📚 Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学：二次方程求解
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from basic factorisation to real-life problems involving area, motion and finance. This revision guide explains the three main algebraic methods - factorisation, completing the square and the quadratic formula - and shows how to choose the most efficient method under exam conditions.
二次方程贯穿 IGCSE 数学课程，从基本的因式分解到涉及面积、运动和金融的实际问题。本复习指南讲解三种主要代数方法——因式分解法、配方法和求根公式法，并说明如何在考试条件下选择最高效的方法。
1. Recognising a Quadratic Equation | 识别二次方程
earrangement, such as 3x² = 12 or x(x - 5) = 0, but they all reduce to this standard form.
一元二次方程是任何可写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。关键特征是含有 x² 项；若 a = 0，方程就变成一次方程。二次方程在整理前可能形式不同，例如 3x² = 12 或 x(x - 5) = 0，但它们都能化简为上述标准形式。
thod. For example, x² = 5x - 6 should be rearranged to x² - 5x + 6 = 0 before factorising.
在 IGCSE 考试中，你可能会遇到 x² 项在右边、带括号或含分数的二次方程。选择解法之前，一定要先把方程整理成标准形式。例如，x² = 5x - 6 应整理为 x² - 5x + 6 = 0，再进行因式分解。
2. Standard Form and Key Terms | 标准形式与关键术语
a is called the leading coefficient, b is the linear coefficient, and c is the constant term. The values of x that satisfy the equation are called roots, solutions or x-intercepts because they are the points where the graph crosses the x-axis.
二次方程的标准形式为 ax² + bx + c = 0。系数 a 称为首项系数，b 称为一次项系数，c 称为常数项。使方程成立的 x 值称为根、解或 x 轴截距，因为它们是图像与 x 轴相交的点。
| Term | Meaning | Example for 2x² - 3x + 1 = 0 |
|---|---|---|
| Leading coefficient | Coefficient of x², cannot be zero | a = 2 |
| Linear coefficient | Coefficient of x | b = -3 |
| Constant term | Term without x | c = 1 |
| Roots or solutions | Values of x that satisfy the equation | x = 1 and x = 0.5 |
atic formula. Watch the signs carefully: in 2x² - 3x + 1 = 0, the value of b is -3, not 3.
你应该能够熟练识别任意二次方程中的 a、b 和 c，尤其是在使用求根公式时。要特别注意符号：在 2x² - 3x + 1 = 0 中，b 的值是 -3，而不是 3。
3. Solving by Factorisation | 因式分解法求解
equals zero, then at least one bracket must equal zero, so you can solve two simple linear equations.
当二次方程具有简单的整数根时，因式分解法通常是最快的方法。其思路是把 ax² + bx + c 写成两个一次因式的乘积：(px + q)(rx + s) = 0。如果乘积为零，那么至少有一个因式等于零，因此你可以解两个简单的一次方程。
ct is 10 and whose sum is -7. The numbers are -5 and -2, so x² - 7x + 10 = (x - 5)(x - 2).
当 a = 1 时，因式分解变为 x² + bx + c = (x + m)(x + n)，其中 m + n = b，mn = c。例如，要对 x² - 7x + 10 进行因式分解，需要找到两个数，它们的积为 10，和为 -7。这两个数是 -5 和 -2，因此 x² - 7x + 10 = (x - 5)(x - 2)。
IGCSE answers, both roots must be stated, so do not stop after finding only one solution.
因式分解后，令每个括号等于零：(x - 5)(x - 2) = 0 得到 x - 5 = 0 或 x - 2 = 0，因此 x = 5 或 x = 2。在 IGCSE 答案中，必须写出两个根，所以不要只找到一个解就停止。
4. Worked Example: Factorisation | 因式分解例题
6 and whose sum is b = 5. The numbers are 6 and -1 because 6 × (-1) = -6 and 6 + (-1) = 5.
用因式分解法解 2x² + 5x - 3 = 0。由于 a 不等于 1，需要找到两个数，它们的积为 a × c = 2 × (-3) = -6，和为 b = 5。这两个数是 6 和 -1，因为 6 × (-1) = -6，且 6 + (-1) = 5。
x + 3) = 0. Since both groups contain (x + 3), the expression becomes (2x - 1)(x + 3) = 0.
