📚 Quadratic Equations: Solving and Graphing | 二次方程：求解与图像
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, geometry, and real-world problems. This article explains the standard forms, solving methods, graphical properties, and common pitfalls, with clear worked examples.
二次方程是 IGCSE 数学中最重要的主题之一。它出现在代数、图像、几何和实际应用题中。这篇文章将解释标准形式、求解方法、图像性质以及常见易错点，并配有清晰的例题。
1. Standard Form and Key Terms | 标准形式与关键术语
e constants, and a must not be zero. If a = 0, the equation becomes linear, not quadratic.
二次方程是次数为 2 的多项式方程，其一般形式写作 ax² + bx + c = 0，其中 a、b、c 是常数，且 a 不能为 0。如果 a = 0，方程就变成了一次方程，而不是二次方程。
The values of x that make the equation true are called roots, solutions or zeros. A quadratic equation can have two real roots, one repeated real root, or no real roots, depending on the value of the discriminant.
使方程成立的 x 值称为根、解或零点。根据判别式的不同，二次方程可以有两个不相等的实数根、一个重根，或者没有实数根。
ax² + bx + c = 0, a ≠ 0
2. Solving by Factorisation | 因式分解法
0, look for two numbers whose product is 6 and whose sum is 5. These numbers are 2 and 3.
当二次表达式有简单的整数因式时，因式分解是最快的方法。解 x² + 5x + 6 = 0，需要找到两个数，乘积为 6，且和为 5。这两个数是 2 和 3。
(x + 2)(x + 3) = 0
Then set each factor equal to zero and solve. This uses the zero product property: if the product of two expressions is zero, then at least one of them must be zero.
然后让每个因式分别等于 0 并求解。这里用到零乘积性质：如果两个表达式的乘积为 0，那么其中至少有一个必须为 0。
x + 2 = 0 → x = −2
x + 3 = 0 → x = −3
Always expand your factored form to check that it matches the original expression.
始终将因式展开，检查是否与原表达式一致。
3. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This method is especially useful for finding the turning point of a graph and for solving equations that do not factorise easily.
配方法将二次表达式改写为 a(x + p)² + q 的形式。这种方法特别适用于求图像的顶点，以及解那些不容易因式分解的方程。
move the constant to the other side, then add the square of half the coefficient of x to both sides.
例如，解 x² + 6x + 1 = 0。先将常数项移到另一边，然后两边加上 x 系数一半的平方。
x² + 6x = −1
x² + 6x + 9 = −1 + 9
(x + 3)² = 8
Now take the square root on both sides, remembering both positive and negative roots.
然后两边开平方根，注意正负两个根。
x + 3 = ±√8 → x = −3 ± 2√2
Thus the exact solutions are x = −3 + 2√2 and x = −3 − 2√2.
因此精确解为 x = −3 + 2√2 和 x = −3 − 2√2。
4. The Quadratic Formula | 求根公式
onal or complex roots. For ax² + bx + c = 0, the solutions are given by the formula below.
求根公式适用于所有二次方程，包括有无理数根或复数根的情况。对于 ax² + bx + c = 0，解由下面的公式给出。
x = (−b ± √b² − 4ac) / (2a)
This formula is derived from completing the square. You are expected to memorise it for the IGCSE exam. The expression under the square root, b² − 4ac, is called the discriminant.
这个公式是由配方法推导出来的。在 IGCSE 考试中，你必须记住它。根号下的表达式 b² − 4ac 称为判别式。
Example: solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Substitute into the formula:
例：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入公式得：
x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (2 ± √10) / 2
So the two roots are (2 + √10)/2 and (2 − √10)/2. These are approximately 2.58 and −0.58.
所以两个根是 (2 + √10)/2 和 (2 − √10)/2，近似值分别为 2.58 和 −0.58。
5. The Discriminant | 判别式
ant Δ = b² − 4ac determines the nature of the roots without solving the equation entirely.
判别式 Δ = b² − 4ac 可以在不完全解方程的情况下判断根的性质。
| Discriminant | Nature of roots | Graphical meaning |
| Δ > 0 | Two distinct real roots | Graph crosses the x-axis at two points |
| Δ = 0 | One repeated real root | Graph touches the x-axis at one point |
| Δ < 0 | No real roots | Graph does not cross the x-axis |
Since Δ > 0, there are two distinct real roots. Always check the discriminant before deciding which solving method to use.
