📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in Paper 2, Paper 4, and in practical problem-solving questions. A solid understanding of how to solve them is essential for achieving a high grade.
二次方程是IGCSE数学中最重要的主题之一。它们出现在试卷2、试卷4以及实际应用题中。扎实掌握解法对于取得高分至关重要。
1. What Is a Quadratic Equation | 什么是二次方程
A quadratic equation is an equation where the highest power of the unknown variable is 2. It contains an x² term, and it may also contain an x term and a constant term.
二次方程是未知数的最高次数为2的方程。它包含x²项，也可能包含x项和常数项。
- Examples: x² − 5x + 6 = 0 and 2x² + 3x − 1 = 0 are quadratic equations.
- 例子：x² − 5x + 6 = 0 和 2x² + 3x − 1 = 0 都是二次方程。
- x³ + 2x = 0 is a cubic equation, not quadratic.
- x³ + 2x = 0 是三次方程，不是二次方程。
Notice that the coefficient of x² cannot be zero; if it were zero, the equation would become linear.
注意x²的系数不能为零；如果为零，方程就变成了线性方程。
2. Standard Form of a Quadratic Equation | 二次方程的标准形式
adratic equation is written as ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程的标准形式写为 ax² + bx + c = 0，其中a、b、c是常数，且a ≠ 0。
ax² + bx + c = 0 (a ≠ 0)
he equation into this form. For example, x² = 4x + 5 must be rewritten as x² − 4x − 5 = 0.
在解方程之前，应始终将方程整理成这种形式。例如，x² = 4x + 5 必须改写为 x² − 4x − 5 = 0。
| Equation | 方程 | a | b | c |
| x² − 5x + 6 = 0 | 1 | −5 | 6 |
| 2x² + 3x − 1 = 0 | 2 | 3 | −1 |
| x² = 4x + 5 → x² − 4x − 5 = 0 | 1 | −4 | −5 |
3. Solving by Factorisation | 因式分解法
Factorisation is the quickest method when the quadratic expression can be written as a product of two linear brackets. The rule is: if the product of two expressions is zero, then at least one of them must be zero.
当二次表达式可以写成两个一次括号的乘积时，因式分解是最快的方法。规则是：如果两个表达式的乘积为零，那么其中至少有一个必须为零。
Example | 例子: Solve x² − 5x + 6 = 0.
- Find two numbers that multiply to give 6 and add to give −5: they are −2 and −3.
- 找到两个数，乘积为6，和为−5：它们是−2和−3。
- Write the brackets: (x − 2)(x − 3) = 0.
- 写出括号：(x − 2)(x − 3) = 0。
- Set each bracket to zero: x − 2 = 0 or x − 3 = 0.
- 令每个括号为零：x − 2 = 0 或 x − 3 = 0。
- Solutions: x = 2 or x = 3.
- 解为：x = 2 或 x = 3。
difference of two squares: x² − 9 = (x − 3)(x + 3). This gives solutions x = 3 and x = −3.
另一个常见模式是平方差：x² − 9 = (x − 3)(x + 3)。这给出解 x = 3 和 x = −3。
x² − 9 = 0 → (x − 3)(x + 3) = 0 → x = 3 or x = −3
4. Solving by the Quadratic Formula | 求根公式法
When factorisation is difficult or impossible, use the quadratic formula. This formula works for any quadratic equation in standard form.
当因式分解困难或不适用时，使用求根公式。该公式适用于任何标准形式的二次方程。
x = (−b ± √b² − 4ac) / (2a)
Example | 例子: Solve 2x² + 3x − 2 = 0 using the formula.
- Here a = 2, b = 3, c = −2.
- 这里 a = 2，b = 3，c = −2。
- Substitute into the formula: x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2).
- 代入公式：x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2)。
- Simplify: x = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4 = (−3 ± 5) / 4.
- 化简：x = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4 = (−3 ± 5) / 4。
- So x = (2) / 4 = 0.5 or x = (−8) / 4 = −2.
- 所以 x = 2/4 = 0.5 或 x = −8/4 = −2。
Always write down the substitution clearly; many exam marks are given for method, not just the final answer.
务必清楚地写出代入过程；考试中很多分数是给方法的，而不只是最终答案。
5. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This method is especially useful for finding the vertex of a parabola or solving equations without using the formula.
