Mastering Quadratic Equations for IGCSE | IGCSE 二次方程完全攻略

📚 Mastering Quadratic Equations for IGCSE | IGCSE 二次方程完全攻略

A quadratic equation is one of the most important topics in the IGCSE Mathematics syllabus. It appears not only as a standalone question but also inside graphs, inequalities, sequences and real-life modelling problems. Understanding quadratics gives you a strong foundation for further study in algebra and calculus.

二次方程是 IGCSE 数学大纲中最重要的主题之一。它不仅作为独立考题出现，还会出现在图像、不等式、数列和实际建模问题中。理解二次方程可以为你进一步学习代数和微积分打下坚实基础。

1. What is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is a polynomial equation in which the highest power of the variable is 2. The general form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

二次方程是变量的最高次数为 2 的多项式方程。其一般形式为 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。

If a = 0, the equation becomes linear, not quadratic. This condition is important because it keeps the x² term present.

如果 a = 0，方程就变成一次方程，而不再是二次方程。这个条件很重要，因为它确保 x² 项存在。

Examples of quadratic equations include x² + 5x + 6 = 0, 2x² – 3x + 1 = 0 and x² – 9 = 0.

二次方程的例子包括 x² + 5x + 6 = 0、2x² – 3x + 1 = 0 和 x² – 9 = 0。

2. Standard Form and Coefficients | 标准形式与系数

The standard form of a quadratic equation is ax² + bx + c = 0. Here a is the coefficient of x², b is the coefficient of x, and c is the constant term.

二次方程的标准形式是 ax² + bx + c = 0。其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。

Before solving, always rearrange the equation so that one side is zero. For example, 3x² = 5x – 2 should be written as 3x² – 5x + 2 = 0.

在求解之前，一定要把方程整理成一边为零的形式。例如，3x² = 5x – 2 应写成 3x² – 5x + 2 = 0。

Identifying a, b and c correctly is essential when using the quadratic formula or the discriminant.

正确识别 a、b、c 对于使用二次公式或判别式非常重要。

3. Solving by Factorising | 因式分解法

Factorising is often the quickest method when the quadratic has simple integer roots. We look for two numbers that multiply to give ac and add to give b.

当二次方程有简单的整数根时，因式分解通常是最快的方法。我们要寻找两个数，它们的乘积等于 ac，和等于 b。

For x² + 5x + 6 = 0, we need two numbers that multiply to 6 and add to 5. The numbers are 2 and 3, so the equation becomes (x + 2)(x + 3) = 0.

对于 x² + 5x + 6 = 0，我们需要两个数，乘积为 6，和为 5。这两个数是 2 和 3，因此方程变为 (x + 2)(x + 3) = 0。

Using the zero-product property, set each factor equal to zero: x + 2 = 0 or x + 3 = 0. Therefore x = -2 or x = -3.

利用零乘积性质，令每个因式等于零：x + 2 = 0 或 x + 3 = 0。因此 x = -2 或 x = -3。

Always check by expanding: (x + 2)(x + 3) = x² + 5x + 6.
务必通过展开检验：(x + 2)(x + 3) = x² + 5x + 6。

4. Solving by Completing the Square | 配方法

Completing the square transforms a quadratic into the form (x + p)² = q. This is useful when the quadratic cannot be factorised easily.

配方法将二次方程转化为 (x + p)² = q 的形式。当二次方程不容易因式分解时，这种方法非常有用。

For x² + 6x + 2 = 0, first move the constant: x² + 6x = -2. Then add (6/2)² = 9 to both sides to get x² + 6x + 9 = 7, which is (x + 3)² = 7.

对于 x² + 6x + 2 = 0，首先移常数项：x² + 6x = -2。然后两边加上 (6/2)² = 9，得到 x² + 6x + 9 = 7，即 (x + 3)² = 7。

Taking the square root gives x + 3 = ±√7, so x = -3 ± √7.

开平方得到 x + 3 = ±√7，因此 x = -3 ± √7。

This method also reveals the turning point of a quadratic graph, which is why it appears frequently in IGCSE questions.

这种方法还能揭示二次函数图像的顶点，因此它在 IGCSE 考题中经常出现。

5. The Quadratic Formula | 二次公式

The quadratic formula solves any quadratic equation ax² + bx + c = 0. It is given by:

二次公式可以求解任何形如 ax² + bx + c = 0 的二次方程。公式如下：

x = (-b ± √(b² – 4ac)) / 2a

For 2x² – 4x – 6 = 0, identify a = 2, b = -4, c = -6. Then x = (4 ± √((-4)² – 4 × 2 × (-6))) / (2 × 2) = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4.

