Mastering Quadratic Equations | 掌握二次方程 (IGCSE Mathematics)

📚 Mastering Quadratic Equations | 掌握二次方程 (IGCSE Mathematics)

Quadratic equations sit at the heart of IGCSE Mathematics. They appear in Paper 2 and Paper 4, and they are frequently combined with graphs, algebraic manipulation and real-life problem solving. This revision guide walks you through every method you need: factorising, the quadratic formula, completing the square, and the discriminant.

二次方程是 IGCSE 数学的核心内容之一。它出现在试卷二和试卷四中，并且经常与函数图像、代数变形以及实际情境问题结合考查。本复习指南将带你逐一掌握所有必备方法：因式分解法、求根公式法、配方法以及判别式的应用。

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

are constants and a ≠ 0. The highest power of the variable x is 2. If a were zero, the equation would become linear, not quadratic. Depending on the value of the discriminant, a quadratic equation can have two real solutions, one repeated solution, or no real solutions.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数且 a ≠ 0。变量 x 的最高次数为 2。如果 a 为零，方程就变成一次方程，而不是二次方程。根据判别式的值，二次方程可以有两个实数解、一个重根解或没有实数解。

ax² + bx + c = 0,   a ≠ 0

Always rewrite the equation in the form ax² + bx + c = 0 before choosing a solution method, and always identify the values of a, b and c carefully, including their signs.

即使二次方程没有写成标准形式，你也需要能识别它们。例如，3x² = 5x − 2 可以整理为 3x² − 5x + 2 = 0。在选择解法之前，一定要先把方程写成 ax² + bx + c = 0 的形式，并仔细确认 a、b、c 的数值及其正负号。

Remember that a, b and c do not have to be positive integers. They can be negative, fractional or even irrational, although in IGCSE questions they are usually integers or simple fractions.

请记住，a、b、c 不一定是正整数，它们可以是负数、分数，甚至无理数；不过在 IGCSE 题目中，它们通常是整数或简单分数。

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2. Standard Form and Hidden Quadratics | 标准形式与隐藏的二次方程

Most marks are lost before the algebra even starts, simply because the equation was not rearranged correctly. Expand any brackets first, collect all terms on one side so that the other side equals zero, and then simplify.

很多分数其实在代数运算开始之前就已经丢掉了，原因仅仅是方程没有正确变形。先展开所有括号，把所有项移到同一侧，使另一侧等于零，然后再化简。

expands to x² − 3x − 4 = 0. This is already in standard form, so a = 1, b = −3 and c = −4.

例如，(x + 1)(x − 4) = 0 展开后得到 x² − 3x − 4 = 0。这已经是标准形式，所以 a = 1，b = −3，c = −4。

ul: multiplying by an expression containing x can introduce extra solutions, so check each answer in the original equation.

有些方程把二次的本质隐藏起来了。如果你看到 x + 6 ÷ x = 5，就把每一项都乘以 x，得到 x² + 6 = 5x，整理后为 x² − 5x + 6 = 0。要小心：乘以含 x 的式子可能引入多余的解，所以每个答案都要代回原方程检验。

n convert back and solve for x. Always give all possible values of x, including negatives.

含有 x⁴ 和 x² 的方程，例如 x⁴ − 5x² + 4 = 0，也是伪装的二次方程。令 u = x²，得到 u² − 5u + 4 = 0，先解出 u，再换回 x 求解。一定要写出 x 的所有可能取值，包括负值。

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3. Solving by Factorising | 因式分解法

add to give b. For example, x² + 5x + 6 = 0 gives (x + 2)(x + 3) = 0, so x = −2 or x = −3.

当二次式可以容易地因式分解时，因式分解法通常是最快的方法。对于形如 x² + bx + c 的二次式，寻找两个数，它们相乘得到 c，相加得到 b。例如，x² + 5x + 6 = 0 可以写成 (x + 2)(x + 3) = 0，所以 x = −2 或 x = −3。

After factorising, always set each bracket equal to zero, because if a product is zero, at least one factor must be zero. This is called the zero product principle, and it is the reason we are allowed to split the equation into two simple linear equations.

