Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are among the most frequently tested topics in IGCSE Mathematics. Whether they appear as direct solving questions, graph sketching exercises, or real-world word problems, mastering quadratics is essential for achieving a high grade.

二次方程是 IGCSE 数学中考查频率最高的内容之一。无论是直接求解、图像作图，还是实际应用题，掌握二次方程都是冲击高分的关键基础。

1. The General Form of a Quadratic Equation | 二次方程的一般形式

A quadratic equation is an equation in which the highest power of the unknown variable is 2. Its general form is written as:

二次方程是未知数的最高次数为 2 的方程，其一般形式为：

ax² + bx + c = 0, where a ≠ 0

In this form, a, b and c are constants, and a cannot be zero. If a = 0, the equation degenerates into a linear equation. The constants b or c may be zero, but the quadratic term ax² must always be present.

在此形式中，a、b、c 是常数，且 a 不能为 0。如果 a = 0，方程就退化为一次方程。常数 b 或 c 可以等于 0，但二次项 ax² 必须始终存在。

Equation	Quadratic?	Reason
3x² − 2x + 1 = 0	Yes	Highest power is 2
x² − 9 = 0	Yes	b = 0 is allowed
2x + 5 = 0	No	No x² term
x³ − x² = 0	No	Highest power is 3

Before solving, you should always rearrange the equation so that one side equals zero and the terms are in descending order of power.

在求解之前，务必把方程整理成一边等于 0 的形式，并按次数从高到低排列各项。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer solutions. It relies on the zero product property: if the product of two expressions is zero, then at least one of them must be zero.

当二次方程存在简单的整数解时，因式分解是最快捷的方法。它利用零积性质：若两个表达式的乘积等于 0，则其中至少有一个必须等于 0。

If A × B = 0, then A = 0 or B = 0

Example: Solve x² − 5x + 6 = 0.

例：解方程 x² − 5x + 6 = 0。

Look for two numbers that multiply to give 6 and add to give −5. These numbers are −2 and −3. Hence the factorisation is (x − 2)(x − 3) = 0.

寻找两个数，它们的乘积为 6，且和为 −5。这两个数是 −2 和 −3。因此分解因式为 (x − 2)(x − 3) = 0。

x − 2 = 0 or x − 3 = 0 → x = 2 or x = 3

Always expand your answer to verify the factorisation. A quick check: (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓

完成因式分解后应展开验证：(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓

For quadratics with a leading coefficient of 1, the method is: find two numbers whose product is c and whose sum is b when the equation is x² + bx + c = 0.

对于首项系数为 1 的二次方程 x² + bx + c = 0，方法是：找到两个数，使其乘积为 c，且和为 b。

When the coefficient of x² is not 1, use the method of splitting the middle term or apply the quadratic formula directly.

当 x² 的系数不是 1 时，可以使用十字相乘法（裂项法），或直接使用求根公式。

3. The Quadratic Formula | 求根公式

The quadratic formula can solve any quadratic equation, including those that cannot be factorised. It is derived from completing the square and is one of the most important formulas in the IGCSE syllabus.

求根公式可以解任何二次方程，包括无法因式分解的方程。它由配方法推导而来，是 IGCSE 考纲中最重要的公式之一。

x = (−b ± √(b² − 4ac)) / (2a)

Example: Solve 2x² − 4x − 3 = 0 using the quadratic formula.

例：用求根公式解 2x² − 4x − 3 = 0。

Here a = 2, b = −4 and c = −3. Substitute these values carefully:

这里 a = 2，b = −4，c = −3。小心代入：

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

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4

Since √40 = 2√10, we can simplify to x = (4 ± 2√10) / 4 = (2 ± √10) / 2. Therefore x = (2 + √10) / 2 or x = (2 − √10) / 2.

由于 √40 = 2√10，可以化简为 x = (4 ± 2√10) / 4 = (2 ± √10) / 2。因此 x = (2 + √10) / 2 或 x = (2 − √10) / 2。

When using the quadratic formula, pay special attention to negative signs. Writing b² − 4ac with the correct brackets avoids sign errors.

使用求根公式时，要特别注意负号。将 b² − 4ac 用括号正确括起来，可以避免符号错误。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This form reveals the turning point of the parabola and is especially useful in sketching graphs and solving equations without using the formula.

配方法将二次方程改写为 (x + p)² + q 的形式。这种形式能直接显示抛物线的顶点坐标，在作图和解方程中非常实用。

Example: Solve x² + 6x + 2 = 0 by completing the square.

