Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, functions, and even in examination questions on sequences and problem solving. A secure understanding of quadratics will directly raise your grade across multiple paper sections.

二次方程是 IGCSE 数学中考核频率最高的考点之一。它不仅出现在代数部分，还渗透到坐标几何、函数、数列以及综合性应用题中。扎实掌握二次方程，能直接提升你整张试卷多个板块的得分率。

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1. The Standard Form | 标准形式

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the unknown variable is 2. The general form is written as:

二次方程是最高次数为 2 的多项式方程，即未知数的最高次幂为 2。其一般形式写作：

ax² + bx + c = 0   (a ≠ 0)

Here, a, b and c are real numbers, and a cannot be zero. If a = 0, the equation becomes linear because the x² term vanishes.

其中 a、b、c 为常数，且 a 不能等于 0。若 a = 0，x² 项消失，方程就退化为一次方程。

For example, x² − 5x + 6 = 0 and 2x² + 3x − 4 = 0 are quadratics, while x + 7 = 0 is not. Before solving, always rearrange the equation so that one side equals zero.

例如，x² − 5x + 6 = 0 和 2x² + 3x − 4 = 0 都是二次方程，而 x + 7 = 0 不是。解题前务必先将方程整理为一边等于 0 的形式。

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2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer roots. The principle is the zero product property: if the product of two expressions is zero, then at least one of them must be zero.

当方程存在简单的整数根时，因式分解法是速度最快的方法。其核心是零乘积性质：若两个表达式的乘积为 0，则其中至少有一个表达式等于 0。

Consider x² − 5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5. The numbers −2 and −3 work, so:

考虑 x² − 5x + 6 = 0。我们需要找到两个数，使其乘积为 6、和为 −5。显然 −2 和 −3 满足条件，因此：

(x − 2)(x − 3) = 0

x = 2  or  x = 3

When the coefficient of x² is not 1, the method is slightly longer. For 2x² + 7x + 3 = 0, multiply a and c to get 6, then find two numbers multiplying to 6 and adding to 7: those are 1 and 6. Split the middle term and factor by grouping:

当 x² 的系数不为 1 时，过程会稍长一些。以 2x² + 7x + 3 = 0 为例，先将 a 与 c 相乘得到 6，再找乘积为 6、和为 7 的两个数，即 1 和 6。然后拆分中间项并分组因式分解：

2x² + x + 6x + 3 = 0

x(2x + 1) + 3(2x + 1) = 0

(x + 3)(2x + 1) = 0

x = −3  or  x = −½

Always expand your factorised answer to check that it recreates the original equation.

完成因式分解后，务必展开验证，确保结果与原方程一致。

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3. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This method is essential for finding the turning point of a parabola and for solving equations that do not factorise neatly.

配方法将二次方程改写为 a(x + p)² + q 的形式。这种方法对于求抛物线顶点，以及解那些无法整齐分解的方程至关重要。

For the expression x² + 6x + 5, take half of 6 to get 3, then square it to get 9. Write:

对于表达式 x² + 6x + 5，取 6 的一半得 3，再平方得 9。于是可以写成：

x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4

Now solve (x + 3)² − 4 = 0:

现在解方程 (x + 3)² − 4 = 0：

(x + 3)² = 4

x + 3 = ± 2

x = −1  or  x = −5

Notice that taking the square root introduces both a positive and a negative solution. For equations with an a coefficient, factor a out first, then complete the square on the remaining x² term.

注意，开平方时会同时引入正负两个解。若 a 不为 1，应先将 a 提出，再对剩下的 x² 项配方。

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4. The Quadratic Formula | 求根公式

The quadratic formula works for every quadratic equation, including those that cannot be factorised. It is derived from completing the square on the general form ax² + bx + c = 0.