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因式分解之后，一定要让每个括号等于零，因为如果乘积为零，至少有一个因式必须为零。这叫做零积原理，也是我们可以把方程拆成两个简单一次方程的原因。

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

multiply to 6 and add to 7, namely 6 and 1. So 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3), giving x = −1/2 or x = −3.

当 x² 的系数不是 1 时，例如 2x² + 7x + 3，你需要拆分中间项或使用试错法。这里 ac = 2 × 3 = 6，需要找到相乘得 6、相加得 7 的两个数，即 6 和 1。于是 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)，得到 x = −1/2 或 x = −3。

re gives a² + 2ab + b² = (a + b)². Recognising these instantly is a genuine exam advantage.

有两个特殊模式值得背下来，因为它们能节省大量时间。平方差公式为 a² − b² = (a + b)(a − b)，完全平方公式为 a² + 2ab + b² = (a + b)²。能在考场上瞬间识别它们，是真正的优势。

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4. Solving by the Quadratic Formula | 求根公式法

When a quadratic does not factorise nicely, use the quadratic formula. It works for every quadratic equation, so it is the safest fallback option when you are stuck.

当二次式无法漂亮地因式分解时，就使用求根公式。它对所有二次方程都适用，因此是遇到困难时最稳妥的备选方法。

x = (−b ± √b² − 4ac) ÷ (2a)

Substitute the values of a, b and c carefully, keeping the signs attached. If b is negative then −b is positive, and if c is negative then −4ac becomes positive. A very large number of marks are lost each year through sign slips here.

代入 a、b、c 时一定要仔细，连正负号一起带上。如果 b 是负数，那么 −b 就是正数；如果 c 是负数，那么 −4ac 就变成正数。每年都有大量分数因为这里符号写错而丢分。

For example, in 2x² − 4x − 3 = 0, we have a = 2, b = −4 and c = −3, so x = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4. Since √40 = 2√10, this simplifies to x = (2 ± √10) ÷ 2, which is approximately 2.58 or −0.58.

例如，对于 2x² − 4x − 3 = 0，有 a = 2，b = −4，c = −3，所以 x = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4。由于 √40 = 2√10，化简得 x = (2 ± √10) ÷ 2，约等于 2.58 或 −0.58。

If the square root is not a perfect square, leave the answer in exact surd form unless the question asks for a decimal. When rounding is required, work with full calculator accuracy and round only at the very last step, usually to 2 or 3 significant figures.

如果根号内的数不是完全平方数，除非题目要求小数，否则应保留精确的根式形式。需要四舍五入时，计算过程要使用计算器的完整精度，只在最后一步取近似值，通常保留 2 或 3 位有效数字。

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5. Solving by Completing the Square | 配方法

Completing the square rewrites x² + bx + c in the form (x + p)² + q. Use the rule that half of the coefficient of x goes inside the bracket, then subtract the square of that half.

配方法把 x² + bx + c 改写成 (x + p)² + q 的形式。方法是：括号里写 x 的系数的一半，再减去这一半的平方。

(x + b ÷ 2)² − (b ÷ 2)² + c

Example: x² + 6x + 1 = 0 becomes (x + 3)² − 9 + 1 = 0, so (x + 3)² = 8, giving x = −3 ± √8 = −3 ± 2√2. This is exact, and the method never fails, which is why some exam boards reward it specifically.

例如：x² + 6x + 1 = 0 化为 (x + 3)² − 9 + 1 = 0，所以 (x + 3)² = 8，得到 x = −3 ± √8 = −3 ± 2√2。这是精确解，而且该方法从不失效，因此某些考纲会专门为此给分。

first two terms only. For 2x² + 8x + 3 = 0, divide by 2 to get x² + 4x + 1.5 = 0, then (x

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