Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear throughout the IGCSE Mathematics syllabus, whether in pure algebra, graph sketching, or real-life problem solving. In the Edexcel IGCSE (4MA1) examination, students are expected to solve quadratic equations confidently using several methods, interpret the discriminant, and link solutions to graphs.

一元二次方程贯穿 IGCSE 数学的全部内容，无论是纯代数、图像绘制还是实际应用问题。在 Edexcel IGCSE（4MA1）考试中，学生需要熟练掌握解一元二次方程的多种方法，理解判别式的意义，并将解与图像联系起来。

1. What is a Quadratic Equation? | 什么是一元二次方程？

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its general form is:

一元二次方程是最高次数为 2 的多项式方程，即变量的最高次幂为 2。它的一般形式为：

ax² + bx + c = 0, where a ≠ 0

Here a, b, and c are constants, and x is the unknown variable. If a = 0, the equation becomes linear, not quadratic. For example, x² – 4x + 3 = 0 is a quadratic equation, while 2x + 1 = 0 is not.

其中 a、b、c 为常数，x 为未知数。若 a = 0，则方程变为一次方程，而非二次方程。例如，x² – 4x + 3 = 0 是一元二次方程，而 2x + 1 = 0 不是。

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

Before solving, you must be fluent in expanding double brackets and factorising quadratics. Expanding (x + 2)(x + 3) gives x² + 5x + 6. Factorising is the reverse process: x² + 5x + 6 = (x + 2)(x + 3).

在求解之前，你需要熟练展开双括号以及对二次式进行因式分解。展开 (x + 2)(x + 3) 得到 x² + 5x + 6。因式分解是相反的过程：x² + 5x + 6 = (x + 2)(x + 3)。

For a quadratic in the form x² + bx + c, look for two numbers that multiply to give c and add to give b. For example, in x² – 7x + 12, the numbers -3 and -4 multiply to 12 and add to -7, so x² – 7x + 12 = (x – 3)(x – 4).

对于形式为 x² + bx + c 的二次式，需要找到两个数，它们相乘等于 c，相加等于 b。例如，在 x² – 7x + 12 中，-3 和 -4 相乘等于 12，相加等于 -7，因此 x² – 7x + 12 = (x – 3)(x – 4)。

When the coefficient of x² is not 1, such as 2x² + 7x + 3, you may need to factorise by grouping or use trial and improvement. Here, 2x² + 7x + 3 = (2x + 1)(x + 3).

当 x² 系数不为 1 时，例如 2x² + 7x + 3，你可能需要使用分组分解法或试错法。这里，2x² + 7x + 3 = (2x + 1)(x + 3)。

3. Solving by Factorising | 用因式分解法求解

Once the quadratic is factorised, use the zero product property: if the product of two expressions is zero, then at least one of them must be zero.

一旦二次式完成了因式分解，就可以使用零积性质：如果两个表达式的乘积为零，那么其中至少有一个必须为零。

For example, solve x² – 5x + 6 = 0. Factorise to get (x – 2)(x – 3) = 0. Then set each factor to zero: x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.

例如，解 x² – 5x + 6 = 0。因式分解得到 (x – 2)(x – 3) = 0。然后令每个因式为零：x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。

Always rearrange the equation into the standard form ax² + bx + c = 0 before factorising. For instance, x² = 9 – 2x must first be rewritten as x² + 2x – 9 = 0. This quadratic does not factorise nicely, so another method is needed.

在因式分解之前，务必先将方程整理成标准形式 ax² + bx + c = 0。例如，x² = 9 – 2x 必须首先改写为 x² + 2x – 9 = 0。这个二次式不能方便地因式分解，因此需要使用其他方法。

4. Solving by Completing the Square | 用配方法求解

Completing the square rewrites a quadratic as (x + p)² + q. This method is useful when factorising is difficult and also helps find the turning point of a graph.

配方法将二次式改写为 (x + p)² + q 的形式。当因式分解困难时，此方法非常有用，同时也有助于求图像的最高点或最低点。

For x² + 6x + 7, take half of 6, which is 3, and write (x + 3)² – 9 + 7 = (x + 3)² – 2. Therefore, x² + 6x + 7 = (x + 3)² – 2.

对于 x² + 6x + 7，取 6 的一半为 3，写成 (x + 3)² – 9 + 7 = (x + 3)² – 2。因此，x² + 6x + 7 = (x + 3)² – 2。

To solve (x + 3)² – 2 = 0, add 2 to both sides: (x + 3)² = 2. Then take the square root of both sides: x + 3 = ±√2. Hence x = -3 ± √2.

