Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in many areas of IGCSE Mathematics, from factorisation to graph sketching. This revision guide will help you master every key method step by step.

二次方程在 IGCSE 数学中无处不在，从因式分解到图像草图都会涉及。本复习指南将帮助你逐步掌握每一种关键方法。

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

A quadratic equation can be written in the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The highest power of the variable x is 2.

二次方程可以写成 ax² + bx + c = 0 的形式，其中 a、b、c 是实数，且 a ≠ 0。变量 x 的最高次数是 2。

For example, 2x² − 5x + 3 = 0 is a quadratic equation. Here a = 2, b = −5, c = 3.

例如，2x² − 5x + 3 = 0 是一个二次方程。这里 a = 2，b = −5，c = 3。

If a = 0, the equation becomes linear, so it is not quadratic. Always check that the coefficient of x² is not zero.

如果 a = 0，方程就变成一次方程，因此它不是二次的。始终要检查 x² 的系数不为零。

2. Solving by Factorisation | 用因式分解法求解

Factorisation is often the fastest method when the quadratic factors neatly into two binomials. The key idea is to use the zero product property: if A × B = 0, then A = 0 or B = 0.

因式分解通常是当二次式能整洁地分解为两个二项式时最快捷的方法。核心思想是零乘积性质：如果 A × B = 0，那么 A = 0 或 B = 0。

Example: Solve x² − 7x + 12 = 0.

例：解 x² − 7x + 12 = 0。

Find two numbers that multiply to 12 and add to −7. They are −3 and −4. So x² − 7x + 12 = (x − 3)(x − 4) = 0.

找出两个数，它们相乘得 12，相加得 −7。这两个数是 −3 和 −4。因此 x² − 7x + 12 = (x − 3)(x − 4) = 0。

x = 3 or x = 4

x = 3 或 x = 4

Always expand your factors to verify the original expression.

始终展开你的因式以验证原表达式。

3. Factorising with a Leading Coefficient | 处理首项系数不为 1 的因式分解

When a ≠ 1, factorisation requires more care. For example, solve 2x² + 7x + 3 = 0.

当 a ≠ 1 时，因式分解需要更加小心。例如，解 2x² + 7x + 3 = 0。

Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: 1 and 6.

将 a 和 c 相乘：2 × 3 = 6。找出两个数，相乘得 6，相加得 7：1 和 6。

Rewrite the middle term: 2x² + x + 6x + 3 = 0.

重写中间项：2x² + x + 6x + 3 = 0。

Group and factor: x(2x + 1) + 3(2x + 1) = 0, so (x + 3)(2x + 1) = 0.

分组并因式分解：x(2x + 1) + 3(2x + 1) = 0，因此 (x + 3)(2x + 1) = 0。

x = −3 or x = −½

x = −3 或 x = −½

This grouping method works for any quadratic that factors over integers.

这种分组方法适用于任何能在整数范围内因式分解的二次方程。

4. Using the Quadratic Formula | 使用求根公式

The quadratic formula solves any quadratic equation, even when factorisation is difficult or impossible. For ax² + bx + c = 0:

求根公式可以解任何二次方程，即使因式分解困难或不可行。对于 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / (2a)

Example: Solve 3x² − 5x − 2 = 0 using the formula.

例：用公式解 3x² − 5x − 2 = 0。

Here a = 3, b = −5, c = −2. Substitute:

这里 a = 3，b = −5，c = −2。代入：

x = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6

So x = 2 or x = −⅓. Always simplify the square root fully.

因此 x = 2 或 x = −⅓。始终要完全化简平方根。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form p(x + q)² + r. This is especially useful for finding turning points and solving equations.

配方法将二次式改写为 p(x + q)² + r 的形式。这对于求顶点和解方程特别有用。

Example: Solve x² + 6x − 7 = 0 by completing the square.

例：用配方法解 x² + 6x − 7 = 0。

Take half of 6, square it: (6/2)² = 9. Rewrite:

取 6 的一半，再平方：(6/2)² = 9。改写：

(x + 3)² − 9 − 7 = 0

Therefore (x + 3)² = 16, so x + 3 = ±4.

因此 (x + 3)² = 16，所以 x + 3 = ±4。

x = 1 or x = −7

x = 1 或 x = −7

For a ≠ 1, factor out a first: a(x + b/2a)² + c − b²/4a.

当 a ≠ 1 时，先提出 a：a(x + b/2a)² + c − b²/4a。

6. The Discriminant | 判别式

The discriminant Δ = b² − 4ac tells us how many real roots a quadratic equation has.

判别式 Δ = b² − 4ac 告诉我们二次方程有多少个实数根。

If Δ > 0: two distinct real roots.

如果 Δ > 0：两个不同的实数根。
If Δ = 0: one repeated real root.

如果 Δ = 0：一个重根（两个相等的实数根）。
If Δ < 0: no real roots.

如果 Δ < 0：没有实数根。

Example: For 2x² − 3x + 5 = 0, Δ = (−3)² − 4×2×5 = 9 − 40 = −31 < 0, so there are no real roots.

例：对于 2x² − 3x + 5 = 0，Δ = (−3)² − 4×2×5 = 9 − 40 = −31 < 0，因此没有实数根。

The discriminant is also useful when a quadratic equation has a parameter, such as k, and we want a condition for the number of roots.

