📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear throughout the IGCSE syllabus, from pure algebra to applied problem solving. This teacher's guide provides a structured review of every essential method, common pitfalls, and exam-style strategies.
二次方程贯穿整个 IGCSE 课程，从纯代数到实际应用题。本教师用书为你提供每个关键方法的系统回顾、常见易错点以及针对考试的解题策略。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation in one variable can be written in the general form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. The term ax² is called the quadratic term, bx the linear term, and c the constant term.
一元二次方程的一般形式为 ax² + bx + c = 0，其中 a、b、c 为实数，且 a ≠ 0。ax² 称为二次项，bx 称为一次项，c 称为常数项。
To solve a quadratic equation, we must find all values of x that make the equation true. These values are called roots or solutions.
解二次方程就是求出所有能使方程成立的 x 值，这些值称为方程的根或解。
ax² + bx + c = 0
e equation is incomplete but still quadratic. For example, 2x² − 8 = 0 has no linear term.
如果 b 或 c 为零，方程仍属于二次方程，只是不完整。例如 2x² − 8 = 0 就没有一次项。
2. Solving by Factorisation | 因式分解法
Factorisation is often the fastest method when the equation has simple integer roots. It relies on rewriting ax² + bx + c as a product of two linear factors.
当方程具有简单整数根时，因式分解通常是最快的方法。它把 ax² + bx + c 改写为两个一次因式的乘积。
and whose sum is −5. These are −2 and −3. Therefore (x − 2)(x − 3) = 0, so x = 2 or x = 3.
示例：解 x² − 5x + 6 = 0。找两个数，乘积为 6，和为 −5，这两个数是 −2 和 −3。因此 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。
x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 或 x = 3
coefficient not equal to 1, such as 2x² + 5x − 3 = 0, we can use the method of grouping or "ac" method.
当二次项系数不为 1 时，例如 2x² + 5x − 3 = 0，我们可以使用分组法或 ac 法。
Always set the expression equal to zero before factorising, and remember that the product of two factors is zero only when at least one factor is zero.
因式分解前务必把方程整理为等于零的形式，并记住两个因式的乘积为零时，至少有一个因式为零。
3. The Quadratic Formula | 二次公式
s difficult or impossible. For ax² + bx + c = 0, the roots are given by the formula below.
二次公式可以直接解任何二次方程，即使因式分解很困难或不可能。对于 ax² + bx + c = 0，其根由下面的公式给出。
x = (−b ± √b² − 4ac) / 2a
To use the formula, substitute the values of a, b and c carefully. Be especially careful with negative signs.
使用公式时，要仔细代入 a、b、c 的值，尤其注意负号。
2. The discriminant b² − 4ac = 9 + 16 = 25, so x = (−3 ± 5) / 4, giving x = 0.5 or x = −2.
示例：解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。判别式 b² − 4ac = 9 + 16 = 25，因此 x = (−3 ± 5) / 4，得到 x = 0.5 或 x = −2。
In IGCSE exams, the formula is often the safest method when the question does not specify a method. Write down the substitution clearly to earn method marks.
在 IGCSE 考试中，如果题目没有指定方法，二次公式通常是最稳妥的选择。清晰写出代入过程可以获得方法分。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form reveals the vertex of the parabola and helps solve equations that cannot be factorised.
配方法将二次式改写成 a(x + p)² + q 的形式。这种形式能揭示抛物线的顶点，并帮助解无法因式分解的方程。
x, we add and subtract (b/2)². For example, x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.
对于 x² + bx，我们加上并减去 (b/2)²。例如，x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。
x² + 6x + 2 = (x + 3)² − 7
squared term and then take the square root of both sides. Remember to include the ± sign.
用配方法解方程时，先分离平方项，然后对方程两边开平方。记得包含 ± 号。
xample: Solve x² + 6x + 2 = 0 by completing the square. We have (x + 3)² = 7, so x + 3 = ±√7, hence x = −3 ± √7.
示例：用配方法解 x² + 6x + 2 = 0。得到 (x + 3)² = 7，所以 x + 3 = ±√7，因此 x = −3 ± √7。
5. The Discriminant | 判别式
c = 0 is Δ = b² − 4ac. It tells us how many real roots exist without solving the equation.
二次方程 ax² + bx + c = 0 的判别式为 Δ = b² − 4ac。它无需解方程即可告诉我们方程有几个实数根。
tinct real roots. If Δ = 0, it has one repeated real root. If Δ < 0, it has no real roots.
如果 Δ > 0，方程有两个不同的实数根；如果 Δ = 0，方程有一个重根；如果 Δ < 0，方程没有实数根。
Δ > 0 → 两个实根；Δ = 0 → 一个重根；Δ < 0 → 无实根
The discriminant is also used to determine whether a quadratic expression can be factorised over the integers: it must be a perfect square.
判别式还可以用来判断二次式能否在整数范围内因式分解：此时 Δ 必须是一个完全平方数。
wice, Δ = 0 means it touches the x-axis once, and Δ < 0 means it does not meet the x-axis.
从图像角度看，Δ > 0 时抛物线与 x 轴相交两次，Δ = 0 时与 x 轴相切一次，Δ < 0 时与 x 轴没有交点。
6. Sum and Product of Roots | 根的和与积
product of the roots is c/a. This relationship is useful for checking answers and for constructing equations.