解方程 (x + 3)² – 2 = 0，两边加 2： (x + 3)² = 2。然后两边开平方：x + 3 = ±√2。因此 x = -3 ± √2。

When the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 5, write 2(x² + 4x) + 5, then complete the square inside the bracket: 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3.

当 x² 系数不为 1 时，先将其提出。对于 2x² + 8x + 5，写成 2(x² + 4x) + 5，然后在括号内配方：2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3。

5. The Quadratic Formula | 二次公式

The quadratic formula provides a direct way to solve any quadratic equation. For ax² + bx + c = 0, the solutions are given by:

二次公式为解任何一元二次方程提供了直接的方法。对于 ax² + bx + c = 0，其解为：

x = (-b ± √(b² – 4ac)) / (2a)

This formula works for all quadratics, even when factorisation is impossible. You should memorise it and be able to substitute values accurately.

这个公式适用于所有二次方程，即使无法因式分解也能求解。你应该记住它，并能够准确代入数值。

Example: Solve 2x² – 4x – 6 = 0. Here a = 2, b = -4, c = -6. Substitute into the formula:

例：解 2x² – 4x – 6 = 0。这里 a = 2，b = -4，c = -6。代入公式：

x = (4 ± √((-4)² – 4(2)(-6))) / (2(2)) = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4

So x = (4 + 8)/4 = 3 or x = (4 – 8)/4 = -1. The solutions are x = 3 and x = -1.

因此 x = (4 + 8)/4 = 3 或 x = (4 – 8)/4 = -1。解为 x = 3 和 x = -1。

6. The Discriminant | 判别式

The expression b² – 4ac under the square root in the quadratic formula is called the discriminant, often denoted by Δ (delta). It determines the number and nature of the roots.

在二次公式中根号下的表达式 b² – 4ac 称为判别式，通常用 Δ（大写德尔塔）表示。它决定了根的数量和性质。

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 exactly one real root (a repeated root).

若 Δ = 0，方程只有一个实数根（重根）。
If Δ < 0, the equation has no real roots (two complex roots).

若 Δ < 0，方程没有实数根（有两个复数根）。

For example, in x² – 6x + 9 = 0, Δ = (-6)² – 4(1)(9) = 36 – 36 = 0, so the graph touches the x-axis at one point. The repeated root is x = 3.

例如，在 x² – 6x + 9 = 0 中，Δ = (-6)² – 4(1)(9) = 36 – 36 = 0，因此图像与 x 轴只有一个交点。重根为 x = 3。

7. Sketching Quadratic Graphs | 绘制二次函数图像

The graph of y = ax² + bx + c is a parabola. Its shape depends on the sign of a. If a > 0, the parabola opens upwards (a U-shape). If a < 0, it opens downwards (an n-shape).

y = ax² + bx + c 的图像是一条抛物线。其开口方向由 a 的符号决定。若 a > 0，抛物线开口向上（U 形）。若 a < 0，开口向下（n 形）。

To sketch the graph, find the roots by solving the equation, the y-intercept (when x = 0), and the turning point. The points where the graph crosses the x-axis correspond to the solutions of the quadratic equation.

绘制图像时，需要先通过解方程求出根、y 截距（即 x = 0 时的值）以及顶点。图像与 x 轴的交点就是二次方程的解。

Example: For y = x² – 4x + 3, the roots are x = 1 and x = 3 (since (x – 1)(x – 3) = 0). The y-intercept is 3. The axis of symmetry lies halfway between the roots, at x = 2.

例如，对于 y = x² – 4x + 3，根为 x = 1 和 x = 3（因为 (x – 1)(x – 3) = 0）。y 截距为 3。对称轴位于两个根的正中间，即 x = 2。

8. Turning Point and Axis of Symmetry | 顶点与对称轴

The turning point of a quadratic graph is its minimum (if a > 0) or maximum (if a < 0). The x-coordinate of the turning point is given by x = -b / (2a).

二次函数图像的顶点是它的最低点（若 a > 0）或最高点（若 a < 0）。顶点的 x 坐标为 x = -b / (2a)。

You can find the y-coordinate by substituting this x-value back into the equation. For example, for y = 2x² – 8x + 7, a = 2, b = -8, so x = -(-8) / (2 × 2) = 8/4 = 2. Then y = 2(2)² – 8(2) + 7 = 8 – 16 + 7 = -1. The turning point is (2, -1).

