Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. They appear in algebra, geometry, and even in real-world problem-solving. In this article, we will explore the definition, three standard solving methods, the discriminant, and how to interpret quadratic graphs, with worked examples aligned to the Edexcel IGCSE specification.

二次方程是IGCSE数学大纲中最重要的话题之一。它们出现在代数、几何甚至现实问题解决中。在本文中，我们将探讨二次方程的定义、三种标准解法、判别式以及如何解读二次函数图像，并附带符合Edexcel IGCSE考纲的例题。

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

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b, and c are constants, with a ≠ 0. The highest power of the unknown x is 2, which makes the equation “quadratic”. If a = 0, the equation becomes linear.

二次方程是形如ax² + bx + c = 0的方程，其中a、b、c为常数，且a ≠ 0。未知数x的最高次数为2，因此方程称为“二次的”。若a = 0，方程退化为一次方程。

For example, 2x² − 5x + 3 = 0 is a quadratic equation. Notice that the coefficient a is 2, b is −5, and c is 3. The equation may also be given in expanded form, factorised form, or as a product of two linear factors.

例如，2x² − 5x + 3 = 0是一个二次方程。注意系数a = 2，b = −5，c = 3。方程可能以展开式、因式分解式或两个一次因式的乘积形式给出。

2. Solving by Factorisation | 因式分解法

Factorisation is the quickest method when the quadratic has simple integer factors. The principle is to rewrite ax² + bx + c as (px + q)(rx + s), then set each factor equal to zero. This works because if the product of two numbers is zero, at least one must be zero.

当二次式具有简单的整数因式时，因式分解是最快捷的方法。其原理是将ax² + bx + c改写为(px + q)(rx + s)，然后令每个因式等于零。这是因为如果两个数的乘积为零，则至少有一个数为零。

Example: Solve x² − 7x + 12 = 0.

例：解方程 x² − 7x + 12 = 0。

We look for two numbers that multiply to 12 and add to −7. They are −3 and −4. So x² − 7x + 12 = (x − 3)(x − 4) = 0.

我们寻找两个数，它们相乘得12，相加得−7。这两个数是−3和−4。因此x² − 7x + 12 = (x − 3)(x − 4) = 0。

Thus x − 3 = 0 or x − 4 = 0, giving x = 3 or x = 4.

因此x − 3 = 0 或 x − 4 = 0，得到x = 3 或 x = 4。

When the coefficient of x² is not 1, factorisation can still be done by method of splitting the middle term. For example, 2x² + 5x − 3 = 0. Multiply a and c: 2 × (−3) = −6. Find two numbers whose product is −6 and sum is 5: they are 6 and −1. Rewrite: 2x² + 6x − x − 3 = 0, then factor by grouping: 2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3) = 0. Hence x = ½ or x = −3.

当x²的系数不为1时，仍可使用“拆项法”进行因式分解。例如2x² + 5x − 3 = 0。将a与c相乘：2 × (−3) = −6。寻找两个数，其乘积为−6，和为5：它们是6和−1。重写：2x² + 6x − x − 3 = 0，然后分组分解：2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3) = 0。因此x = ½ 或 x = −3。

3. Solving by the Quadratic Formula | 公式法

The quadratic formula works for any quadratic equation, including those that cannot be factorised. For ax² + bx + c = 0, the solutions are given by:

公式法适用于任何二次方程，包括那些无法因式分解的方程。对于ax² + bx + c = 0，解由下式给出：

x = (−b ± √(b² − 4ac)) / (2a)

This formula comes from completing the square. You must memorise it for the exam. In the Edexcel IGCSE, the formula is provided on the formula sheet, but you still need to know how to apply it correctly.

该公式来源于配方法。考试中必须记住它。在Edexcel IGCSE中，公式会在公式表上给出，但你仍需知道如何正确应用。

Example: Solve 3x² − 4x − 2 = 0 using the formula.

例：用公式法解3x² − 4x − 2 = 0。

Here a = 3, b = −4, c = −2. Substitute into the formula:

这里a = 3，b = −4，c = −2。代入公式：

x = (4 ± √((−4)² − 4×3×(−2))) / (2×3) = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6

Since √40 = 2√10, we can simplify to x = (2 ± √10) / 3. So x ≈ 1.72 or x ≈ −0.39 (to 2 decimal places).

由于√40 = 2√10，可简化为x = (2 ± √10) / 3。因此x ≈ 1.72 或 x ≈ −0.39（保留两位小数）。

Always check if the question asks for exact values or decimal approximations. Use the simplified surd form where required.

