📚 Solving Quadratic Equations | 解二次方程
A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in almost every exam paper, often in both Paper 2 and Paper 4, and forms the foundation for more advanced work in functions, graphs and calculus.
二次方程是 IGCSE 数学中最重要的内容之一。它几乎出现在每一份试卷中，通常 Paper 2 和 Paper 4 都会涉及，并且是函数、图像和微积分等进阶知识的基石。
1. What Is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2. The highest power of the variable is 2, and the general form is:
二次方程是次数为 2 的多项式方程。变量的最高次数是 2，其一般形式为：
ax² + bx + c = 0, where a ≠ 0
n. The condition a ≠ 0 is essential: if a = 0, the equation becomes linear, not quadratic.
其中 a、b、c 是常数，x 是未知数。条件 a ≠ 0 至关重要：如果 a = 0，方程就变成一次方程，而不是二次方程。
For example, x² + 5x + 6 = 0 is quadratic, while 2x + 3 = 0 is linear.
例如，x² + 5x + 6 = 0 是二次方程，而 2x + 3 = 0 是一次方程。
2. The Standard Form | 标准形式
expanding brackets, collecting like terms, and making sure everything is on one side of the equals sign.
在解任何二次方程之前，你应先将它整理成标准形式 ax² + bx + c = 0。这意味着要展开括号、合并同类项，并确保所有项都在等号的一侧。
Consider the equation:
考虑以下方程：
(x + 3)² = 2x + 10
Expand the square: x² + 6x + 9 = 2x + 10.
展开平方：x² + 6x + 9 = 2x + 10。
sides: x² + 4x − 1 = 0. Now the equation is in standard form with a = 1, b = 4 and c = −1.
两边同时减去 2x 和 10：x² + 4x − 1 = 0。此时方程就是标准形式，其中 a = 1，b = 4，c = −1。
3. Solving by Factorisation | 因式分解法
Factorisation is the fastest method when the quadratic has simple integer factors. The key idea is to write ax² + bx + c as a product of two brackets, then set each bracket equal to zero.
当二次方程具有简单的整数因式时，因式分解是最快的方法。核心思想是把 ax² + bx + c 写成两个括号的乘积，然后令每个括号分别等于零。
Worked example 1: Solve x² + 5x + 6 = 0.
例题 1：解 x² + 5x + 6 = 0。
We need two numbers that multiply to 6 and add to 5. The numbers are 2 and 3. Therefore:
我们需要找出两个数，它们相乘得 6，相加得 5。这两个数是 2 和 3。因此：
(x + 2)(x + 3) = 0
Now set each bracket to zero:
现在令每个括号等于零：
x + 2 = 0 or x + 3 = 0
ll factor law: if the product of two expressions is zero, then at least one of them must be zero.
所以 x = −2 或 x = −3。这称为零因子律：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。
Worked example 2: Solve x² − 3x − 10 = 0.
例题 2：解 x² − 3x − 10 = 0。
We need two numbers with product −10 and sum −3. The numbers are −5 and 2:
我们需要两个数，其乘积为 −10，和为 −3。这两个数是 −5 和 2：
(x − 5)(x + 2) = 0
Therefore x = 5 or x = −2.
因此 x = 5 或 x = −2。
Remember: the signs inside the brackets depend on the signs of b and c. Always check your factorisation by expanding the brackets again.
请记住：括号内的符号取决于 b 和 c 的符号。务必通过重新展开括号来检验你的因式分解是否正确。
4. Solving by Completing the Square | 配方法
Completing the square is a powerful technique that works for any quadratic, even when factorisation is difficult. It rewrites the quadratic in the form (x + p)² + q.
配方法是一种强大的技巧，适用于任何二次方程，即使因式分解很困难时也同样有效。它把二次式改写为 (x + p)² + q 的形式。
For a quadratic x² + bx + c, take half of b, square it, and add and subtract this value.
