Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, geometry, and even calculus at higher levels. Mastering the techniques to solve them will give you a strong foundation for all future mathematics.

二次方程是 IGCSE 数学中最重要的话题之一。它出现在代数、图像、几何甚至更高层次的微积分中。掌握解二次方程的方法，将为你今后学习所有数学内容打下坚实的基础。

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

A quadratic equation is an equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The highest power of the variable x is 2, which is why it is called "quadratic" — from the Latin word "quadratus", meaning square.

二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a ≠ 0。变量 x 的最高次数是 2，因此被称为 "二次"，源自拉丁语 "quadratus"，意为 "平方"。

se the highest power is 3. In IGCSE, you will usually deal with equations where a, b and c are integers or simple fractions.

例如：x² − 5x + 6 = 0 是一个二次方程，其中 a = 1，b = −5，c = 6。但 x³ − 2x = 0 不是二次方程，因为它最高次数是 3。在 IGCSE 中，我们通常处理 a、b、c 为整数或简单分数的方程。

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2. Solving by Factorisation | 因式分解法

le is simple: if p × q = 0, then p = 0 or q = 0. This is called the zero product property.

因式分解通常是当二次式能够写成两个一次因式乘积时最快捷的方法。原理很简单：如果 p × q = 0，那么 p = 0 或 q = 0。这称为零乘积性质。

. So x² − 5x + 6 = (x − 2)(x − 3) = 0. Then x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

我们用因式分解法解 x² − 5x + 6 = 0。寻找两个数，它们相乘得 6，相加得 −5。这两个数是 −2 和 −3。因此 x² − 5x + 6 = (x − 2)(x − 3) = 0。于是 x − 2 = 0 或 x − 3 = 0，得到 x = 2 或 x = 3。

. If the coefficient of x² is not 1, for example 2x² + 5x + 2 = 0, you may need to use the "ac method" or trial and error: (2x + 1)(x + 2) = 0, giving x = −½ or x = −2.

记住平方差公式的特殊情况：x² − 9 = (x + 3)(x − 3) = 0，所以 x = ±3。如果 x² 的系数不为 1，例如 2x² + 5x + 2 = 0，你可能需要使用 "ac 方法" 或试错法：(2x + 1)(x + 2) = 0，得到 x = −½ 或 x = −2。

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3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This is useful for solving equations and also for finding the turning point of a curve. To complete the square for x² + bx, add and subtract (b/2)².

配方法将二次式改写成 (x + p)² + q 的形式。这对于解方程以及求曲线顶点非常有用。对于 x² + bx，需要加上并减去 (b/2)²。

= 0, that is (x + 3)² − 1 = 0. Then (x + 3)² = 1, so x + 3 = ±1, giving x = −2 or x = −4.

例：解 x² + 6x + 8 = 0。首先，x² + 6x = (x + 3)² − 9。因此方程变为 (x + 3)² − 9 + 8 = 0，即 (x + 3)² − 1 = 0。于是 (x + 3)² = 1，所以 x + 3 = ±1，得到 x = −2 或 x = −4。

. Used this way, completing the square always works, even when factorisation is difficult.

对于 ax² + bx + c，可以先提取 a。例如 2x² − 12x + 10 = 0 变为 2(x² − 6x) + 10 = 0，然后 2[(x − 3)² − 9] + 10 = 0，最终得到 x = 1 或 x = 5。这样使用配方法总是有效，即使因式分解很困难。

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4. The Quadratic Formula | 求根公式

that solves any quadratic equation. If ax² + bx + c = 0, then the solutions are given by:

求根公式是解任意二次方程的通用方法。若 ax² + bx + c = 0，则解由下式给出：

x = (−b ± √b² − 4ac) / 2a

= 2, b = 3, c = −5. First calculate b² − 4ac = 9 − 4(2)(−5) = 9 + 40 = 49. Then x = (−3 ± √49) / 4 = (−3 ± 7) / 4. So x = (4)/4 = 1 or x = (−10)/4 = −2.5.

