastering the three main solving techniques – factorisation, the quadratic formula and completing the square – will allow you to tackle any quadratic problem with confidence. This revision guide breaks down every method step by step and highlights the traps examiners love to set.

二次方程贯穿 IGCSE 数学考纲的始终，从纯代数到画图再到应用题，无处不在。熟练掌握三大解法——因式分解、求根公式和配方法——你就能从容应对任何二次方程问题。本复习指南将逐步拆解每一种方法，并指出考官最爱设置的陷阱。

1. What Is a Quadratic Equation? | 什么是二次方程

A quadratic equation is a polynomial equation in which the highest power of the unknown variable is 2. Its general form is written as ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. If a were zero, the equation would become linear, not quadratic.

二次方程是未知量最高次数为 2 的多项式方程。它的一般形式写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。如果 a 等于零，方程就会退化为一次（线性）方程，而不是二次方程。

For example, in 3x² – 5x + 2 = 0 we have a = 3, b = -5 and c = 2. Notice that x² = -4x – 3 can be rearranged to x² + 4x + 3 = 0, so it is also a quadratic equation. Always rearrange all terms onto one side before identifying a, b and c.

例如，在 3x² – 5x + 2 = 0 中，a = 3，b = -5，c = 2。注意 x² = -4x – 3 可以移项重排为 x² + 4x + 3 = 0，因此它同样是二次方程。在确定 a、b、c 之前，务必先把所有项移到等号的一侧。

2. Expanding Double Brackets | 展开双括号

Before we can factorise, we need to be fluent in expanding. To expand (x + 3)(x + 5), multiply each term in the first bracket by each term in the second bracket. This is often taught with the acronym FOIL: First, Outer, Inner, Last.

在学会因式分解之前，我们必须熟练展开。展开 (x + 3)(x + 5) 时，要用第一个括号中的每一项分别乘以第二个括号中的每一项。常用首字母缩写 FOIL 来记忆：First（首项）、Outer（外项）、Inner（内项）、Last（尾项）。

(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15

Always simplify by collecting like terms: 5x + 3x = 8x. In general, (x + p)(x + q) = x² + (p + q)x + pq. This pattern is the key to factorising simple quadratics.

一定要合并同类项：5x + 3x = 8x。一般地，(x + p)(x + q) = x² + (p + q)x + pq。这个规律正是分解简单二次式的关键。

For harder cases, such as (2x + 3)(x – 4), use the same method: 2x × x = 2x², 2x × (-4) = -8x, 3 × x = 3x, and 3 × (-4) = -12. This gives 2x² – 5x – 12.

对于更复杂的情形，如 (2x + 3)(x – 4)，方法相同：2x × x = 2x²，2x × (-4) = -8x，3 × x = 3x，3 × (-4) = -12。最终得到 2x² – 5x – 12。

3. Factorising Quadratics | 因式分解二次式

Factorising is the reverse of expanding. To factorise x² + 7x + 12, look for two integers whose product is 12 and whose sum is 7. The numbers 3 and 4 work because 3 × 4 = 12 and 3 + 4 = 7. Hence x² + 7x + 12 = (x + 3)(x + 4).

因式分解是展开的逆运算。要分解 x² + 7x + 12，需要找到两个整数，使它们的乘积为 12，和为 7。3 和 4 满足条件，因为 3 × 4 = 12 且 3 + 4 = 7。因此 x² + 7x + 12 = (x + 3)(x + 4)。

Be careful with negative signs. To factorise x² – 5x + 6, find two numbers with product 6 and sum -5: the numbers are -2 and -3, so x² – 5x + 6 = (x – 2)(x – 3). Two special cases are worth memorising:

处理负号时要格外小心。分解 x² – 5x + 6 时，要找乘积为 6、和为 -5 的两个数：答案是 -2 和 -3，所以 x² – 5x + 6 = (x – 2)(x – 3)。有两个特殊形式值得牢记：

Difference of two squares: x² – a² = (x – a)(x + a), for example x² – 9 = (x – 3)(x + 3).
两平方差：x² – a² = (x – a)(x + a)，例如 x² – 9 = (x – 3)(x + 3)。
Perfect square: x² ± 2ax + a² = (x ± a)², for example x² + 6x + 9 = (x + 3)².
完全平方：x² ± 2ax + a² = (x ± a)²，例如 x² + 6x + 9 = (x + 3)²。

4. Solving by Factorisation | 用因式分解法求解

The central idea is simple: if the product of two factors is zero, then at least one of the factors must equal zero. This is called the zero-product property.

核心思想非常简单：如果两个因式的乘积等于零，那么至少有一个因式必须等于零。这称为零乘积性质。

If AB = 0, then A = 0 or B = 0.

若 AB = 0，则 A = 0 或 B = 0。

Solve x² – 5x + 6 = 0. Factorise first: (x – 2)(x – 3) = 0. Therefore x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3. Always state both solutions clearly.

