Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper. Whether you are sitting the Core or Extended paper, mastering the three main solution methods – factorisation, the quadratic formula and completing the square – is essential for securing top marks.

二次方程几乎出现在每一份 IGCSE 数学试卷中。无论你参加 Core 还是 Extended 卷，掌握三种主要解法——因式分解法、二次公式和配方法——都是获得高分的关键。

1. Standard Form | 标准形式

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 highest power of x is 2, which is why it is called “quadratic” (from the Latin quadratus, meaning square).

二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数且 a ≠ 0。x 的最高次数是 2，这就是为什么它被称为”二次”（quadratic，源自拉丁语 quadratus，意为”平方”）。

For example, x² − 5x + 6 = 0 has a = 1, b = −5 and c = 6. The coefficient a is often 1 in exam questions, but you must also be comfortable with examples where a is any non-zero number, such as 2x² + 7x + 3 = 0.

例如，在 x² − 5x + 6 = 0 中，a = 1，b = −5，c = 6。考试题中 a 通常为 1，但你也必须熟悉 a 为其他非零数的情况，例如 2x² + 7x + 3 = 0。

Not all quadratic equations are given in standard form. Sometimes you must expand and rearrange first. For instance, (x − 3)(x + 2) = 4 expands to x² − x − 6 = 4, then subtracting 4 from both sides gives x² − x − 10 = 0. Always rearrange into the form ax² + bx + c = 0 before choosing a solution method.

并非所有二次方程都是以标准形式给出的。有时你需要先展开并移项。例如 (x − 3)(x + 2) = 4 展开得 x² − x − 6 = 4，两边减去 4 得 x² − x − 10 = 0。在选择解法之前，务必先将方程整理为 ax² + bx + c = 0 的形式。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when a = 1. To solve x² + bx + c = 0, look for two numbers that multiply to give c and add to give b.

当 a = 1 时，因式分解通常是最快的方法。要解 x² + bx + c = 0，找两个数，使它们相乘等于 c、相加等于 b。

Take x² − 5x + 6 = 0. The two numbers are −2 and −3, because (−2) × (−3) = 6 and (−2) + (−3) = −5. Hence the equation becomes (x − 2)(x − 3) = 0, so x = 2 or x = 3.

以 x² − 5x + 6 = 0 为例。这两个数是 −2 和 −3，因为 (−2) × (−3) = 6 且 (−2) + (−3) = −5。因此方程化为 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。

The key principle here is the zero product property: if the product of two factors is zero, then at least one of the factors must be zero. This is why setting each bracket equal to zero gives valid solutions.

这里的关键原理是零积性质：如果两个因式的乘积为零，那么至少有一个因式必须为零。这就是为什么将每个括号分别等于零就能得到有效的解。

When a ≠ 1, we recommend the “factor by grouping” method. For example, solve 2x² + 7x + 3 = 0. First multiply a and c: 2 × 3 = 6. Then find two numbers that multiply to 6 and add to 7: these are 6 and 1. Rewrite the middle term: 2x² + 6x + x + 3 = 0. Now group: 2x(x + 3) + 1(x + 3) = 0, which gives (2x + 1)(x + 3) = 0. Therefore x = −½ or x = −3.

当 a ≠ 1 时，推荐使用”分组分解法”。例如解 2x² + 7x + 3 = 0。先将 a 和 c 相乘：2 × 3 = 6。然后找两个相乘为 6、相加为 7 的数：它们是 6 和 1。重写中间项：2x² + 6x + x + 3 = 0。再分组：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。

3. Using the Quadratic Formula | 二次公式

Not every quadratic expression can be factorised easily, especially when the roots are irrational or involve decimals. The quadratic formula works for all quadratic equations and is given in the IGCSE formula booklet.

并非所有二次式都能轻松因式分解，尤其是当根为无理数或小数时。二次公式适用于所有二次方程，并且会在 IGCSE 公式册中给出。

x = (−b ± √(b² − 4ac)) / 2a

To use it, identify a, b and c carefully, substitute them into the formula, then simplify. Pay special attention to negative signs – this is where many marks are lost each year.

