Solving Quadratic Equations | 二次方程求解

📚 Solving Quadratic Equations | 二次方程求解

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. Mastering them not only scores you marks directly but also builds a foundation for functions, graphs, and coordinate geometry.

二次方程是 IGCSE 数学中最常考的知识点之一。掌握它不仅能直接获得分数，更是学习函数、图像和解析几何的基础。

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1. Standard Form of a Quadratic Equation | 二次方程的标准形式

Any quadratic equation can be written in the standard form:

ax² + bx + c = 0

Here, a, b and c are constants, with a ≠ 0. The coefficient a determines the curvature of the associated parabola, while b and c affect its position on the coordinate plane.

任何二次方程都可以写成标准形式：

ax² + bx + c = 0

其中 a、b、c 为常数，且 a ≠ 0。系数 a 决定对应抛物线的弯曲程度，b 与 c 则影响其在坐标平面上的位置。

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2. Solving by Factorisation | 因式分解法

When the quadratic expression can be factored into two linear brackets, the solutions are found by setting each factor equal to zero. For example, consider x² − 5x + 6 = 0. This can be written as (x − 2)(x − 3) = 0.

当二次表达式可分解为两个一次因式相乘时，令每个因式等于零即可求出解。例如，考虑 x² − 5x + 6 = 0，可写成 (x − 2)(x − 3) = 0。

According to the zero product property, if the product of two expressions is zero, then at least one of them must be zero. Therefore, x − 2 = 0 or x − 3 = 0, giving two roots: x = 2 and x = 3.

根据零乘积性质，若两个表达式的乘积为零，则其中至少有一个为零。因此 x − 2 = 0 或 x − 3 = 0，得到两个根：x = 2 和 x = 3。

• Always rearrange the equation into standard form before factoring.
• Check whether the coefficient of x² is greater than 1 — if so, employ the ac-method or trial and error.

• 在因式分解前，务必将方程整理成标准形式。
• 检查 x² 的系数是否大于 1 — 若是，采用 ac 法或试错法。

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3. The Quadratic Formula | 求根公式

When factorisation is not straightforward, the quadratic formula provides a universal solution. For ax² + bx + c = 0:

因式分解不便时，求根公式提供了通用解法。对于 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / (2a)

This formula is derived by completing the square on the general quadratic equation. It always produces correct real or complex roots, provided you substitute the values accurately.

该公式通过对一般二次方程配方推导得出。只要代入准确，它总能给出正确的实数或复数根。

For instance, solve 2x² + 3x − 2 = 0 using the formula. Here a = 2, b = 3, c = −2. Substitute:

例如，用公式求解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入：

x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √25) / 4

Thus x = (−3 + 5) / 4 = 1/2, or x = (−3 − 5) / 4 = −2.

因此 x = (−3 + 5) / 4 = 1/2，或 x = (−3 − 5) / 4 = −2。

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4. Completing the Square | 配方法

Completing the square is a powerful technique, especially when the coefficient of x² is 1. You rewrite the expression in the form (x + p)² + q. For example, x² + 6x + 5 = 0 becomes (x + 3)² − 9 + 5 = 0, then (x + 3)² = 4.

配方法是一项强大的技巧，尤其在 x² 的系数为 1 时。将表达式改写成 (x + p)² + q 的形式。例如，x² + 6x + 5 = 0 化为 (x + 3)² − 9 + 5 = 0，随后 (x + 3)² = 4。

Taking the square root of both sides yields x + 3 = ±2, hence x = −1 or x = −5. In general, divide b by 2 to find p, and adjust the constant term accordingly.

两边开平方得到 x + 3 = ±2，因此 x = −1 或 x = −5。一般步骤是：将 b 除以 2 得到 p，并相应地调整常数项。

This method also reveals the turning point of the parabola: the vertex is located at (−p, q) when the expression is written as (x + p)² + q.

配方法还能揭示抛物线的顶点：当表达式写成 (x + p)² + q 时，顶点坐标为 (−p, q)。

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5. The Discriminant | 判别式

The expression underneath the square root in the quadratic formula is called the discriminant:

求根公式中被开方数的表达式称为判别式：

Δ = b² − 4ac

It tells us the nature of the roots without solving the full equation:

它能在不解完整方程的情况下告诉我们根的性质：

Discriminant value | 判别式取值	Number and type of roots | 根的个数与类型
Δ > 0	Two distinct real roots | 两个不相等的实数根
Δ = 0	One repeated real root | 一个相等实数根（重根）
Δ < 0	No real roots | 无实数根

For example, x² + 9 = 0 has Δ = 0 − 36 = −36 < 0, so it has no real roots. Such a graph never crosses the x-axis.