配方法将二次式改写为 (x + p)² + q 的形式。这种方法在求抛物线顶点或不用公式解方程时特别有用。
x² + bx + c = (x + b/2)² − (b/2)² + c
Example | 例子: Solve x² + 6x + 4 = 0 by completing the square.
- Take half of 6, which is 3, and write: (x + 3)² − 9 + 4 = 0.
- 取6的一半，即3，写出：(x + 3)² − 9 + 4 = 0。
- Simplify: (x + 3)² − 5 = 0.
- 化简：(x + 3)² − 5 = 0。
- Rearrange: (x + 3)² = 5.
- 移项：(x + 3)² = 5。
- Take the square root: x + 3 = ±√5.
- 取平方根：x + 3 = ±√5。
- Solutions: x = −3 + √5 or x = −3 − √5.
- 解为：x = −3 + √5 或 x = −3 − √5。
Remember that taking a square root always gives two answers, one positive and one negative.
记住，取平方根总是会得到两个答案，一个正一个负。
6. The Discriminant | 判别式
The expression b² − 4ac is called the discriminant, often written as Δ. It tells us how many real roots a quadratic equation has, without solving it completely.
表达式 b² − 4ac 称为判别式，通常写作 Δ。它告诉我们二次方程有多少个实数根，而无需完全求解。
Δ = b² − 4ac
| Discriminant | 判别式 | Number of Real Roots | 实数根个数 | Graph | 图像 |
| Δ > 0 | Two distinct real roots | 两个不相等的实数根 | Crosses the x-axis twice | 与x轴有两个交点 |
| Δ = 0 | One repeated real root | 一个重根 | Touches the x-axis once | 与x轴相切 |
| Δ < 0 | No real roots | 没有实数根 | Does not meet the x-axis | 与x轴无交点 |
For example, the equation x² + 2x + 5 = 0 has Δ = 4 − 20 = −16, so it has no real roots.
例如，方程 x² + 2x + 5 = 0 的判别式 Δ = 4 − 20 = −16，因此它没有实数根。
7. Sum and Product of Roots | 根的和与积
uadratic equation ax² + bx + c = 0 with roots α and β, there are two useful relationships.
对于根为 α 和 β 的二次方程 ax² + bx + c = 0，有两个有用的关系。
Sum of roots: α + β = −b/a
Product of roots: αβ = c/a
These relationships allow you to check your answers or to build a quadratic equation from given roots.
这些关系允许你检查答案，或从已知根构造二次方程。
Example | 例子: A quadratic equation has roots 2 and −3. Find the equation.
- Sum = 2 + (−3) = −1, so −b/a = −1.
- 和 = 2 + (−3) = −1，所以 −b/a = −1。
- Product = 2 × (−3) = −6, so c/a = −6.
- 积 = 2 × (−3) = −6，所以 c/a = −6。
- Taking a = 1 gives b = 1 and c = −6.
- 取 a = 1，得到 b = 1，c = −6。
- The equation is x² + x − 6 = 0.
- 方程为 x² + x − 6 = 0。
8. Word Problems Involving Quadratics | 二次方程应用题
Quadratics are often used to model real-life situations such as areas, projectile motion and number problems. The key step is to translate the words into an equation, then solve it.
二次方程常用于模拟现实情境，如面积、抛体运动和数字问题。关键步骤是将文字转化为方程，然后求解。
Example | 例子: A rectangle is 3 cm longer than it is wide. Its area is 40 cm². Find its width.
- Let the width be x, so the length is x + 3.
- 设宽为 x，则长为 x + 3。
- Area equation: x(x + 3) = 40.
- 面积方程：x(x + 3) = 40。
- Expand: x² + 3x = 40, so x² + 3x − 40 = 0.
- 展开：x² + 3x = 40，所以 x² + 3x − 40 = 0。
- Factorise: (x + 8)(x − 5) = 0.
- 因式分解：(x + 8)(x − 5) = 0。
- Thus x = −8 or x = 5. Since width cannot be negative, the width is 5 cm.
- 因此 x = −8 或 x = 5。由于宽度不能为负，宽度为5厘米。
Always reject negative answers when the context requires a positive quantity.