对于二次方程 ax² + bx + c = 0，两根之和为 −b/a，两根之积为 c/a。这个关系可用于检验答案和构造方程。
根之和: x₁ + x₂ = −b/a 根之积: x₁x₂ = c/a
α and β, then the quadratic equation can be written as x² − (α + β)x + αβ = 0 when a = 1.
若两根为 α 和 β，则当 a = 1 时，方程可写为 x² − (α + β)x + αβ = 0。
ic equation with roots 3 and −2. Sum = 1, product = −6, so the equation is x² − x − 6 = 0.
示例：求以 3 和 −2 为根的二次方程。两根之和为 1，之积为 −6，因此方程为 x² − x − 6 = 0。
7. Solving Quadratic Inequalities | 二次不等式
x + 6 ≤ 0. The solution set is best found by sketching the parabola or using a sign table.
二次不等式涉及如 x² − 5x + 6 > 0 或 x² − 5x + 6 ≤ 0 的表达。解集最好通过画抛物线或使用符号表来确定。
First solve the corresponding equation to find the critical points. Then determine the sign of the quadratic in each interval between these points.
首先解对应的方程以找到临界点，然后判断这些点之间的每个区间内二次式的符号。
For x² − 5x + 6 > 0, the roots are 2 and 3. The parabola opens upward, so the expression is positive when x < 2 or x > 3, and negative when 2 < x < 3. Therefore the solution is x < 2 or x > 3.
对于 x² − 5x + 6 > 0，根为 2 和 3。抛物线开口向上，因此当 x < 2 或 x > 3 时表达式为正，当 2 < x < 3 时为负。所以解为 x < 2 或 x > 3。
x² − 5x + 6 > 0 ⇔ x < 2 或 x > 3
Remember to check whether the inequality is strict or non-strict. Use open circles for strict inequalities and closed circles for non-strict ones on a number line.
注意区分严格不等式和非严格不等式。在数轴上，严格不等式用空心圆，非严格不等式用实心圆。
8. Graphs of Quadratic Functions | 二次函数图像
x² + bx + c is a parabola. The sign of a determines whether it opens upward (a > 0) or downward (a < 0).
二次函数 y = ax² + bx + c 的图像是一条抛物线。a 的符号决定开口方向：a > 0 开口向上，a < 0 开口向下。
ts are the roots of the equation ax² + bx + c = 0. The y-intercept is the constant term c.
抛物线与 x 轴的交点就是方程 ax² + bx + c = 0 的根，与 y 轴的交点是常数项 c。
find the y-coordinate. In completed square form y = a(x + p)² + q, the vertex is (−p, q).
抛物线的顶点横坐标为 −b/(2a)，将其代入函数可得到纵坐标。在配方法形式 y = a(x + p)² + q 中，顶点坐标为 (−p, q)。
顶点横坐标 = −b/(2a)
Understanding the graph helps visualise the discriminant, the range of the function, and the solution of inequalities.
理解图像有助于直观认识判别式、函数的值域以及不等式的解集。
9. Applications in Problem Solving | 应用题
Quadratic equations arise in geometry, physics, economics and many other areas. Common IGCSE problems include area, projectile motion, and consecutive integers.
二次方程出现在几何、物理、经济以及许多其他领域。IGCSE 常见题型包括面积问题、抛体运动以及连续整数问题。
x² + 5x − 24 = 0. Solving gives x = 3 or x = −8. Since width cannot be negative, x = 3 cm.
示例：一个矩形的面积为 24 cm²，它的长比宽长 5 cm。设宽为 x cm，则 x(x + 5) = 24，即 x² + 5x − 24 = 0。解得 x = 3 或 x = −8。宽度不能为负，因此 x = 3 cm。
x(x + 5) = 24 → x = 3 (舍弃 x = −8)
Always define a variable for the unknown quantity, form a quadratic equation, solve it, and then check whether each solution is valid in the original context.
解题时先为未知量设变量，建立二次方程，求解之后检查每个解在原情境中是否有意义。
10. Common Mistakes and Tips | 常见错误与技巧
Students often make the following errors when solving quadratic equations. Teachers should highlight these during review.
学生在解二次方程时经常犯下列错误。教师在复习时应重点强调。
- o set the equation to zero before factorising. Always rearrange to ax² + bx + c = 0 first.
- 在因式分解前忘记把方程化为等于零的形式。务必先整理成 ax² + bx + c = 0。
- Misapplying the quadratic formula with negative signs. Write down a, b, c explicitly and substitute carefully.
- 在代入二次公式时弄错负号。先把 a、b、c 明确写出，再仔细代入。
- ng the ± sign when taking square roots. For example, from x² = 9, write x = ±3, not x = 3.
- 开平方时遗漏 ± 号。例如由 x² = 9 应得到 x = ±3，而不是 x = 3。
- Confusing the discriminant with the equation itself. The discriminant is an expression, not a solution.
- 混淆判别式和方程本身。判别式是一个表达式，不是方程的解。
- In inequality problems, testing only one interval instead of all. Always test a value from each interval.
- 在解不等式时只检验一个区间而不是所有区间。务必从每个区间各取一个值进行检验。
A good tip is to substitute your answers back into the original equation to verify them. This takes only seconds and prevents careless mistakes.
一个实用技巧是把求得的解代回原方程进行验证。这只需几秒钟，可以避免粗心错误。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图