对于二次式 x² + bx + c，取 b 的一半，将其平方，然后加上再减去这个值。
Worked example 3: Solve x² + 6x + 2 = 0.
例题 3：解 x² + 6x + 2 = 0。
Half of 6 is 3, and 3² = 9. So:
6 的一半是 3，且 3² = 9。所以：
x² + 6x + 9 − 9 + 2 = 0
Group the first three terms as a perfect square:
将前三项组合成一个完全平方：
(x + 3)² − 7 = 0
Now solve: (x + 3)² = 7, so x + 3 = ±√7. Therefore:
现在解方程：(x + 3)² = 7，所以 x + 3 = ±√7。因此：
x = −3 ± √7
This gives x ≈ −3 + 2.646 = −0.354 or x ≈ −3 − 2.646 = −5.646 (to 3 decimal places).
这给出 x ≈ −3 + 2.646 = −0.354 或 x ≈ −3 − 2.646 = −5.646（保留 3 位小数）。
Completing the square also helps you find the turning point (vertex) of a quadratic graph and is the basis for deriving the quadratic formula.
配方法还能帮助你找到二次函数图像的顶点（转向点），同时也是推导二次公式的基础。
5. Solving by the Quadratic Formula | 二次公式法
The quadratic formula can solve any quadratic equation. You should memorise it:
二次公式可以解任何二次方程。你应该牢记它：
x = (−b ± √b² − 4ac) / 2a
Worked example 4: Solve 2x² + 3x − 5 = 0 using the quadratic formula.
例题 4：使用二次公式解 2x² + 3x − 5 = 0。
Identify a = 2, b = 3 and c = −5. Substitute into the formula:
确定 a = 2，b = 3，c = −5。代入公式：
x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2)
Simplify inside the square root: 9 + 40 = 49. So:
化简根号内的部分：9 + 40 = 49。所以：
x = (−3 ± 7) / 4
Thus x = (−3 + 7)/4 = 1 or x = (−3 − 7)/4 = −2.5.
因此 x = (−3 + 7)/4 = 1 或 x = (−3 − 7)/4 = −2.5。
In the IGCSE exam, the quadratic formula is printed in the formula booklet, but you must practise substituting and simplifying accurately, especially with negative signs.
在 IGCSE 考试中，二次公式会印在公式表中，但你必须练习准确地代入和化简，尤其是处理负号时。
6. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression b² − 4ac is called the discriminant. It tells us how many solutions a quadratic equation has without solving it fully.
表达式 b² − 4ac 称为判别式。它无需完整求解就能告诉我们二次方程有多少个解。
If b² − 4ac > 0: two distinct real roots (the graph cuts the x-axis at two points).
如果 b² − 4ac > 0：有两个不同的实根（图像与 x 轴交于两点）。
b² − 4ac = 0: exactly one repeated real root (the graph touches the x-axis at one point).
如果 b² − 4ac = 0：恰好有一个重根（图像与 x 轴相切于一点）。
If b² − 4ac < 0: no real roots (the graph does not cross the x-axis).
如果 b² − 4ac < 0：没有实根（图像不与 x 轴相交）。
Worked example 5: Determine the nature of the roots of x² − 4x + 4 = 0.
例题 5：判断 x² − 4x + 4 = 0 的根的性质。
Here a = 1, b = −4 and c = 4. The discriminant is:
这里 a = 1，b = −4，c = 4。判别式为：
(−4)² − 4 × 1 × 4 = 16 − 16 = 0
has one repeated real root. Indeed, x² − 4x + 4 = (x − 2)², so x = 2 is the only solution.
因为判别式等于 0，方程有一个重根。事实上，x² − 4x + 4 = (x − 2)²，所以 x = 2 是唯一解。
Discriminant questions are very common in IGCSE papers, especially in Paper 4, and often appear with the phrase 'state the nature of the roots'.
判别式问题在 IGCSE 试卷中非常常见，尤其是在 Paper 4 中，通常以"说明根的性质"的形式出现。
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