当题目情境要求正数时，务必舍去负数解。
9. Graphs of Quadratic Functions | 二次函数图像
bola. When a is positive, the curve opens upwards; when a is negative, it opens downwards.
y = ax² + bx + c 的图像是抛物线。当 a 为正时，曲线开口向上；当 a 为负时，开口向下。
- The x-intercepts are the roots of the equation ax² + bx + c = 0.
- 与x轴的交点是方程 ax² + bx + c = 0 的根。
- The y-intercept is the value of c.
- 与y轴的交点是 c 的值。
- The axis of symmetry is the vertical line x = −b/(2a).
- 对称轴是垂直线 x = −b/(2a)。
- The vertex lies on the axis of symmetry.
- 顶点位于对称轴上。
− 4x + 3 has roots x = 1 and x = 3, a y-intercept of 3, and its axis of symmetry is x = 2.
例如，y = x² − 4x + 3 的根为 x = 1 和 x = 3，与y轴交点为3，对称轴为 x = 2。
10. Intersection of a Line and a Quadratic | 直线与二次曲线的交点
substitute the line equation into the parabola equation and solve the resulting quadratic.
要求直线 y = mx + k 与抛物线 y = ax² + bx + c 的交点，将直线方程代入抛物线方程，然后解所得的二次方程。
ax² + bx + c = mx + k → ax² + (b − m)x + (c − k) = 0
- If the discriminant is positive, the line cuts the curve at two points.
- 若判别式为正，直线与曲线相交于两点。
- If the discriminant is zero, the line touches the curve at one point.
- 若判别式为零，直线与曲线相切于一点。
- If the discriminant is negative, the line does not meet the curve.
- 若判别式为负，直线与曲线没有交点。
These ideas also connect to the tangent condition in coordinate geometry, which is frequently tested.
这些想法还与坐标几何中的切线条件相关，这是常考的考点。
11. Common Exam Mistakes | 常见考试错误
Many students lose marks through small but avoidable errors. Watch out for the following.
许多学生因为细小但可避免的错误而失分。请注意以下几点。
- Forgetting to rearrange the equation to the form ax² + bx + c = 0 before factorising.
- 在因式分解前忘记将方程整理为 ax² + bx + c = 0 的形式。
- Making sign errors when substituting negative values into the quadratic formula.
- 将负值代入求根公式时出现符号错误。
- Dividing both sides of the equation by x, which loses the solution x = 0.
- 方程两边同时除以x，从而丢失了 x = 0 这个解。
- Forgetting the ± sign when taking a square root.
- 取平方根时忘记 ± 号。
- Taking the square root of a negative number when the quadratic has no real roots.
- 当二次方程没有实数根时，对负数取平方根。
Write every step clearly and check your solutions by substituting them back into the original equation.
清晰地写出每一步，并通过将解代回原方程来检验答案。
12. Practice Questions | 练习
Try these questions yourself. Solutions are given, but attempt them before checking.
请自己尝试以下题目。答案已给出，但请先尝试再核对。
| Question | 题目 | Answer | 答案 |
| Solve x² − 7x + 10 = 0 | 解方程 x² − 7x + 10 = 0 | x = 2 or x = 5 | x = 2 或 x = 5 |
| Solve 3x² + 5x − 2 = 0 | 解方程 3x² + 5x − 2 = 0 | x = 1/3 or x = −2 | x = 1/3 或 x = −2 |
| Find the discriminant of x² + 6x + 9 = 0 | 求 x² + 6x + 9 = 0 的判别式 | Δ = 0, one repeated root | Δ = 0，一个重根 |
| A number squared plus 5 times the number equals 24. Find the positive number. | 一个数的平方加上这个数的5倍等于24。求这个正数。 | x = 3 | x = 3 |
t x² + 5x = 24 becomes (x + 8)(x − 3) = 0, giving x = −8 or x = 3. Only x = 3 is positive.
对于最后一题，x² + 5x = 24 变为 (x + 8)(x − 3) = 0，得到 x = −8 或 x = 3。只有 x = 3 是正数。
Quadratic equations are a bridge between algebra and geometry. Master the factorisation method, the quadratic formula, and the discriminant, and you will be well prepared for both calculator and non-calculator papers.
二次方程是代数与几何之间的桥梁。掌握因式分解法、求根公式和判别式，你就能为非计算器和计算器试卷做好充分准备。
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