对于方程 3x² + 6x + 2 = 0，判别式 Δ = 6² − 4 × 3 × 2 = 36 − 24 = 12。因为 Δ > 0，所以有两个不相等的实数根。在决定使用哪种求解方法之前，先检查判别式总是好的做法。
6. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards, like a U shape. If a < 0, it opens downwards, like an upside-down U. This is the first thing you should identify when sketching.
y = ax² + bx + c 的图像是抛物线。如果 a > 0，抛物线开口向上，像字母 U。如果 a < 0，抛物线开口向下，像倒过来的 U。这是画草图时首先要判断的信息。
en x = 0, y = c. The roots of the quadratic equation give the x-intercepts, if they exist.
y 截距始终是点 (0, c)，因为当 x = 0 时，y = c。如果存在实数根，二次方程的根给出 x 截距。
To sketch the graph, you need three or four key points: the y-intercept, the x-intercepts (if any), and the turning point. Plot these points and draw a smooth curve through them.
要画草图，你需要三个或四个关键点：y 截距、x 截距（如果有）以及顶点。标出这些点，然后画一条平滑曲线穿过它们。
7. Roots, Turning Points and Symmetry | 根、顶点与对称性
vertex is x = −b / (2a). This can be found by completing the square or using the formula.
二次图像的顶点也称为驻点或极值点。对于形式为 y = ax² + bx + c 的抛物线，顶点的 x 坐标为 x = −b / (2a)。这可以通过配方法或公式求得。
e through the vertex, given by x = −b / (2a). The parabola is symmetrical about this line.
对称轴是过顶点且竖直的直线，方程为 x = −b / (2a)。抛物线关于这条线对称。
2. Substitute this x-value back into the equation to find the y-coordinate of the vertex.
如果两个根是 r₁ 和 r₂，那么顶点的 x 坐标也是两根的中点：x = (r₁ + r₂) / 2。将这个 x 值代回方程，即可求出顶点的 y 坐标。
f symmetry is x = (1 + 3)/2 = 2. When x = 2, y = 4 − 8 + 3 = −1. So the vertex is (2, −1).
例如，y = x² − 4x + 3 的根为 1 和 3。对称轴为 x = (1 + 3)/2 = 2。当 x = 2 时，y = 4 − 8 + 3 = −1。所以顶点是 (2, −1)。
8. Solving Quadratic Inequalities | 二次不等式
Quadratic inequalities are solved by first finding the critical values (the roots), then testing intervals on a number line or using the graph of the parabola.
解二次不等式，首先要求出临界值（即根），然后在数轴上测试区间，或者借助抛物线的图像来判断。
values are x = 5 and x = −2. These two values divide the number line into three intervals.
以不等式 x² − 3x − 10 < 0 为例。先因式分解：(x − 5)(x + 2) < 0。临界值为 x = 5 和 x = −2。这两个数把数轴分成三个区间。
x < −2: both factors negative → product positive
−2 < x < 5: one factor negative, one positive → product negative
x > 5: both factors positive → product positive
x < 5. If the inequality had been ≤ 0 or ≥ 0, include the endpoints with a closed bracket.
因为我们需要乘积小于 0，所以解集为 −2 < x < 5。如果原不等式是 ≤ 0 或 ≥ 0，那么端点要用闭区间包含。
If a > 0 and the inequality is '< 0', the solution is between the roots. If it is '> 0', the solution is outside the roots. This rule helps you check your answer quickly.
如果 a > 0，且不等式是 '< 0'，解集在两个根之间。如果是 '> 0'，解集在两个根之外。利用这个规律可以快速检查答案。
9. Applications in Problem Solving | 实际应用
Quadratic equations are used to model many real-world situations: the path of a thrown object, the area of a rectangle with fixed perimeter, and profit functions in business.
二次方程可以用来建立许多实际问题的模型：抛体运动的路径、固定周长下矩形的面积，以及商业中的利润函数。
Worked example: A rectangle has length (x + 4) cm and width x cm. Its area is 45 cm². Find the value of x.