例：用配方法解 x² + 6x + 2 = 0。

Step 1: Halve the coefficient of x. Half of 6 is 3, so we write (x + 3)².

步骤 1：将 x 的系数除以 2。6 的一半是 3，所以我们写成 (x + 3)²。

(x + 3)² = x² + 6x + 9

Step 2: Adjust the constant term. Since we added 9 but the original constant is 2, we subtract 7.

步骤 2：调整常数项。因为我们加了 9，而原常数项是 2，所以要减去 7。

x² + 6x + 2 = (x + 3)² − 7

Step 3: Solve the equation (x + 3)² − 7 = 0:

步骤 3：解方程 (x + 3)² − 7 = 0：

(x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7

Completing the square also gives the minimum point of the curve y = x² + 6x + 2 as (−3, −7). This is a common IGCSE question.

配方法还能直接给出曲线 y = x² + 6x + 2 的最低点 (−3, −7)，这也是 IGCSE 的常见考点。

5. The Discriminant | 判别式

The discriminant is the expression b² − 4ac in the quadratic formula. It determines the nature of the roots without actually solving the equation.

判别式是求根公式中的表达式 b² − 4ac。它无需实际解方程，就能判断根的性质。

Value of b² − 4ac	Nature of Roots	Graph Interpretation
Positive (Δ > 0)	Two distinct real roots	Curve cuts the x-axis twice
Zero (Δ = 0)	One repeated real root	Curve touches the x-axis once
Negative (Δ < 0)	No real roots	Curve does not meet the x-axis

Example: Determine the number of real roots of 3x² + 2x + 5 = 0.

例：判断方程 3x² + 2x + 5 = 0 的实根个数。

Here a = 3, b = 2, c = 5, so b² − 4ac = 4 − 60 = −56, which is negative. Hence the equation has no real roots.

这里 a = 3，b = 2，c = 5，所以 b² − 4ac = 4 − 60 = −56，为负数。因此方程没有实数根。

The discriminant also appears in problems asking you to find unknown constants. For example, if the equation kx² + 4x + 1 = 0 has exactly one real root, then set the discriminant equal to zero and solve the resulting equation in k.

判别式还常用于求未知常数。例如，若方程 kx² + 4x + 1 = 0 只有一个实根，则令判别式等于 0，再解关于 k 的方程即可。

6. Graphical Solutions | 图像解法

The roots of a quadratic equation ax² + bx + c = 0 correspond to the x-intercepts of the curve y = ax² + bx + c. This connection is heavily tested in IGCSE graph questions.

二次方程 ax² + bx + c = 0 的根，对应曲线 y = ax² + bx + c 与 x 轴的交点。这一联系在 IGCSE 作图题中考查频繁。

When the coefficient a is positive, the parabola opens upwards and has a minimum point. When a is negative, it opens downwards and has a maximum point. The x-coordinate of the turning point is given by x = −b / (2a).

当系数 a 为正时，抛物线开口向上并存在最小值点；当 a 为负时，抛物线开口向下并存在最大值点。顶点横坐标为 x = −b / (2a)。

Example: Sketch y = x² − 4x + 3, labelling the roots and turning point.

例：画出 y = x² − 4x + 3 的草图，并标出根与顶点。

Factorising gives y = (x − 1)(x − 3), so the roots are x = 1 and x = 3. The turning point lies halfway between the roots, at x = 2. Substituting x = 2 gives y = 4 − 8 + 3 = −1, so the minimum point is (2, −1).

因式分解得 y = (x − 1)(x − 3)，根为 x = 1 和 x = 3。顶点位于两根正中间，即 x = 2 处。代入 x = 2 得 y = 4 − 8 + 3 = −1，因此最低点为 (2, −1)。

Graphically, the equation x² − 4x + 3 = 0 has two x-intercepts, the curve crosses the x-axis at 1 and 3, and the vertex sits below the x-axis.

从图像上看，方程 x² − 4x + 3 = 0 有两个 x 轴交点，曲线在 1 和 3 处穿过 x 轴，而顶点位于 x 轴下方。

7. Word Problems | 应用题

Quadratic equations frequently arise in geometry and motion problems. The key steps are: define the unknown, translate the conditions into a quadratic equation, solve it, and check the reasonableness of each answer.

二次方程常出现在几何和运动问题中。关键步骤是：设未知数、将条件转化为二次方程、解方程，并检验每个答案的合理性。

Example: A rectangle has a length that is 3 cm longer than its width. Its area is 40 cm². Find the dimensions.