判别式在方程含参数（如 k）并且我们需要根的个数条件时也非常有用。

7. Graphs of Quadratic Functions | 二次函数的图像

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upward; if a < 0, it opens downward.

y = ax² + bx + c 的图像是抛物线。如果 a > 0，开口向上；如果 a < 0，开口向下。

The x-intercepts are the real roots of the equation, if they exist. The y-intercept is c.

与 x 轴的交点就是方程的实数根（如果存在）。与 y 轴的交点是 c。

The axis of symmetry is the vertical line x = −b/(2a), and the turning point lies on this axis.

对称轴是竖直线 x = −b/(2a)，顶点位于这条对称轴上。

Vertex: ( −b/(2a) , c − b²/(4a) )

顶点：( −b/(2a) , c − b²/(4a) )

Sketching the graph requires the roots, the y-intercept and the vertex.

画图需要知道根、y 轴截距和顶点。

8. Roots and Turning Points | 根与顶点

If a quadratic is written in completed square form y = p(x + q)² + r, then the vertex is at (−q, r). The value of p tells us the direction and width of the parabola.

如果二次函数写成完全平方形式 y = p(x + q)² + r，那么顶点在 (−q, r)。p 的值告诉我们抛物线的开口方向和宽度。

Example: y = 2(x − 3)² − 8 has vertex at (3, −8) and opens upward. Setting y = 0:

例：y = 2(x − 3)² − 8 的顶点在 (3, −8)，开口向上。令 y = 0：

2(x − 3)² = 8 ⇒ (x − 3)² = 4 ⇒ x = 5 or x = 1

2(x − 3)² = 8 ⇒ (x − 3)² = 4 ⇒ x = 5 或 x = 1

The roots are 1 and 5, and they are symmetric about x = 3.

根是 1 和 5，它们关于 x = 3 对称。

9. Word Problems with Quadratics | 二次方程应用题

Many IGCSE problems translate real-life situations into quadratic equations. For example, the area of a rectangle with length (x + 4) and width (x − 1) is 30 cm².

许多 IGCSE 题目将现实情境转化为二次方程。例如，一个长方形，长为 (x + 4)，宽为 (x − 1)，面积为 30 cm²。

Set up the equation: (x + 4)(x − 1) = 30.

建立方程：(x + 4)(x − 1) = 30。

Expand and rearrange: x² + 3x − 4 = 30 ⇒ x² + 3x − 34 = 0.

展开并移项：x² + 3x − 4 = 30 ⇒ x² + 3x − 34 = 0。

Use the quadratic formula to solve, then reject any negative value if length must be positive.

用求根公式求解，如果长度必须为正，则舍去负值。

Always check that your answer makes sense in the original context.

始终检查你的答案在原始情境中是否合理。

10. Common Mistakes | 常见错误

Students often make avoidable errors when solving quadratic equations. Here are the most common ones.

学生在解二次方程时常犯一些可以避免的错误。以下是最常见的几种。

Forgetting to set the equation to zero before factorising or using the formula.

在因式分解或使用公式前忘记将方程化为等于零的形式。
Miscalculating the discriminant: remember b² − 4ac, not b² − 4a.

计算判别式出错：记住是 b² − 4ac，而不是 b² − 4a。
Losing the ± sign in the quadratic formula.

在求根公式中丢掉 ± 号。
Incorrectly simplifying √(p²) as p instead of |p|.

错误地将 √(p²) 简化为 p 而不是 |p|。
Dividing both sides of an equation by x when x might be zero.

当 x 可能为零时，两边同除以 x。

To avoid these, always write out every step and test your solutions by substituting back.

为避免这些错误，务必写出每一步，并通过代回检验你的解。

11. Practice Questions | 练习题目

Try these questions before checking the answers.

先尝试以下题目，再核对答案。

Solve x² − 9x + 20 = 0.

解 x² − 9x + 20 = 0。
Solve 3x² + 5x − 2 = 0 using the quadratic formula.

用求根公式解 3x² + 5x − 2 = 0。
Complete the square for x² − 4x + 1, then solve.

对 x² − 4x + 1 配方，然后求解。
Find the discriminant of 2x² − 4x + 5 and state the number of real roots.

求 2x² − 4x + 5 的判别式，并说明实数根的个数。

Answers: 1) x = 5 or x = 4. 2) x = ⅓ or x = −2. 3) (x − 2)² − 3, so x = 2 ± √3. 4) Δ = −24, no real roots.

答案：1) x = 5 或 x = 4。2) x = ⅓ 或 x = −2。3) (x − 2)² − 3，所以 x = 2 ± √3。4) Δ = −24，没有实数根。

12. Summary | 总结

Quadratic equations can be solved by factorisation, the quadratic formula, or completing the square. The discriminant predicts the number of real roots, and the graph of a quadratic is a parabola with a clear vertex.

二次方程可以通过因式分解、求根公式或配方法求解。判别式预测实数根的个数，二次函数的图像是带有明确顶点的抛物线。

For IGCSE Edexcel, you must be fluent in all methods and know when to use each one. Practice with past papers to build speed and accuracy.

对于 IGCSE Edexcel，你必须熟练掌握所有方法，并知道何时使用哪一种。通过练习历年真题来提高速度和准确性。

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Solving Quadratic Equations | 解二次方程