Solving Quadratic Equations by Factorising, Completing the Square and the Quadratic Formula | 二次方程求解:因式分解、配方法与求根公式

📚 Solving Quadratic Equations by Factorising, Completing the Square and the Quadratic Formula | 二次方程求解:因式分解、配方法与求根公式

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation to graph sketching and real-world modelling. A quadratic equation is an equation of the form ax² + bx + c = 0, where a ≠ 0. In this revision guide, you will learn three core methods for solving quadratics: factorising, completing the square, and using the quadratic formula. You will also explore the discriminant, graphical meaning, and common exam pitfalls.

二次方程贯穿 IGCSE 数学课程始终,从代数运算到图像绘制和实际问题建模都会出现。二次方程的一般形式为 ax² + bx + c = 0,其中 a ≠ 0。本复习指南将学习三种核心求解方法:因式分解法、配方法和求根公式法。同时还会探讨判别式、图像意义以及常见的考试陷阱。


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

An equation is quadratic if the highest power of the unknown variable is 2. The standard form is ax² + bx + c = 0, where a, b and c are constants and a cannot be zero. If a were zero, the equation would become linear. Quadratic equations can have two real solutions, one repeated solution, or no real solutions depending on the discriminant.

如果未知数的最高次数是 2,则该方程为二次方程。标准形式为 ax² + bx + c = 0,其中 a、b、c 为常数且 a 不能为零。若 a 为零,方程就变成一次方程。二次方程可能有两个实数解、一个重根或没有实数解,这取决于判别式。

Some examples of quadratic equations are shown below. Notice that each equation can be rearranged into the standard form with a non-zero coefficient of x².

以下是一些二次方程的例子。请注意,每个方程都可以整理成标准形式,且 x² 的系数不为零。

  • 2x² + 3x − 5 = 0
  • x² − 4x + 4 = 0
  • 3x² + x + 7 = 0

These equations all have degree 2, even if some terms are missing or the coefficients are negative.

这些方程的次数都是 2,即使某些项缺失或系数为负数也是如此。


2. Standard Form and Key Terms | 标准形式与关键术语

Before solving, always rearrange the equation into standard form ax² + bx + c = 0. The coefficient a is the quadratic coefficient, b is the linear coefficient, and c is the constant term. Solutions to the equation are also called roots or x-intercepts of the graph y = ax² + bx + c.

在求解之前,一定要把方程整理成标准形式 ax² + bx + c = 0。系数 a 是二次项系数,b 是一次项系数,c 是常数项。方程的解也叫作根,或者图像 y = ax² + bx + c 的 x 轴截距。

For example, in the equation 2x² − 5x + 3 = 0, we have a = 2, b = −5 and c = 3. The leading coefficient a tells you whether the parabola opens upwards or downwards, while c gives the y-intercept of the graph.

例如,在方程 2x² − 5x + 3 = 0 中,a = 2,b = −5,c = 3。首项系数 a 决定抛物线开口向上还是向下,而 c 给出图像与 y 轴的截距。

Remember that rearranging terms is often necessary. You may need to bring all terms to one side, combine like terms, and place them in descending powers of x.

请记住,经常需要整理项。你可能需要把所有项移到一边,合并同类项,并按 x 的降幂排列。


3. Solving by Factorising | 因式分解法

Factorising is the fastest method when the quadratic can be expressed as a product of two linear factors. For x² + bx + c, look for two numbers p and q such that p + q = b and pq = c. Then write (x + p)(x + q) = 0 and use the zero product property: if AB = 0, then A = 0 or B = 0.

当二次式可以写成两个一次因式的乘积时,因式分解法最快。对于 x² + bx + c,寻找两个数 p 和 q,使得 p + q = b 且 pq = c。然后写成 (x + p)(x + q) = 0,并利用零乘积性质:如果 AB = 0,则 A = 0 或 B = 0。

Example: x² + 5x + 6 = (x + 2)(x + 3) = 0 ⇒ x = −2 or x = −3

In this example, the numbers 2 and 3 add to 5 and multiply to 6. Writing the equation as a product of two factors allows us to set each factor equal to zero and solve directly.