用这两个数改写中间项：2x² + 6x - x - 3 = 0。然后分组分解：2x(x + 3) - 1(x + 3) = 0。由于两组都含有 (x + 3)，表达式变为 (2x - 1)(x + 3) = 0。
or x = -3. You can check these roots by substituting them back into the original equation.
现在令每个括号为零：2x - 1 = 0 得到 x = 0.5，x + 3 = 0 得到 x = -3。因此解为 x = 0.5 或 x = -3。你可以将这些根代回原方程进行检验。
5. Completing the Square | 配方法
Completing the square is a powerful method that always works and is especially useful for finding the vertex of a quadratic graph or when the roots are irrational. The goal is to rewrite x² + bx as a perfect square plus or minus a constant. The key identity is:
配方法是一种始终有效的方法，尤其适用于求二次函数图像的顶点或当根是无理数时。其目标是将 x² + bx 改写为一个完全平方加上或减去一个常数。关键的恒等式为：
x² + bx = (x + ½b)² - (½b)²
anding (x + 3)² = x² + 6x + 9, so we must subtract 9 to return to the original expression.
例如，x² + 6x 可以写成 (x + 3)² - 9，因为 ½ × 6 = 3，且 3² = 9。这个技巧来自展开式 (x + 3)² = x² + 6x + 9，所以我们必须减去 9 才能回到原式。
To solve a quadratic equation by completing the square, first move the constant term to the right-hand side. Then add the square of half the x-coefficient to both sides. Finally take the square root of both sides and remember to include both the positive and negative square roots.
要用配方法解二次方程，首先把常数项移到方程右边。然后在两边同时加上一次项系数一半的平方。最后对两边开平方，并记住要同时取正平方根和负平方根。
6. Worked Example: Completing the Square | 配方法例题
s: x² + 6x = -2. Half of 6 is 3, and 3² = 9, so add 9 to both sides: x² + 6x + 9 = -2 + 9.
用配方法解 x² + 6x + 2 = 0。首先从两边减去 2：x² + 6x = -2。6 的一半是 3，3² = 9，所以两边同时加上 9：x² + 6x + 9 = -2 + 9。
side is now a perfect square: (x + 3)² = 7. Take the square root of both sides: x + 3 = ±√7. Finally subtract 3 from both sides to isolate x: x = -3 ± √7. These two irrational roots are exact and cannot be simplified further.
左边现在是一个完全平方：(x + 3)² = 7。对两边开平方：x + 3 = ±√7。最后从两边减去 3，得到 x = -3 ± √7。这两个无理数根是精确值，不能再进一步化简。
In an IGCSE exam, leaving the answer as x = -3 + √7 or x = -3 - √7 is normally acceptable when the question asks for exact solutions. If the question requires answers to a given number of decimal places, use a calculator to approximate √7.
在 IGCSE 考试中，当题目要求精确解时，通常可以保留 x = -3 + √7 或 x = -3 - √7 的形式。如果题目要求答案保留指定的小数位数，可用计算器求出 √7 的近似值。
7. The Quadratic Formula | 求根公式
hod for solving quadratics. For any equation ax² + bx + c = 0, the solutions are given by:
求根公式是解二次方程最通用的方法。对于任意方程 ax² + bx + c = 0，其解为：
x = [-b ± √b² - 4ac] ÷ (2a)
t. If Δ < 0, there are no real roots, although the equation still has complex roots at a more advanced level.
平方根下的表达式 b² - 4ac 称为判别式，通常记作 Δ。判别式可以在不解方程的情况下判断根的性质。如果 Δ > 0，方程有两个不同的实根；如果 Δ = 0，方程有一个重根；如果 Δ < 0，方程没有实根，不过在更高阶的学习中方程仍有复数根。
e. It is especially useful when factorisation is difficult, when coefficients are fractions or decimals, or when you need to find roots quickly under exam pressure.
该公式由对一般二次方程 ax² + bx + c = 0 进行配方推导而来，因此在所有情况下都适用。当因式分解困难、系数为分数或小数、或者考试时间紧张需要快速求根时，求根公式特别有用。
8. Worked Example: Quadratic Formula and Discriminant | 求根公式与判别式例题
the signs: b is negative and c is negative. Now substitute these values into the formula.