这个公式在 IGCSE 公式表中提供，但你必须知道如何正确应用。我们来解 2x² + 3x − 5 = 0。其中 a = 2，b = 3，c = −5。先计算 b² − 4ac = 9 − 4(2)(−5) = 9 + 40 = 49。然后 x = (−3 ± √49) / 4 = (−3 ± 7) / 4。因此 x = 4/4 = 1 或 x = −10/4 = −2.5。

ndle a negative c; substitute carefully with brackets, for example −4ac = −4(2)(−5) = +40.

在考试中一定要一步一步写出公式。常见错误是忘记 ± 号，从而丢失一个根。另一个错误是误处理负的 c；代入时要小心使用括号，例如 −4ac = −4(2)(−5) = +40。

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5. The Discriminant | 判别式

root. It tells us how many real roots a quadratic equation has, without solving it fully.

判别式是根号内的表达式 Δ = b² − 4ac。它告诉我们二次方程有多少个实数根，而无需完整求解。

• If Δ > 0, the equation has two distinct real roots — the graph crosses the x-axis at two points.

  若 Δ > 0，方程有两个不同的实数根——图像与 x 轴相交于两点。

• If Δ = 0, the equation has exactly one real root (a repeated root) — the graph touches the x-axis at one point.

  若 Δ = 0，方程恰好有一个实数根（重根）——图像与 x 轴相切于一点。

• If Δ < 0, the equation has no real roots — the graph does not cross the x-axis.

  若 Δ < 0，方程没有实数根——图像不与 x 轴相交。

s no real solutions. The discriminant is frequently tested both for solving and for graph questions, so memorise the three cases.

例如，x² − 4x + 4 = 0 的 Δ = 16 − 16 = 0，因此有一个重根 x = 2。而 x² + 1 = 0 的 Δ = 0 − 4 = −4，因此没有实数解。判别式在解方程和图像题中经常被考查，请熟记这三种情况。

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6. Sketching Quadratic Graphs | 二次函数图像

ction y = ax² + bx + c always produces a parabola. If a > 0, the parabola opens upwards (a "smile"); if a < 0, it opens downwards (a "frown").

二次函数 y = ax² + bx + c 的图像总是一条抛物线。若 a > 0，抛物线开口向上（"微笑"形）；若 a < 0，开口向下（"皱眉"形）。

/ 2a, and the y-coordinate is found by substitution. The turning point is the minimum when a > 0 and the maximum when a < 0.

画图时需要三个关键特征：(1) 根，令 y = 0 求得；(2) y 轴截距，即 c；(3) 顶点。顶点的 x 坐标为 x = −b / 2a，y 坐标通过代入求得。当 a > 0 时顶点为最小值，当 a < 0 时顶点为最大值。

The parabola opens upwards, crossing (−1, 0) and (3, 0), and passing through (0, −3) with minimum at (1, −4).

考虑 y = x² − 2x − 3。根：x² − 2x − 3 = (x − 3)(x + 1) = 0，所以 x = 3 和 x = −1。y 轴截距：(0, −3)。对称轴：x = 1。顶点：x = 1，y = 1 − 2 − 3 = −4。抛物线开口向上，穿过 (−1, 0) 和 (3, 0)，经过 (0, −3)，最小值为 (1, −4)。

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7. Quadratic Inequalities | 二次不等式

solve x² − 5x + 6 > 0. First factorise: (x − 2)(x − 3) > 0. The roots are x = 2 and x = 3.

解二次不等式需要求出使二次式为正或负的 x 范围。例如，解 x² − 5x + 6 > 0。先因式分解：(x − 2)(x − 3) > 0。根为 x = 2 和 x = 3。

Sketch the parabola mentally: it opens upwards. The product is positive when x is outside the interval between the roots. Therefore the solution is x < 2 or x > 3. For the inequality (x − 2)(x − 3) < 0, the solution is 2 < x < 3.

在头脑中画出抛物线：开口向上。当 x 在两个根之间的区间之外时乘积为正。因此解为 x < 2 或 x > 3。对于不等式 (x − 2)(x − 3) < 0，解为 2 < x < 3。

ct > or < sign, use open circles and round brackets. For example, x² ≤ 9 gives −3 ≤ x ≤ 3.

