Mastering Quadratic Equations | 二次方程全面突破

📚 Mastering Quadratic Equations | 二次方程全面突破

Quadratic equations are among the most important topics in IGCSE Mathematics. They appear in pure algebra, graph work, and real-world problem solving. 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. The name comes from the Latin word “quadratus”, meaning square, because the highest power of the variable is 2. Every IGCSE paper contains at least one question on quadratics, and mastery of this topic is essential for a high grade.

二次方程是 IGCSE 数学中最重要的专题之一。它出现在纯代数、函数图像以及实际应用题中。二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数，且 a ≠ 0。其名称源自拉丁语 “quadratus”，意为”正方形”，因为变量的最高次数是 2。每一份 IGCSE 试卷至少有一道关于二次方程的题目，掌握这一专题是取得高分的关键。

---

1. Expanding and Factorising | 展开与因式分解

Before solving quadratics, you need two fundamental skills: expanding and factorising. Expanding means removing brackets by multiplying every term inside the bracket by every term outside. For example, (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15. Factorising is the reverse process: you start with x² + 8x + 15 and return to (x + 3)(x + 5).

在求解二次方程之前，你需要掌握两个基本技能：展开与因式分解。展开是指通过将括号外的每一项乘以括号内的每一项来去掉括号。例如，(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15。因式分解是相反的过程：你从 x² + 8x + 15 出发，还原为 (x + 3)(x + 5)。

To factorise a quadratic of the form x² + bx + c, follow these steps:

要对形如 x² + bx + c 的二次式进行因式分解，请按以下步骤操作：

• Identify the values of b and c. | 确定 b 和 c 的值。
• List all factor pairs of c. | 列出 c 的所有因数对。
• Choose the pair whose sum is b. | 选择和为 b 的那一对。
• Write the answer as (x + p)(x + q). | 将答案写成 (x + p)(x + q) 的形式。

For example, factorise x² – 7x + 12. We need two numbers whose product is 12 and whose sum is -7. The numbers are -3 and -4, so x² – 7x + 12 = (x – 3)(x – 4).

例如，因式分解 x² – 7x + 12。我们需要找到两个数，其乘积为 12 且和为 -7。这两个数是 -3 和 -4，因此 x² – 7x + 12 = (x – 3)(x – 4)。

When the coefficient of x² is not 1, factorising requires extra care. Consider 2x² + 7x + 3. We split the middle term: find two numbers whose product is 2 × 3 = 6 and whose sum is 7; the numbers are 6 and 1. Then 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1). This method is called factorising by grouping.

当 x² 的系数不是 1 时，因式分解需要格外小心。考虑 2x² + 7x + 3。我们拆分中间项：找到两个数，其乘积为 2 × 3 = 6，且和为 7；这两个数是 6 和 1。于是 2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)。这种方法称为分组因式分解。

---

2. Solving by Factorisation | 因式分解法求解

Once a quadratic is factored, you can solve it using the zero product property: if the product of two expressions is zero, then at least one of them must be zero. Consider x² – 5x + 6 = 0. Factorising gives (x – 2)(x – 3) = 0. Therefore either x – 2 = 0, so x = 2, or x – 3 = 0, so x = 3. The solution is x = 2 or x = 3.

一旦完成因式分解，就可以利用零积性质来求解：如果两个表达式的乘积为零，那么至少其中一个必须为零。考虑 x² – 5x + 6 = 0。因式分解得 (x – 2)(x – 3) = 0。因此要么 x – 2 = 0，即 x = 2；要么 x – 3 = 0，即 x = 3。解为 x = 2 或 x = 3。

Always rearrange the equation so that the right-hand side is 0 before factorising. If the equation is given as x² – 5x = -6, you must first add 6 to both sides to obtain x² – 5x + 6 = 0. A common mistake is to factorise before moving everything to one side, which leads to incorrect solutions.

在因式分解之前，务必先将方程整理为右边等于 0 的标准形式。如果题目给出 x² – 5x = -6，你必须先将两边同时加上 6，得到 x² – 5x + 6 = 0。一个常见错误是在把各项移到同一边之前就进行因式分解，这会导致错误的解。

Here is a second example with a leading coefficient not equal to 1. Solve 2x² + 7x + 3 = 0. From the previous section, 2x² + 7x + 3 = (x + 3)(2x + 1), so (x + 3)(2x + 1) = 0. Thus x + 3 = 0, giving x = -3, or 2x + 1 = 0, giving x = -1/2.

下面是一个首项系数不为 1 的例子。解方程 2x² + 7x + 3 = 0。根据上一节，2x² + 7x + 3 = (x + 3)(2x + 1)，所以 (x + 3)(2x + 1) = 0。于是 x + 3 = 0，得 x = -3；或者 2x + 1 = 0，得 x = -1/2。

---

3. Solving by Completing the Square | 配方法求解

Completing the square is a powerful technique that also helps with graph sketching and finding maximum or minimum values. The key identity is:

配方法是一种非常实用的技巧，它还能帮助绘制函数图像以及求最大值或最小值。关键恒等式是：

x² + bx = (x + b/2)² – (b/2)²

For example, x² + 6x = (x + 3)² – 9. To solve x² + 6x + 2 = 0, rewrite the first two terms: x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7. The equation becomes (x + 3)² – 7 = 0, so (x + 3)² = 7. Taking the square root of both sides gives x + 3 = ±√7, so x = -3 + √7 or x = -3 – √7.