求根公式适用于所有二次方程，包括那些无法因式分解的情形。它由对一般形式 ax² + bx + c = 0 进行配方推导而来。

x = (−b ± √(b² − 4ac)) / 2a

Let us apply it to 2x² − 4x − 1 = 0. Here a = 2, b = −4, c = −1. Substitute carefully, paying special attention to signs:

我们以 2x² − 4x − 1 = 0 为例。这里 a = 2，b = −4，c = −1。代入时要特别小心符号：

x = (4 ± √(16 + 8)) / 4

x = (4 ± √24) / 4

x = (4 ± 2√6) / 4  =  1 ± (√6)/2

These are the two exact solutions. Leave your answer in surd form unless the question asks for a decimal. The formula is reliable, but it is slower than factorisation, so reserve it for equations that do not factorise easily.

这就是两个精确解。除非题目要求小数，否则应保留根号形式。求根公式非常可靠，但比因式分解法慢，因此建议留作无法轻松分解时的备用方法。

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5. The Discriminant | 判别式

The expression b² − 4ac inside the square root is called the discriminant, often denoted by the symbol Δ. It tells us the number and type of roots without solving the full equation.

根号内的表达式 b² − 4ac 称为判别式，通常用符号 Δ 表示。它无需完整解方程，就能告诉我们根的数量与类型。

Δ = b² − 4ac

Discriminant Δ | 判别式	Nature of Roots | 根的性质
Δ > 0	Two distinct real roots | 两个不相等的实数根
Δ = 0	One repeated real root | 两个相等的实数根（重根）
Δ < 0	No real roots | 无实数根

For example, in x² − 4x + 4 = 0, Δ = 16 − 16 = 0, so there is exactly one repeated root. In x² + 2x + 5 = 0, Δ = 4 − 20 = −16, so there are no real roots.

例如，在 x² − 4x + 4 = 0 中，Δ = 16 − 16 = 0，因此只有一个重根。而在 x² + 2x + 5 = 0 中，Δ = 4 − 20 = −16，所以无实数根。

Exam questions often ask you to find the range of k for which a quadratic has two real roots. Write the discriminant in terms of k, then solve the resulting inequality.

考试常要求学生求 k 的取值范围，使某个二次方程有两个实数根。此时只需用 k 表示判别式，再解对应不等式即可。

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6. Sum and Product of Roots | 根的和与积

If α and β are the two roots of ax² + bx + c = 0, then there are two elegant relationships that allow us to construct quadratics quickly:

如果 α 与 β 是 ax² + bx + c = 0 的两个根，那么存在两个简洁的关系，可以帮我们快速构造二次方程：

α + β = −b / a

αβ = c / a

Suppose a question asks for a quadratic equation whose roots are 3 and −2. Here the sum is 1 and the product is −6. Using x² − (sum)x + (product) = 0, we obtain:

假设题目要求一个以 3 和 −2 为根的二次方程。此时根的和为 1，乘积为 −6。利用 x² − (和)x + (积) = 0，可得：

x² − x − 6 = 0

This technique is especially useful in papers that test manipulation before factorisation, and it saves time in multiple-choice questions.

这一技巧在考查代数变形能力的题目中尤为实用，也能在选择题中显著节省时间。

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7. Sketching Quadratic Graphs | 二次函数图像

The graph of y = ax² + bx + c is a parabola. Its shape is determined by the sign of a: if a > 0, the parabola opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point.

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，存在最小值点；若 a < 0，抛物线开口向下，存在最大值点。

The key features to mark when sketching are:

画草图时需要标注的关键特征有：

• The y-intercept, found by setting x = 0, which equals c.
• y 轴截距：令 x = 0 求得，即 c 的值。
• The x-intercepts, found by solving ax² + bx + c = 0.
• x 轴截距：通过解 ax² + bx + c = 0 求得。
• The turning point, read directly from the completed square form a(x + p)² + q as (−p, q).
• 顶点坐标：从配方法形式 a(x + p)² + q 中直接读出为 (−p, q)。

For example, y = (x − 1)² − 4 has its vertex at (1, −4), y-intercept at −3, and x-intercepts at x = 3 and x = −1, since (x − 1)² = 4 gives x = 1 ± 2.

例如，y = (x − 1)² − 4 的顶点为 (1, −4)，y 轴截距为 −3，而由 (x − 1)² = 4 得 x = 1 ± 2，所以 x 轴截距为 3 和 −1。

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8. Quadratic Inequalities | 二次不等式

Solving a quadratic inequality such as x² − 5x + 6 < 0 requires a sign diagram. First factorise to find the critical values, then test intervals between them.