始终检查题目要求精确值还是近似值。如需要，使用简化的根式形式。

4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This is especially useful for finding the turning point of a quadratic graph and for solving equations without the quadratic formula.

配方法将二次式改写为a(x + p)² + q的形式。这在寻找二次函数图像的顶点以及不使用公式法求解方程时特别有用。

For a quadratic x² + bx + c, take half of b, square it, and adjust the constant. For example, x² + 6x + 1 = 0. The term (x + 3)² = x² + 6x + 9, so we write x² + 6x + 1 = (x + 3)² − 8.

对于二次式x² + bx + c，取b的一半，平方，并调整常数。例如x² + 6x + 1 = 0。因为(x + 3)² = x² + 6x + 9，所以x² + 6x + 1 = (x + 3)² − 8。

Now solve (x + 3)² − 8 = 0. Then (x + 3)² = 8, so x + 3 = ±√8, hence x = −3 ± 2√2.

现在解(x + 3)² − 8 = 0。于是(x + 3)² = 8，因此x + 3 = ±√8，所以x = −3 ± 2√2。

When the coefficient of x² is not 1, first factor out a. For example, 2x² + 8x + 5 = 0 becomes 2(x² + 4x) + 5 = 2((x + 2)² − 4) + 5 = 2(x + 2)² − 3. Setting this equal to zero gives 2(x + 2)² = 3, so (x + 2)² = 1.5, and x = −2 ± √(3/2).

当x²的系数不为1时，先提出a。例如2x² + 8x + 5 = 0变为2(x² + 4x) + 5 = 2((x + 2)² − 4) + 5 = 2(x + 2)² − 3。令其等于零得2(x + 2)² = 3，所以(x + 2)² = 1.5，即x = −2 ± √(3/2)。

5. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: b² − 4ac. It tells us how many real solutions a quadratic equation has, without fully solving it.

判别式是公式法中根号内的部分：b² − 4ac。它告诉我们二次方程有多少个实数解，而无需完整求解。

If b² − 4ac > 0 – there are two distinct real roots.
如果b² − 4ac > 0 – 有两个不同的实数根。
If b² − 4ac = 0 – there is one repeated real root (or two equal roots).
如果b² − 4ac = 0 – 有一个重根（两个相等的实数根）。
If b² − 4ac < 0 – there are no real roots (the roots are complex/imaginary).
如果b² − 4ac < 0 – 没有实数根（根为复数/虚数）。

For example, for 2x² − 3x + 1 = 0, the discriminant is (−3)² − 4(2)(1) = 9 − 8 = 1 > 0, so there are two distinct real roots. For x² + 4x + 4 = 0, the discriminant is 16 − 16 = 0, so there is one repeated root.

例如，对于2x² − 3x + 1 = 0，判别式为(−3)² − 4(2)(1) = 9 − 8 = 1 > 0，因此有两个不同的实数根。对于x² + 4x + 4 = 0，判别式为16 − 16 = 0，所以有一个重根。

The discriminant also tells us whether the quadratic graph crosses the x-axis, touches it tangentially, or never touches it.

判别式还告诉我们二次函数图像与x轴相交、相切，还是永不相交。

6. Solving Quadratic Inequalities | 解二次不等式

Though not always tested as a separate topic, quadratic inequalities build on quadratic equations. To solve x² − 5x + 6 > 0, first solve x² − 5x + 6 = 0 to get x = 2 or x = 3. The graph of y = x² − 5x + 6 is a U-shaped parabola. It is above the x-axis when x < 2 or x > 3. So the solution is x < 2 or x > 3.

虽然二次不等式不总是单独考查，但它建立在二次方程之上。要解x² − 5x + 6 > 0，先解x² − 5x + 6 = 0得x = 2或x = 3。函数y = x² − 5x + 6的图像是开口向上的抛物线。当x < 2或x > 3时，图像在x轴上方。因此解集为x < 2或x > 3。

For x² − 5x + 6 ≤ 0, the graph is below or on the x-axis between the roots, so 2 ≤ x ≤ 3. Always sketch a simple graph or test intervals to confirm the inequality sign.