始终注意不等号方向以及根是否包含在内。若不等式带 ≥ 或 ≤，使用实心圆和方括号；若带严格 > 或 <，使用空心圆和圆括号。例如，x² ≤ 9 给出 −3 ≤ x ≤ 3。

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8. Applications and Word Problems | 应用与应用题

= 5 (we reject x = −6 because a length cannot be negative). The rectangle is 8 cm by 3 cm.

二次方程常出现在面积、数字和抛体运动问题中。一个经典面积问题：一个长方形的长为 (x + 3) cm，宽为 (x − 2) cm，面积为 24 cm²。则 (x + 3)(x − 2) = 24，展开得 x² + x − 6 = 24，即 x² + x − 30 = 0。因式分解得 (x + 6)(x − 5) = 0，所以 x = 5（舍弃 x = −6，因为长度不能为负）。这个长方形为 8 cm × 3 cm。

olve 20t − 5t² = 15, which rearranges to 5t² − 20t + 15 = 0, giving t = 1 or t = 3 seconds — on the way up and on the way down.

在物理中，向上抛出的物体高度为 h = ut − ½gt²。若 u = 20 m/s 且 g = 10 m/s²，则 h = 20t − 5t²。求何时落地（h = 0），解 20t − 5t² = 0，即 5t(4 − t) = 0，得到 t = 0 或 t = 4 秒。求何时到达 15 m，解 20t − 5t² = 15，整理为 5t² − 20t + 15 = 0，得到 t = 1 或 t = 3 秒——上升途中和下降途中各一次。

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9. Practice Questions | 练习题

Try these questions on your own before checking the answers. They cover all the methods from this article.

请先独立完成以下题目，再核对答案。它们涵盖本文介绍的所有方法。

• Solve x² − 7x + 12 = 0 by factorisation.

  用因式分解法解 x² − 7x + 12 = 0。

  Answer: (x − 3)(x − 4) = 0, so x = 3 or x = 4.

  答案：(x − 3)(x − 4) = 0，所以 x = 3 或 x = 4。

• Solve x² + 10x + 25 = 0 by completing the square.

  用配方法解 x² + 10x + 25 = 0。

  Answer: (x + 5)² = 0, so x = −5 (repeated root).

  答案：(x + 5)² = 0，所以 x = −5（重根）。

• uadratic formula to solve 3x² − 4x − 2 = 0. Give your answers correct to 2 decimal places.

  用求根公式解 3x² − 4x − 2 = 0，答案保留两位小数。

  Answer: x = (4 ± √40) / 6 = (2 ± √10) / 3, so x ≈ 1.72 or x ≈ −0.39.

  答案：x = (4 ± √40) / 6 = (2 ± √10) / 3，因此 x ≈ 1.72 或 x ≈ −0.39。

• Find the set of values of x for which x² − 4x − 5 ≤ 0.

  求满足 x² − 4x − 5 ≤ 0 的 x 取值范围。

  Answer: (x − 5)(x + 1) ≤ 0, so −1 ≤ x ≤ 5.

  答案：(x − 5)(x + 1) ≤ 0，所以 −1 ≤ x ≤ 5。

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10. Summary | 总结

Quadratic equations can be solved by four main methods: factorisation, completing the square, the quadratic formula, and the graphical method. Each has its place, and exam questions will often let you choose. Factorisation is fastest when roots are integers; completing the square also gives the turning point; the quadratic formula always works; the graph method reveals the overall structure of the parabola.

二次方程可以通过四种主要方法求解：因式分解、配方、求根公式和图像法。每种方法各有其适用场景，考试中通常可以自行选择。根为整数时因式分解最快；配方法还能给出顶点；求根公式始终有效；图像法则揭示抛物线的整体结构。

of roots; the turning point has x-coordinate −b / 2a; quadratics appear in factorisation, inequalities, graph sketching, and word problems. With regular practice, solving quadratic equations will become second nature.

记住关键要点：判别式 Δ = b² − 4ac 决定根的个数；顶点的 x 坐标为 −b / 2a；二次方程出现在因式分解、不等式、图像画法和应用题中。通过定期练习，解二次方程将变得得心应手。

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