解二次不等式（如 x² − 5x + 6 < 0）需要借助符号图。先因式分解求出临界值，再在各区间内试值判断符号。

For x² − 5x + 6 < 0, the roots are 2 and 3. The parabola y = x² − 5x + 6 opens upwards, so it is below the x-axis between the roots:

对于 x² − 5x + 6 < 0，根为 2 和 3。抛物线 y = x² − 5x + 6 开口向上，因此在两根之间位于 x 轴下方：

2 < x < 3

If the inequality sign is reversed, x² − 5x + 6 > 0, the solution is the union of the two outer regions:

如果不等号方向相反，即 x² − 5x + 6 > 0，则解为两个外部区域的并集：

x < 2  or  x > 3

Remember that strict inequalities use open circles on a number line, while non-strict inequalities such as ≤ or ≥ use filled circles.

注意，严格不等式在数轴上用空心圆圈，而 ≤ 或 ≥ 这类非严格不等式则用实心圆圈。

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9. Applications in Word Problems | 实际应用题

Quadratic equations often arise in geometry and motion problems. A classic exam question involves a rectangle: the length of a rectangle is 3 cm more than its width, and its area is 28 cm². Find the width.

二次方程常出现在几何与运动类应用题中。一个经典考题是：矩形的长比宽多 3 cm，面积为 28 cm²，求宽。

Let the width be x. Then the length is x + 3, so x(x + 3) = 28. Expand and rearrange:

设宽为 x，则长为 x + 3，因此 x(x + 3) = 28。展开并整理：

x² + 3x − 28 = 0

(x + 7)(x − 4) = 0

This gives x = −7 or x = 4. Since a length cannot be negative, the width is 4 cm. Always reject any negative root that does not make sense in the context of the problem, and state your final answer with units.

解得 x = −7 或 x = 4。由于长度不能为负，所以宽为 4 cm。务必舍去不符合题目实际意义的负根，并在最终答案中带上单位。

In projectile problems, the height of an object after t seconds may be modelled by h = −5t² + 20t + 1. To find when the object reaches the ground, set h = 0 and solve the resulting quadratic using the formula or factorisation.

在抛体运动中，物体 t 秒后的高度可用 h = −5t² + 20t + 1 表示。要求物体落地的时间，只需令 h = 0 并利用公式法或因式分解法解对应方程。

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10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Many students lose marks on quadratics through avoidable errors. Here are the most common pitfalls and how to avoid them.

许多学生因可避免的错误在二次方程上丢分。以下是最常见的失分点及应对策略。

• Failing to rearrange the equation to standard form before identifying a, b and c. Always move all terms to one side first.
• 没有先将方程整理为标准形式就急着确定 a、b、c 的值。务必先把所有项移到同一边。
• Forgetting the ± sign when taking square roots, or losing negative solutions when factorising.
• 开平方时漏写 ± 号，或因式分解时遗漏负数解。
• Misapplying signs in the quadratic formula, especially when b is negative.
• 代入求根公式时符号出错，尤其是当 b 为负数时。
• Concluding that an equation has no roots when Δ < 0, instead of stating that there are no real roots.
• 当 Δ < 0 时直接断言方程无根，而没有准确表述为“无实数根”。

To minimise mistakes, always check your solutions by substituting them back into the original equation. For quick verification in a calculator paper, store the coefficients and evaluate both roots.

为减少失误，务必将解代回原方程进行验证。在允许使用计算器的试卷中，可将系数存入计算器并快速检验两个根。

Finally, practise recognising which method to use: factorise when the roots are integers, complete the square when the turning point is needed, and use the formula when nothing else works. With consistent practice, quadratics will become one of the most reliable scoring areas in your IGCSE Mathematics paper.

最后，练习快速判断应选用哪种方法：整数根时用因式分解；需要顶点坐标时用配方法；除此之外用求根公式。只要坚持练习，二次方程将成为你 IGCSE 数学考试中最稳定的得分点之一。

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