对于 2x² – 4x – 6 = 0，识别 a = 2、b = -4、c = -6。那么 x = (4 ± √((-4)² – 4 × 2 × (-6))) / (2 × 2) = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4。

This gives x = 3 or x = -1. The formula is especially useful when factorising is difficult or when coefficients are not integers.

这样得到 x = 3 或 x = -1。当因式分解困难或系数不是整数时，该公式特别有用。

6. The Discriminant | 判别式

The discriminant is the expression inside the square root of the quadratic formula: Δ = b² – 4ac. It tells us how many real solutions a quadratic equation has.

判别式是二次公式中平方根号内的表达式：Δ = b² – 4ac。它告诉我们二次方程有多少个实数解。

If Δ > 0, the equation has two distinct real roots.
如果 Δ > 0，方程有两个不同的实数根。
If Δ = 0, the equation has exactly one repeated real root.
如果 Δ = 0，方程有一个重复的实数根。
If Δ < 0, the equation has no real roots, but two complex roots.
如果 Δ < 0，方程没有实数根，但有两个复数根。

For x² + 4x + 5 = 0, Δ = 4² – 4 × 1 × 5 = 16 – 20 = -4. Since Δ < 0, the equation has no real solutions.

对于 x² + 4x + 5 = 0，Δ = 4² – 4 × 1 × 5 = 16 – 20 = -4。由于 Δ < 0，该方程没有实数解。

IGCSE exam questions often ask you to find the range of k for which a quadratic has two real roots. This requires solving an inequality involving the discriminant.

IGCSE 考试题经常要求你求出使二次方程有两个实数根的 k 的取值范围。这需要解一个涉及判别式的不等式。

7. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a parabola. 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。如果 a > 0，抛物线开口向上，有一个最低点。如果 a < 0，抛物线开口向下，有一个最高点。

The y-intercept is the constant term c, because when x = 0, y = c. The x-intercepts are the roots of the equation ax² + bx + c = 0.

y 轴截距是常数项 c，因为当 x = 0 时，y = c。x 轴截距是方程 ax² + bx + c = 0 的根。

The axis of symmetry is a vertical line that passes through the turning point. Its equation is x = -b / 2a.

对称轴是一条经过顶点的竖直线。它的方程是 x = -b / 2a。

Sketching a quadratic graph usually requires finding the intercepts, the turning point and the direction of opening.

绘制二次函数图像通常需要找到截距、顶点和开口方向。

8. Turning Point and Symmetry | 顶点与对称性

The turning point is the point where the parabola changes direction. Its x-coordinate is given by x = -b / 2a.

顶点是抛物线改变方向的点。它的 x 坐标由 x = -b / 2a 给出。

To find the y-coordinate, substitute x = -b / 2a back into the equation y = ax² + bx + c.

要求 y 坐标，将 x = -b / 2a 代回方程 y = ax² + bx + c。

If the quadratic is written in completed-square form y = a(x – h)² + k, the turning point is simply (h, k).

如果二次函数写成完全平方形式 y = a(x – h)² + k，顶点就是 (h, k)。

For y = 2(x – 3)² – 5, the turning point is (3, -5). Since a = 2 > 0, it is a minimum point.

对于 y = 2(x – 3)² – 5，顶点是 (3, -5)。因为 a = 2 > 0，它是一个最低点。

Symmetry means that points equidistant from the axis of symmetry have the same y-value. This helps when reading values from a table or graph.

对称性意味着与对称轴距离相等的点具有相同的 y 值。这有助于从表格或图像中读取数值。

9. Roots and Coefficients | 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, the sum of the roots is α + β = -b/a and the product is αβ = c/a.

对于有根 α 和 β 的二次方程 ax² + bx + c = 0，根的和为 α + β = -b/a，根的积为 αβ = c/a。

If the roots of x² – 7x + 10 = 0 are 2 and 5, then the sum is 2 + 5 = 7 = -(-7)/1 and the product is 2 × 5 = 10 = 10/1.

如果 x² – 7x + 10 = 0 的根是 2 和 5，那么和为 2 + 5 = 7 = -(-7)/1，积为 2 × 5 = 10 = 10/1。

These relationships can be used to form a quadratic equation from given roots. If the roots are p and q, the equation is x² – (p + q)x + pq = 0.