使用求根公式解 2x² - 4x - 3 = 0。首先确定 a = 2，b = -4，c = -3。要非常注意符号：b 为负，c 也为负。现在将这些值代入公式。
40. Since Δ > 0, the equation has two distinct real roots. The formula gives x = [-(-4) ± √40] ÷ (2 × 2) = [4 ± √40] ÷ 4.
判别式为 Δ = b² - 4ac = (-4)² - 4 × 2 × (-3) = 16 + 24 = 40。由于 Δ > 0，方程有两个不同的实根。公式给出 x = [-(-4) ± √40] ÷ (2 × 2) = [4 ± √40] ÷ 4。
Simplify the surd: √40 = √4 × 10 = 2√10. Therefore x = (4 ± 2√10) ÷ 4. Divide both terms in the numerator by 4 to obtain x = (2 ± √10) ÷ 2. The two exact solutions are x = (2 + √10) ÷ 2 or x = (2 - √10) ÷ 2.
化简二次根式：√40 = √4 × 10 = 2√10。因此 x = (4 ± 2√10) ÷ 4。将分子中的两项都除以 4，得到 x = (2 ± √10) ÷ 2。两个精确解为 x = (2 + √10) ÷ 2 或 x = (2 - √10) ÷ 2。
9. Choosing the Best Method | 选择最佳方法
No single method is always the best; your choice should depend on the form of the quadratic and the question requirements. Use factorisation when the coefficients are small integers and the roots are likely to be integers or simple fractions. Use completing the square when you need the vertex form or when the question specifically asks for this method. Use the quadratic formula when factorisation is not obvious or when coefficients involve decimals, fractions or irrational numbers.
没有一种方法始终是最好的；你的选择应取决于二次方程的形式和题目要求。当系数为较小的整数且根很可能是整数或简单分数时，使用因式分解法。当需要顶点式或题目明确要求使用配方法时，使用配方法。当因式分解不明显或系数含有小数、分数或无理数时，使用求根公式。
| Method | Best for | Limitation |
|---|---|---|
| Factorisation | Simple integer roots, a = 1 or easy factor pairs | Not always possible with simple brackets |
| Completing the square | Vertex form, exact surd answers, a = 1 and b even | Fractions can make working messy |
| Quadratic formula | All quadratics, decimal or fractional coefficients | More steps and easy to make sign errors |
In an IGCSE exam, if the question does not specify a method, choose the one that you find fastest and most accurate for that particular equation. Even if you prefer the formula, checking for simple factorisation first can save valuable time.
在 IGCSE 考试中，如果题目没有指定方法，就选择你对那个特定方程而言最快且最准确的方法。即使你偏好求根公式，先检查是否存在简单因式分解也可以节省宝贵时间。
10. Common Mistakes and Exam Tips | 常见错误与考试提示
so the minus signs are handled correctly. A second common mistake is forgetting that a must not be zero; if the x² term vanishes, the equation is linear and the quadratic formula no longer applies.
许多失分都是由于可避免的符号错误造成的。使用求根公式时，一定要把负值放在括号里：例如 b = -4 时，要写成 -(-4)，这样才能正确处理负号。第二个常见错误是忘记 a 不能为零；如果 x² 项消失，方程就变成一次方程，求根公式不再适用。
s while completing the square, always include the ± sign to produce both possible answers.
另一个常见错误是遗漏根。二次方程通常有两个解，因此在因式分解后，一定要令每一个括号等于零。此外，配方法中开平方时，要始终加上 ± 号，以得到两个可能的答案。
Finally, always check your answers by substitution if time allows. Substitute each solution back into the original equation; if the left-hand side does not equal zero, you have made an error. In graph problems, remember that the roots are the x-intercepts, and the discriminant tells you whether the graph touches, crosses or never meets the x-axis.
最后，如果时间允许，一定要通过代入来检验答案。将每个解代回原方程；如果左边不等于零，就说明计算有误。在函数图像问题中，请记住根就是 x 轴截距，而判别式可以告诉你图像是与 x 轴相切、相交，还是完全不相交。
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