例如，x² + 9 = 0 中 Δ = 0 − 36 = −36 < 0，所以没有实数根。其图像不与 x 轴相交。

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6. Roots and Coefficients | 根与系数的关系

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

对于具有根 α 和 β 的二次方程 ax² + bx + c = 0，成立如下关系：

α + β = −b / a, αβ = c / a

These relations are extremely useful when checking solutions or constructing equations from given roots. If the roots are 2 and 3, then the equation can be built as x² − (2 + 3)x + (2 × 3) = 0, i.e. x² − 5x + 6 = 0.

这些关系式在检验求解结果或根据已知根构造方程时极为有用。若根为 2 和 3，则可构造方程 x² − (2 + 3)x + (2 × 3) = 0，即 x² − 5x + 6 = 0。

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7. Solving by Graphs | 图像法求解

Graphically, the roots of a quadratic equation correspond to the x-coordinates where the parabola y = ax² + bx + c crosses the x-axis. If the parabola touches the axis at one point, there is exactly one repeated root; if it does not touch the axis, there are no real roots.

从图像上看，二次方程的根对应于抛物线 y = ax² + bx + c 与 x 轴交点的 x 坐标。若抛物线与 x 轴相切于一点，则恰有一个重根；若不相交，则无实数根。

When examining the graph, the vertex represents the maximum or minimum value of the quadratic function. This is especially important in optimisation problems.

观察图像时，顶点代表二次函数的最大值或最小值。这在最优化问题中尤其重要。

In examination settings, you may be asked to estimate roots from a given sketch or to identify how many solutions exist based on the position of the turning point.

在考试中，题目可能要求你从给定草图估计根，或根据顶点位置判断解的个数。

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8. Common Pitfalls | 常见易错点

Students often lose marks through avoidable mistakes. Here are the most frequent ones:

学生常因可避免的错误失分。以下是最高频的几类：

• Forgetting to rearrange the equation to equal zero before applying any method.
• Misplacing signs when substituting into the quadratic formula.
• Dividing both sides by a variable that might be zero — a typical but serious mistake.
• Writing only one solution when two roots exist.

• 在采用任何方法前，忘记将方程整理为等于零的形式。
• 代入求根公式时符号出错。
• 方程两边同时除以可能为零的变量 — 这是常见但严重的错误。
• 当存在两个根时只写出一个解。

Always test your solutions by substituting them back into the original equation. This single step can save precious marks.

务必通过代回原方程检验解的正确性。这一步就能挽救宝贵的分数。

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9. Exam-Style Questions | 典型考题

Here is a typical exam-style question: Solve x² − 7x + 10 = 0. First, factorise: (x − 2)(x − 5) = 0. Hence x = 2 or x = 5. Notice how factorisation provides the quickest route here.

以下是一道典型考题：解 x² − 7x + 10 = 0。先因式分解：(x − 2)(x − 5) = 0。因此 x = 2 或 x = 5。注意因式分解在此题中是最快的路径。

Another common type: find the value of k for which the equation kx² + 4x + 1 = 0 has exactly one solution. Since there is one repeated root, set the discriminant to zero: 4² − 4 × k × 1 = 0, so 16 − 4k = 0, giving k = 4.

另一类常见题型：求常数 k 的值，使方程 kx² + 4x + 1 = 0 有唯一解。由于只有一个重根，令判别式等于零：4² − 4 × k × 1 = 0，即 16 − 4k = 0，解得 k = 4。

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10. Revision Strategy | 复习策略

To perform well, practise each method until fluency is achieved. Start with factorisation, then move to completing the square, and finally master the quadratic formula. Use the discriminant to predict the nature of roots quickly.

要想考得好，须将每种方法练至熟练。从因式分解开始，再到配方法，最后熟练掌握求根公式。运用判别式快速预判根的性质。

Work through past papers under timed conditions, because speed and accuracy are both scored. Model the thinking process aloud at first, then attempt full papers silently.

在限时条件下做历年真题，因为速度和准确性同等计入得分。起初可以边做边讲解思路，然后逐步过渡到全卷默做。

Finally, keep a dedicated notebook for the mistakes you make in practice. Review it weekly until those patterns are firmly corrected in your mind.

最后，准备一本专门的错题本，记录练习中的错误。每周复习一次，直至这些错误模式在脑海中彻底纠正。

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By mastering these ten areas, you can approach any quadratic equation problem with confidence and precision.

掌握以上十个板块，你就能自信且精准地处理任何二次方程问题。

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