使用时，先仔细确定 a、b、c，将它们代入公式，然后化简。要特别注意负号——这是每年许多学生失分的地方。

Worked example: solve 2x² − 3x − 2 = 0. Here a = 2, b = −3 and c = −2. Substituting gives:

例题：解 2x² − 3x − 2 = 0。这里 a = 2，b = −3，c = −2。代入得：

x = (3 ± √(9 + 16)) / 4 = (3 ± 5) / 4

So x = (3 + 5)/4 = 2 or x = (3 − 5)/4 = −½. Notice that when b is negative, the “−b” in the formula becomes positive: −(−3) = 3. This is a common trap, so always double-check your substitution.

所以 x = (3 + 5)/4 = 2 或 x = (3 − 5)/4 = −½。注意当 b 为负数时，公式中的 “−b” 会变成正数：−(−3) = 3。这是一个常见的陷阱，所以务必检查你的代入过程。

4. Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This form is especially useful for finding the turning point of a graph and for solving equations that cannot be factorised.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。这种形式在求图像顶点以及解无法因式分解的方程时特别有用。

For an expression of the form x² + bx, take half of b and square it: add and subtract (b/2)². For example:

对于形如 x² + bx 的表达式，取 b 的一半并平方：加上并减去 (b/2)²。例如：

x² − 6x + 5 = [x² − 6x + 9] − 9 + 5 = (x − 3)² − 4

From this completed square form, the minimum point of the curve y = x² − 6x + 5 is (3, −4). We can also solve the equation: (x − 3)² − 4 = 0, so (x − 3)² = 4. Taking square roots gives x − 3 = ±2, hence x = 5 or x = 1. Remember the ± sign: it produces both solutions.

从这个配方的形式中，曲线 y = x² − 6x + 5 的最小值点为 (3, −4)。我们也可以解方程：(x − 3)² − 4 = 0，所以 (x − 3)² = 4。两边开平方得 x − 3 = ±2，因此 x = 5 或 x = 1。记住 ± 号：它产生两个解。

When a ≠ 1, factor a out of the x terms first. For example, 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 5. Here the vertex is (−2, −5).

当 a ≠ 1 时，先把 a 从含 x 的项中提出来。例如 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 5。此时顶点为 (−2, −5)。

5. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root sign: Δ = b² − 4ac. It tells us how many real roots a quadratic equation has without actually solving it.

判别式是二次公式中根号内的部分：Δ = b² − 4ac。它无需真正解方程就能告诉我们一个二次方程有多少个实数根。

If Δ > 0: two distinct real roots. The graph crosses the x-axis at two points.
如果 Δ > 0：两个不同的实数根。图像与 x 轴相交于两个点。
If Δ = 0: one repeated real root. The graph touches the x-axis at exactly one point.
如果 Δ = 0：一个重根。图像与 x 轴相切于一个点。
If Δ < 0: no real roots. The graph never meets the x-axis.
如果 Δ < 0：没有实数根。图像与 x 轴没有交点。

Example: for 3x² + 6x + 3 = 0, we have Δ = 36 − 36 = 0, so the equation has exactly one repeated root. Indeed, 3x² + 6x + 3 = 3(x + 1)², confirming that x = −1 is a repeated root.

例如：对于 3x² + 6x + 3 = 0，Δ = 36 − 36 = 0，所以方程只有一个重根。实际上 3x² + 6x + 3 = 3(x + 1)²，验证了 x = −1 是重根。

Exam questions often ask: “Find the values of k for which x² + kx + 9 = 0 has equal roots.” Since equal roots means Δ = 0, we set k² − 36 = 0, giving k = ±6. Both values are valid because symmetry means the parabola can be shifted left or right.

考试题经常这样问：”求使 x² + kx + 9 = 0 有等根的 k 值。”因为等根意味着 Δ = 0，我们令 k² − 36 = 0，得到 k = ±6。两个值都有效，因为抛物线的对称性意味着它可以向左或向右平移。

6. Roots and the Graph | 根与图像

The roots of the equation ax² + bx + c = 0 are exactly the x-intercepts of the curve y = ax² + bx + c. This connection

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Solving Quadratic Equations | 解二次方程