然后把这个 x 值代回原方程即可求出 y 坐标。例如，对于 y = 2x² – 8x + 7，a = 2，b = -8，所以 x = -(-8) / (2 × 2) = 8/4 = 2。然后 y = 2(2)² – 8(2) + 7 = 8 – 16 + 7 = -1。顶点为 (2, -1)。

Alternatively, completing the square gives the turning point directly. If y = (x – h)² + k, then the turning point is (h, k).

另外，配方法可以直接得出顶点。如果 y = (x – h)² + k，则顶点为 (h, k)。

9. Quadratic Inequalities | 二次不等式

Solving quadratic inequalities involves finding ranges of x that make the expression positive or negative. First solve the corresponding quadratic equation, then sketch the parabola to decide the range.

解二次不等式需要求出使表达式为正或为负的 x 的取值范围。首先解对应的二次方程，然后画出抛物线来确定范围。

Example: Solve x² – 3x + 2 < 0. First solve x² - 3x + 2 = 0, which factorises to (x - 1)(x - 2) = 0, so x = 1 or x = 2. Since the parabola opens upwards, the expression is negative between the roots. Therefore the solution is 1 < x < 2.

例：解 x² – 3x + 2 < 0。首先解 x² - 3x + 2 = 0，因式分解为 (x - 1)(x - 2) = 0，所以 x = 1 或 x = 2。由于抛物线开口向上，表达式在两根之间为负。因此解为 1 < x < 2。

For x² – 2x – 8 > 0, the roots are x = -2 and x = 4. The graph opens upwards, so the expression is positive to the left of -2 and to the right of 4. The solution is x < -2 or x > 4.

对于 x² – 2x – 8 > 0，根为 x = -2 和 x = 4。图像开口向上，因此表达式在 -2 的左侧和 4 的右侧为正。解为 x < -2 或 x > 4。

Remember to use ≤ or ≥ when the problem asks for “greater than or equal to” or “less than or equal to”, and include the roots as endpoints accordingly.

当题目要求“大于等于”或“小于等于”时，使用 ≤ 或 ≥，并将根作为端点包含在内。

10. Word Problems Leading to Quadratics | 二次方程应用题

Many real-world problems can be modelled by quadratic equations. Common examples include calculating areas, projectile motion, and number problems.

许多实际问题可以用二次方程建模。常见类型包括面积计算、抛体运动和数字问题。

Example: A rectangular garden has length 3 metres more than its width. Its area is 40 m². Let the width be x metres. Then the length is x + 3, and the equation is x(x + 3) = 40, so x² + 3x – 40 = 0.

例：一个矩形花园的长比宽多 3 米，面积为 40 平方米。设宽为 x 米，则长为 x + 3，方程为 x(x + 3) = 40，即 x² + 3x – 40 = 0。

Factorise: (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since width cannot be negative, x = 5. The width is 5 m and the length is 8 m.

因式分解： (x + 8)(x – 5) = 0，所以 x = -8 或 x = 5。由于宽不可能为负，因此 x = 5。宽为 5 米，长为 8 米。

Always check your solutions in the original problem and discard any that are impossible (such as negative lengths or times).

务必在原始问题中检验解，并舍弃任何不可能的答案（例如负长度或负时间）。

11. Common Mistakes and Tips | 常见错误与技巧

Students often lose marks on quadratic equations due to small errors. Here are some tips to avoid them.

学生在解二次方程时常常因为小错误而丢分。以下是一些避免错误的技巧。

Always rearrange the equation to the form ax² + bx + c = 0 before factorising or using the quadratic formula.

在因式分解或使用二次公式之前，务必先将方程整理成 ax² + bx + c = 0 的形式。
When using the quadratic formula, carefully substitute negative values for b and c. Write down all steps.

使用二次公式时，仔细代入 b 和 c 的负值。写下所有步骤。
Do not forget the ± sign when taking square roots in completing the square.

在配方法中开平方时，不要忘记正负号 ±。
If the discriminant is negative, state that there are no real roots.

如果判别式为负，需说明没有实数根。
In word problems, reject extraneous solutions that do not make sense in the real-world context.

在应用题中，舍弃在实际情境中没有意义的增根。

Mastering quadratic equations is essential for IGCSE Mathematics success. Practice each method until you can choose the most efficient one for any given equation. Remember: factorise when possible, complete the square for vertex problems, and rely on the quadratic formula as a universal tool.

掌握一元二次方程对于 IGCSE 数学的成功至关重要。反复练习每种方法，直到你能针对任何给定方程选择最有效的方法。记住：能因式分解时优先分解，求顶点时用配方法，而二次公式则是万能工具。

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Solving Quadratic Equations | 解一元二次方程