解方程 x² – 5x + 6 = 0。先分解因式：(x – 2)(x – 3) = 0。于是 x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。两个解都要清晰写出。

When the coefficient of x² is not 1, factorise by grouping or by trial. For 2x² + 7x + 3 = 0, the factorisation is (2x + 1)(x + 3) = 0, so x = -½ or x = -3. Check by expanding the brackets to confirm.

当 x² 的系数不为 1 时，可用分组分解或试凑法。对 2x² + 7x + 3 = 0，分解结果为 (2x + 1)(x + 3) = 0，所以 x = -½ 或 x = -3。可通过展开括号进行验证。

5. The Quadratic Formula | 求根公式

Not every quadratic can be factorised easily. The quadratic formula solves any quadratic equation of the form ax² + bx + c = 0 in one step:

并非所有二次方程都能轻松分解。求根公式可以一步解出任意形如 ax² + bx + c = 0 的二次方程：

x = (-b ± √(b² – 4ac)) / 2a

Substitute the values of a, b and c carefully, then simplify. For 2x² + 3x – 5 = 0, we have a = 2, b = 3 and c = -5:

小心代入 a、b、c 的值，然后化简。对于 2x² + 3x – 5 = 0，a = 2，b = 3，c = -5：

x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2) = (-3 ± √49) / 4 = (-3 ± 7) / 4

This gives x = (-3 + 7)/4 = 1 or x = (-3 – 7)/4 = -5/2. Notice that a negative sign inside the square root must be handled carefully; here 4 × 2 × (-5) = -40 and subtracting it gives +40.

由此得到 x = (-3 + 7)/4 = 1 或 x = (-3 – 7)/4 = -5/2。注意根号内的负号要小心处理：这里 4 × 2 × (-5) = -40，减去它相当于加上 40。

6. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. Since (x + p)² = x² + 2px + p², the x² coefficient becomes 1 and the x coefficient is 2p. For x² + 6x + 2, half of 6 is 3, so:

配方法将二次式改写为 (x + p)² + q 的形式。因为 (x + p)² = x² + 2px + p²，x² 系数为 1，x 系数为 2p。对于 x² + 6x + 2，6 的一半是 3，于是：

x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7

To solve (x + 3)² – 7 = 0, rearrange to (x + 3)² = 7, then take the square root of both sides, remembering the ± sign:

解 (x + 3)² – 7 = 0 时，先移项得到 (x + 3)² = 7，然后两边开平方，记得加上 ± 号：

x + 3 = ±√7 → x = -3 ± √7

When the coefficient of x² is not 1, factor it out first. For example, 2x² + 8x – 5 = 2[(x + 2)² – 4] – 5 = 2(x + 2)² – 13. This form is especially useful for finding the turning point of a graph.

当 x² 系数不为 1 时，先把它提出来。例如，2x² + 8x – 5 = 2[(x + 2)² – 4] – 5 = 2(x + 2)² – 13。这种形式在求图像顶点时尤其有用。

7. The Discriminant | 判别式

The expression b² – 4ac inside the quadratic formula is called the discriminant. It tells us how many real roots the equation ax² + bx + c = 0 has, without solving it fully.

求根公式中的 b² – 4ac 称为判别式。不解方程，它就能告诉我们 ax² + bx + c = 0 有多少个实数根。

Discriminant Δ = b² – 4ac	Number of real roots	实数根个数
Δ > 0	Two distinct real roots	两个不相等的实数根
Δ = 0	One repeated real root	两个相等的实数根（重根）
Δ < 0	No real roots	无实数根

For x² – 4x + 4 = 0, the discriminant is (-4)² – 4 × 1 × 4 = 16 – 16 = 0, so the equation has one repeated root. Indeed, (x – 2)² = 0 gives x = 2 only. For x² + 2x + 5 = 0, the discriminant is 4 – 20 = -16, which is negative, so there are no real solutions.

对 x² – 4x + 4 = 0，判别式为 (-4)² – 4 × 1 × 4 = 16 – 16 = 0，所以方程只有一个重根。事实上，(x – 2)² = 0 只给出 x = 2。对 x² + 2x + 5 = 0，判别式为 4 – 20 = -16，是负数，因此没有实数解。

8. Quadratic Graphs | 二次函数图像

The graph of y = ax² + bx + c is always a curve called a parabola. Its shape depends on the sign of a:

函数 y = ax² + bx + c 的图像总是一条称为抛物线的曲线。它的开口方向取决于 a 的符号：

If a > 0, the parabola opens upwards, like a U shape, and has a minimum point.
若 a > 0，抛物线开口向上，形如 U，存在最低点（最小值点）。
If a < 0, the parabola opens downwards, like an inverted U, and has a maximum point.
若 a < 0，抛物线开口向下，形如倒 U，存在最高点（最大值点）。

The roots of the equation ax² + bx + c = 0 are exactly the x-intercepts of the graph. The axis of symmetry is the vertical line through the vertex, given by x = -b/(2a). In the completed-square form y = a(x – h)² + k, the vertex is at the point (h, k).

方程 ax² + bx + c = 0 的根恰好

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Solving Quadratic Equations | 解一元二次方程