对于x² − 5x + 6 ≤ 0，图像在两根之间位于x轴下方或轴上，所以2 ≤ x ≤ 3。始终画一个简单的草图或测试区间来确认不等式符号。

7. Sketching Quadratic Graphs | 二次函数图像

A quadratic function y = ax² + bx + c has the following features:

二次函数y = ax² + bx + c具有以下特征：

If a > 0, the parabola opens upwards (U-shaped).
如果a > 0，抛物线开口向上（U形）。
If a < 0, it opens downwards (n-shaped).
如果a < 0，抛物线开口向下（∩形）。
The y-intercept is c, at (0, c).
y轴截距为c，即点(0, c)。
The x-intercepts (if any) are the real roots of ax² + bx + c = 0.
x轴截距（若存在）是ax² + bx + c = 0的实数根。
The vertex (turning point) has x-coordinate −b/(2a).
顶点（转折点）的x坐标为−b/(2a)。

For example, to sketch y = x² − 2x − 3, note it opens upwards. The y-intercept is −3. The x-intercepts come from solving (x − 3)(x + 1) = 0, giving x = 3 and x = −1. The vertex is at x = −(−2)/(2×1) = 1, and y = 1 − 2 − 3 = −4. So the vertex is (1, −4).

例如，要画y = x² − 2x − 3的草图，注意其开口向上。y轴截距为−3。x轴截距由解(x − 3)(x + 1) = 0得x = 3和x = −1。顶点在x = −(−2)/(2×1) = 1处，y = 1 − 2 − 3 = −4。所以顶点为(1, −4)。

8. The Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, the following relationships hold:

对于二次方程ax² + bx + c = 0，其根为α和β，有以下关系：

α + β = −b/a, and αβ = c/a

These are derived from expanding a(x − α)(x − β). They are extremely useful for finding symmetric expressions without solving the equation.

这些关系由展开a(x − α)(x − β)得出。它们在不需要解方程的情况下，对于求对称表达式非常有用。

For example, if α and β are roots of 2x² + 6x − 5 = 0, then α + β = −6/2 = −3 and αβ = −5/2. You can then calculate α² + β² = (α + β)² − 2αβ = 9 − 2(−5/2) = 9 + 5 = 14.

例如，若α和β是2x² + 6x − 5 = 0的根，则α + β = −6/2 = −3，αβ = −5/2。于是可以计算α² + β² = (α + β)² − 2αβ = 9 − 2(−5/2) = 9 + 5 = 14。

9. Word Problems Involving Quadratics | 二次方程应用题

Many exam questions require you to translate a real-world situation into a quadratic equation. For example, the area of a rectangle is 60 cm², and its length is 4 cm longer than its width. Let the width be x cm. Then the length is x + 4 cm. The area is x(x + 4) = 60, so x² + 4x − 60 = 0.

许多考试题目要求你将现实情境转化为二次方程。例如，一个矩形的面积为60 cm²，其长比宽长4 cm。设宽为x cm，则长为x + 4 cm。面积为x(x + 4) = 60，即x² + 4x − 60 = 0。

Factorising gives (x + 10)(x − 6) = 0, so x = −10 or x = 6. Since length cannot be negative, x = 6, so the width is 6 cm and the length is 10 cm.

因式分解得(x + 10)(x − 6) = 0，所以x = −10或x = 6。由于长度不能为负，x = 6，因此宽为6 cm，长为10 cm。

Other common applications include projectile motion (height h = ut − ½gt²), number problems, and geometry problems involving Pythagoras’ theorem. Always check that your answer makes sense in the context.

其他常见应用包括抛体运动（高度h = ut − ½gt²）、数字问题以及涉及勾股定理的几何问题。始终检查你的答案在上下文中有意义。

10. Common Mistakes and Tips | 常见错误与技巧

Students often make mistakes when a ≠ 1, when using the formula with negative b, or when simplifying surds. Write down the values of a, b, and c first, substitute carefully, and simplify step by step.

当a ≠ 1、使用公式时b为负数或化简根式时，学生经常出错。先写下a、b、c的值，仔细代入，再逐步化简。

Another common error is forgetting to set the equation to zero before factorising. For example, x² = 4x must be rewritten as x² − 4x = 0, then x(x − 4) = 0, giving x = 0 or x = 4. Dividing both sides by x would lose the solution x = 0.

另一个常见错误是在因式分解前忘记将方程化为等于零的形式。例如，x² = 4x必须改写为x² − 4x = 0，然后x(x − 4) = 0，得x = 0或x = 4。如果两边同时除以x，会丢失x = 0这个解。

Always check your answers by substituting back into the original equation. On the Edexcel IGCSE calculator paper, you can use the quadratic solver function to verify.

始终通过将答案代回原方程来检查。在Edexcel IGCSE计算器试卷中，你可以使用二次方程求解功能来验证。

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Solving Quadratic Equations | 解二次方程