在这个例子中,2 和 3 相加为 5,相乘为 6。把方程写成两个因式的乘积后,就可以令每个因式等于零并直接求解。

Always check your factorisation by expanding the brackets. This will confirm that the product matches the original quadratic expression.

一定要通过展开括号来检验因式分解。这样可以确认乘积与原来的二次式一致。


4. Factorising When a > 1 | 当二次项系数大于 1 的因式分解

For quadratics of the form ax² + bx + c with a > 1, factorising involves finding two numbers that multiply to give ac and add to give b. Split the middle term, factorise in pairs, and extract the common binomial factor. Example: 2x² + 7x + 3. We need two numbers multiplying to 6 and adding to 7: these are 6 and 1. Write 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

对于 ax² + bx + c 且 a > 1 的二次式,因式分解需要找到两个数,它们的乘积等于 ac,和等于 b。拆开一次项后分组分解,再提取公因式。例如:2x² + 7x + 3。我们需要两个数乘积为 6、和为 7,这两个数是 6 和 1。写出 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

The ac method is reliable but takes practice. Another approach is trial and error with the possible factors of a and c, checking each combination until the middle coefficient is correct.

ac 法可靠但需要练习。另一种方法是尝试 a 和 c 的可能因数组合,逐一检验,直到一次项系数正确为止。

Once factorised, solving is exactly the same: set each factor equal to zero. For (2x + 1)(x + 3) = 0, we get x = −1/2 or x = −3.

一旦完成因式分解,求解方法完全相同:令每个因式等于零。对于 (2x + 1)(x + 3) = 0,得到 x = −1/2 或 x = −3。


5. Solving by Completing the Square | 配方法

Completing the square rewrites the quadratic in the form (x + p)² + q = 0. Start with x² + bx + c = 0. Move the constant to the right, add (b/2)² to both sides, and write the left side as a perfect square. For example, solve x² + 6x + 5 = 0: x² + 6x = −5, add (6/2)² = 9 to both sides, giving (x + 3)² = 4. Then x + 3 = ±2, so x = −1 or x = −5.

配方法把二次式改写为 (x + p)² + q = 0 的形式。从 x² + bx + c = 0 开始,将常数移到右边,两边同时加上 (b/2)²,把左边写成完全平方。例如,解 x² + 6x + 5 = 0:x² + 6x = −5,两边加上 (6/2)² = 9,得到 (x + 3)² = 4。于是 x + 3 = ±2,所以 x = −1 或 x = −5。

The key identity is x² + bx + (b/2)² = (x + b/2)². This method is especially useful when the quadratic cannot be factorised easily or when you need to find the vertex of the parabola.

关键恒等式是 x² + bx + (b/2)² = (x + b/2)²。当二次式不容易因式分解,或者需要求抛物线顶点时,这种方法特别有用。

If the coefficient a is not 1, first divide every term by a, then complete the square. Do not forget to adjust the constant term accordingly.

如果系数 a 不为 1,先让每一项除以 a,然后再配方。不要忘记相应调整常数项。


6. The Quadratic Formula | 求根公式

When factorising is difficult or impossible, the quadratic formula solves any quadratic equation ax² + bx + c = 0. The formula is derived by completing the square on the general form. It states:

当因式分解困难或无法进行时,求根公式可以解任何二次方程 ax² + bx + c = 0。这个公式由配方法推导而来。它给出:

x = (−b ± √(b² − 4ac)) / (2a)

Then plug in the values of a, b and c. For example, solve 2x² − 3x − 2 = 0: a = 2, b = −3, c = −2, so x = (3 ± √(9 + 16)) / 4 = (3 ± 5) / 4. This gives x = 2 or x = −1/2.

代入 a、b、c 的值即可。例如,解 2x² −

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