例：一个长方形的长比宽多 3 cm，面积为 40 cm²。求长和宽。

Let the width be x cm. Then the length is (x + 3) cm. The area is x(x + 3) = 40, which expands to x² + 3x − 40 = 0.

设宽为 x cm，则长为 (x + 3) cm。面积满足 x(x + 3) = 40，展开得 x² + 3x − 40 = 0。

(x + 8)(x − 5) = 0 → x = −8 or x = 5

The negative solution x = −8 is rejected because length cannot be negative. Therefore the width is 5 cm and the length is 8 cm.

负数解 x = −8 不合理，因为长度不可能为负。因此宽为 5 cm，长为 8 cm。

Always check which solution is physically possible. Rejecting invalid roots is an essential exam skill.

务必判断哪个解在实际情况中可行。舍弃不合理解是重要的应考技能。

8. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Many students lose marks on quadratic equations due to small but repeated errors. The table below summarises common pitfalls and their remedies.

很多学生在二次方程上失分，原因往往是一些小但反复出现的错误。下表总结了常见陷阱和解决方法。

Common Mistake	Example	Correction
Solving before rearranging to zero	x² = 3x → x² = 3x → x = 3	Rewrite as x² − 3x = 0 then factorise x(x − 3) = 0
Losing the ± sign in the formula	x = −b + √… only	Always write x = (−b ± √…) / 2a
Sign errors when substituting negatives	a = 1, b = −4 → b² = −16	(−4)² = 16, not −16
Dividing both sides by x	x² = 5x → x = 5	This loses the solution x = 0; factorise instead

The most dangerous habit is dividing both sides of an equation by the unknown x, because this always removes the root x = 0. Factorise rather than divide whenever possible.

最危险的习惯是方程两边同时除以未知数 x，因为这会丢失 x = 0 这个根。只要可能，应优先因式分解而不是除法。

9. Exam-Style Questions | 考试型例题

The following questions reflect the style and difficulty of typical IGCSE examination questions.

以下题目反映了 IGCSE 考试的典型风格和难度。

1. Solve x² − 7x + 10 = 0 by factorisation. (Answer: x = 2 or x = 5)

1. 用因式分解法解 x² − 7x + 10 = 0。（答案：x = 2 或 x = 5）

2. Solve 2x² + 3x − 4 = 0, giving your answers correct to 2 decimal places.

2. 解方程 2x² + 3x − 4 = 0，将答案精确到 2 位小数。

Using the formula with a = 2, b = 3, c = −4 gives x = (−3 ± √41) / 4 = 0.85 or −2.35 (to 2 d.p.).

代入 a = 2，b = 3，c = −4，得 x = (−3 ± √41) / 4 = 0.85 或 −2.35（精确到 2 位小数）。

3. Show that x² + 4x + 7 = 0 has no real roots.

3. 证明方程 x² + 4x + 7 = 0 没有实数根。

The discriminant is 4² − 4 × 1 × 7 = 16 − 28 = −12 < 0, so there are no real roots.

判别式为 4² − 4 × 1 × 7 = 16 − 28 = −12 < 0，因此没有实数根。

4. Express x² − 8x + 5 in the form (x + p)² + q, and hence state the minimum value of x² − 8x + 5.

4. 将 x² − 8x + 5 化为 (x + p)² + q 的形式，并指出 x² − 8x + 5 的最小值。

The answer is (x − 4)² − 11, so the minimum value is −11 when x = 4.

答案为 (x − 4)² − 11，因此当 x = 4 时，最小值为 −11。

10. Summary | 总结

Quadratic equations are solved through four main approaches: factorisation for simple integer roots, the quadratic formula for any equation, completing the square for vertex form, and graphical methods for visual insight. The discriminant quickly tells you the number and type of roots.

二次方程的解法主要有四种：因式分解法适用于简单整数根，求根公式适用于任意方程，配方法用于求顶点式，图像法则提供直观理解。判别式可以快速判断根的个数与类型。

In examinations, always rearrange the equation to the standard form ax² + bx + c = 0 first, check your signs carefully, and verify your solutions by substitution. For word problems, reject any root that does not make sense in the given context.

考试中，务必先将方程整理成标准形式 ax² + bx + c = 0，仔细检查符号，并通过代入验证解。对于应用题，要舍弃在情境中不合理的根。

Practice a variety of quadratics until the methods become automatic. This topic rewards consistent effort and careful attention to detail.

多练习不同类型的二次方程，直到解法成为本能反应。这个知识点会回报你的坚持和对细节的专注。

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Solving Quadratic Equations | 解二次方程