例如，x² + 6x = (x + 3)² – 9。为了求解 x² + 6x + 2 = 0，先改写前两项：x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7。方程变为 (x + 3)² – 7 = 0，于是 (x + 3)² = 7。两边同时开平方得 x + 3 = ±√7，因此 x = -3 + √7 或 x = -3 – √7。

Follow these steps when completing the square:

使用配方时请遵循以下步骤：

• Make sure the coefficient of x² is 1. If not, factor it out first. | 确保 x² 的系数为 1。如果不是，先将其提取出来。
• Halve the coefficient of x and square it. | 将 x 的系数除以 2 并平方。
• Add and subtract this square inside the expression. | 在表达式中同时加上和减去这个平方数。
• Rearrange into the form (x + p)² + q. | 整理为 (x + p)² + q 的形式。
• Solve by taking square roots. | 通过开平方求解。

Completing the square is especially useful when the solutions are surds, because it gives the exact answer without needing a calculator.

配方法在解为无理数时尤其有用，因为它无需计算器就能给出精确答案。

---

4. The Quadratic Formula | 二次求根公式

Some quadratics cannot be factorised easily, especially when the solutions involve irrational numbers. For any quadratic equation ax² + bx + c = 0, the solution is given by the quadratic formula:

有些二次方程不容易因式分解，尤其是当解包含无理数时。对于任意二次方程 ax² + bx + c = 0，其解由二次求根公式给出：

x = (-b ± √(b² – 4ac)) / (2a)

Let us apply this formula to 2x² – 4x – 3 = 0. Here a = 2, b = -4 and c = -3. Substitute these values carefully:

让我们将公式应用于 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。小心代入这些值：

x = (4 ± √((-4)² – 4 × 2 × (-3))) / (2 × 2)

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4

Since √40 = 2√10, we simplify to x = (4 ± 2√10) / 4 = (2 ± √10) / 2. These are the exact solutions. If a decimal answer is requested, a calculator gives x ≈ 2.58 or x ≈ -0.58 correct to 2 decimal places.

因为 √40 = 2√10，化简得 x = (4 ± 2√10) / 4 = (2 ± √10) / 2。这些是精确解。如果题目要求小数答案，用计算器可得 x ≈ 2.58 或 x ≈ -0.58（精确到两位小数）。

When using the quadratic formula, be especially careful with negative values of b. The term -b at the start means the opposite sign of b. For example, if b = -4, then -b = 4. Also remember to multiply 4ac correctly; a common error is to forget the negative sign of c.

使用求根公式时，要特别注意 b 为负数的情况。开头的 -b 表示 b 的相反数。例如，若 b = -4，则 -b = 4。同时要正确计算 4ac；一个常见错误是漏掉 c 的负号。

---

5. The Discriminant and the Nature of Roots | 判别式与根的性质

The quantity under the square root in the quadratic formula, Δ = b² – 4ac, is called the discriminant. It tells you the nature of the roots without solving the equation completely:

求根公式中根号下的表达式 Δ = b² – 4ac 称为判别式。它可以在不完整求解方程的情况下告诉我们根的性质：

• If Δ > 0, the equation has two distinct real roots. | 若 Δ > 0，方程有两个不相等的实数根。
• If Δ = 0, the equation has one repeated real root (a double root). | 若 Δ = 0，方程有一个重根（二重根）。
• If Δ < 0, the equation has no real roots. | 若 Δ < 0，方程没有实数根。

For example, consider 3x² – 2x + 5 = 0. Here Δ = (-2)² – 4 × 3 × 5 = 4 – 60 = -56, which is negative, so this equation has no real roots. The graph of y = 3x² – 2x + 5 never crosses the x-axis.

例如，考虑 3x² – 2x + 5 = 0。这里 Δ = (-2)² – 4 × 3 × 5 = 4 – 60 = -56，是负数，因此该方程没有实数根。函数 y = 3x² – 2x + 5 的图像永远不会与 x 轴相交。

Now consider 4x² – 12x + 9 = 0. Here Δ = (-12)² – 4 × 4 × 9 = 144 – 144 = 0, so there is exactly one repeated root. Factorising confirms this: 4x² – 12x + 9 = (2x – 3)², so x = 3/2.

再看 4x² – 12x + 9 = 0。这里 Δ = (-12)² – 4 × 4 × 9 = 144 – 144 = 0，因此恰好有一个重根。因式分解证实了这一点：4x² – 12x + 9 = (2x – 3)²，所以 x = 3/2。

Questions that ask “find the value of k for which the equation has equal roots” are solved by setting the discriminant equal to zero. For example, for the equation 9x² + kx + 4 = 0 to have a repeated root, we need Δ = k² – 4 × 9 × 4 = 0, so k² = 144, giving k = 12 or k = -12.

题目若问”求 k 的值，使方程有两个相等的根”，只需令判别式等于零即可。例如，使

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）