例题：一个矩形的长为 (x + 4) cm，宽为 x cm，面积为 45 cm²。求 x 的值。
ctorise: (x + 9)(x − 5) = 0. Thus x = −9 or x = 5. Since length cannot be negative, x = 5.
列方程：x(x + 4) = 45，即 x² + 4x − 45 = 0。因式分解得：(x + 9)(x − 5) = 0。所以 x = −9 或 x = 5。因为长度不能为负，所以 x = 5。
When solving applied problems, always reject any negative solution that does not make sense in context. Check whether your final answer satisfies the original condition.
在解应用题时，要舍去在情境中不合理的负解。最后检查答案是否满足原始条件。
10. Common Mistakes and Exam Tips | 常见错误与考试建议
ust first be rewritten as x² − 3x − 4 = 0. Only then can you factorise or use the formula.
一个常见错误是忘记先把方程整理成标准形式。例如，x² = 3x + 4 必须先改写为 x² − 3x − 4 = 0，然后才能因式分解或使用公式。
ve root when solving by taking square roots. Remember that if x² = k and k > 0, then x = ±√k, not just x = √k.
另一个常见错误是两边开平方时丢掉负根。记住，如果 x² = k 且 k > 0，那么 x = ±√k，而不是仅仅 x = √k。
Always check that a ≠ 0 in a quadratic equation.
If the coefficient of x² is not 1, divide carefully or build the factorised form using the correct signs.
adratic formula, be careful with negative b: x = (−(−4) ± ...)/(2 × 2) is easy to miscopy.
For graph sketching, always label the vertex, y-intercept, and x-intercepts clearly.
In the exam, show all working steps. If you use a calculator for the quadratic formula, write down the substituted expression before the final approximate answer. This earns method marks.
考试中要写出完整的步骤。如果你用计算器算求根公式，请先写出代入后的表达式，再写出最终近似值。这样可以获得方法分。
11. Practice Questions | 练习题
Try these questions on your own before checking the solutions. They cover the key skills from this article.
请先独立尝试这些题目，再查看解答。它们涵盖了这篇文章中的核心技能。
Solve x² − 7x + 12 = 0 by factorisation.
Solve 2x² + 8x − 10 = 0 using the quadratic formula.
Find the discriminant of 3x² − 6x + 5 = 0 and state the number of real roots.
Write y = x² + 4x + 1 in the form (x + p)² + q, hence find the vertex.
Solve the inequality x² − 2x − 8 ≥ 0.
12. Solutions and Final Summary | 解答与总结
Solution 1: x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0 → x = 3 or x = 4.
解答 1：x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0 → x = 3 或 x = 4。
Solution 2: a = 2, b = 8, c = −10. x = (−8 ± √64 + 80) / 4 = (−8 ± 12) / 4, so x = 1 or x = −5.
解答 2：a = 2，b = 8，c = −10。x = (−8 ± √64 + 80) / 4 = (−8 ± 12) / 4，所以 x = 1 或 x = −5。
Solution 3: Δ = (−6)² − 4 × 3 × 5 = 36 − 60 = −24. Since Δ < 0, there are no real roots.
解答 3：Δ = (−6)² − 4 × 3 × 5 = 36 − 60 = −24。因为 Δ < 0，所以没有实数根。
Solution 4: y = (x + 2)² − 3, so the vertex is (−2, −3).
解答 4：y = (x + 2)² − 3，所以顶点为 (−2, −3)。
olution 5: x² − 2x − 8 = (x − 4)(x + 2) ≥ 0. Since a > 0, the solution is x ≤ −2 or x ≥ 4.
解答 5：x² − 2x − 8 = (x − 4)(x + 2) ≥ 0。因为 a > 0，所以解集为 x ≤ −2 或 x ≥ 4。
Quadratic equations are a core skill for IGCSE Mathematics. Master the three solving methods, understand the discriminant, and practise sketching graphs. With structured revision and careful algebraic manipulation, this topic becomes a reliable source of marks.
二次方程是 IGCSE 数学的核心技能。掌握三种求解方法，理解判别式的意义，并在画图时多加练习。通过有结构的复习和细心的代数运算，这个主题会成为稳定的得分点。
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