这些关系可以用来根据已知根构造二次方程。如果根是 p 和 q，方程就是 x² – (p + q)x + pq = 0。

This topic connects very well with algebraic manipulation and appears in many IGCSE extended papers.

这一主题与代数变形联系紧密，出现在许多 IGCSE 扩展试卷中。

10. Quadratic Inequalities | 二次不等式

A quadratic inequality such as x² – 4x + 3 > 0 is solved by first factorising to get (x – 1)(x – 3) > 0.

形如 x² – 4x + 3 > 0 的二次不等式可以通过先因式分解为 (x – 1)(x – 3) > 0 来求解。

The critical values are x = 1 and x = 3, where the expression equals zero. These values divide the number line into three intervals: x < 1, 1 < x < 3 and x > 3.

临界值是 x = 1 和 x = 3，此时表达式等于零。这些值将数轴分成三个区间：x < 1、1 < x < 3 和 x > 3。

Test a value from each interval in the inequality. For x = 0, (0 – 1)(0 – 3) = 3 > 0, so x < 1 is part of the solution. For x = 2, (2 - 1)(2 - 3) = -1 < 0, so 1 < x < 3 is not included. For x = 4, (4 - 1)(4 - 3) = 3 > 0, so x > 3 is included.

在每个区间中取一个值代入不等式检验。当 x = 0 时，(0 – 1)(0 – 3) = 3 > 0，所以 x < 1 是解的一部分。当 x = 2 时，(2 - 1)(2 - 3) = -1 < 0，所以 1 < x < 3 不包括在内。当 x = 4 时，(4 - 1)(4 - 3) = 3 > 0，所以 x > 3 包括在内。

Therefore the solution is x < 1 or x > 3. Always write the final answer using inequality signs or interval notation as required.

因此解为 x < 1 或 x > 3。请始终按照题目要求使用不等式符号或区间表示法写出最终答案。

11. Real-Life Applications | 实际应用

Quadratic equations model many real-life situations, such as projectile motion, area optimisation and profit maximisation.

二次方程可以为许多实际情境建模，例如抛体运动、面积优化和利润最大化。

For example, the height h metres of a ball after t seconds can be modelled by h = -5t² + 20t + 1. To find when the ball hits the ground, set h = 0 and solve -5t² + 20t + 1 = 0.

例如，一个球在 t 秒后的高度 h 米可以建模为 h = -5t² + 20t + 1。要求球何时落地，令 h = 0 并解 -5t² + 20t + 1 = 0。

Another common example is a rectangle with perimeter 20 m. If its length is x, the width is 10 – x, so the area is A = x(10 – x) = -x² + 10x. The maximum area occurs at the turning point.

另一个常见例子是周长为 20 m 的长方形。如果长为 x，宽为 10 – x，那么面积 A = x(10 – x) = -x² + 10x。最大面积出现在顶点处。

Using x = -b / 2a = -10 / (2 × -1) = 5, the maximum area is A = 25 m². This shows why quadratic functions are powerful problem-solving tools.

利用 x = -b / 2a = -10 / (2 × -1) = 5，最大面积为 A = 25 m²。这说明二次函数是强大的解题工具。

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

Many marks are lost by simple mistakes. Here are some points to check carefully in IGCSE exams:

许多分数因简单错误而丢失。以下是在 IGCSE 考试中需要仔细检查的要点：

Always rearrange into ax² + bx + c = 0 before identifying a, b and c.
在识别 a、b、c 之前，务必先整理为 ax² + bx + c = 0。
Do not forget the ± sign when taking a square root.
开平方时不要忘记 ± 号。
Check your factorisation by expanding the brackets.
通过展开括号来检查因式分解。
Exact answers should be left in surd form unless the question says otherwise.
除非题目另有说明，精确答案应保留根式形式。

When using the quadratic formula, use brackets around the numerator and denominator to avoid calculator errors.

使用二次公式时，请在分子和分母上使用括号，以避免计算器错误。

Finally, match your final answer to the context. Lengths and times are positive, so reject negative roots that do not make sense.

最后，让你的最终答案与题目情境相符。长度和时间是正数，因此要舍去不合理的负根。

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Mastering Quadratic Equations for IGCSE | IGCSE 二次方程完全攻略