📚 Solving Quadratic Equations | 二次方程解析
A quadratic equation is one of the most important algebraic topics in the IGCSE Mathematics syllabus. Whether you are solving by factorising, using the quadratic formula or completing the square, a solid understanding of quadratics will support your performance in many exam questions.
二次方程是 IGCSE 数学课程中最重要的代数主题之一。无论是通过因式分解、使用二次公式还是配方法来求解，扎实掌握二次方程都能帮助你在许多考试题目中取得更好表现。
1. Standard Form | 标准形式
A quadratic equation is an equation that can be written in the standard form
二次方程是可以用标准形式书写的方程：
ax² + bx + c = 0
where a, b and c are constants, and a is not equal to 0. The coefficient a determines whether the parabola opens upwards or downwards, and the values of b and c affect the position of the curve.
其中 a、b 和 c 是常数，且 a 不等于 0。系数 a 决定抛物线的开口方向，而 b 和 c 的值影响曲线的位置。
has a = 3, b = -2 and c = 1. If a were 0, the equation would become linear, not quadratic.
例如，3x² - 2x + 1 = 0 中 a = 3，b = -2，c = 1。如果 a 为 0，方程将变为一次方程，而不是二次方程。
2. Solving by Factorisation | 因式分解法
r factors. The key principle is the zero-product property: if AB = 0, then A = 0 or B = 0.
当二次表达式可以写成两个一次因式的乘积时，因式分解通常是最快的方法。关键原理是零积性质：如果 AB = 0，那么 A = 0 或 B = 0。
d -3, so the factors are (x - 2)(x - 3). Setting each factor to zero gives x = 2 or x = 3.
例如，解 x² - 5x + 6 = 0。我们寻找两个相乘等于 6 且相加等于 -5 的数。这两个数是 -2 和 -3，因此因式为 (x - 2)(x - 3)。令每个因式等于零，得到 x = 2 或 x = 3。
xample, 2x² + 5x - 3 = 0 can be factorised as (2x - 1)(x + 3) = 0, giving x = ½ or x = -3.
当 x² 的系数不为 1 时，因式分解需要格外小心。例如，2x² + 5x - 3 = 0 可以分解为 (2x - 1)(x + 3) = 0，得到 x = ½ 或 x = -3。
- Always expand your factors to check they produce the original expression.
- 务必展开因式，检查是否得到原表达式。
- If the expression cannot be factorised, use another method.
- 如果表达式无法因式分解，请使用其他方法。
3. The Quadratic Formula | 二次公式
g those that cannot be factorised easily. For ax² + bx + c = 0, the solutions are given by
二次公式适用于所有二次方程，包括那些不容易因式分解的方程。对于 ax² + bx + c = 0，解为：
x = (-b ± √b² - 4ac) / (2a)
mbol ± means that we calculate two values: one with a plus sign and one with a minus sign.
在这个公式中，符号 ± 表示我们需要计算两个值：一个用加号，一个用减号。
- 4x - 3 = 0. Here a = 2, b = -4 and c = -3. Substituting into the formula gives x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4, which simplifies to x ≈ 2.58 or x ≈ -0.58.
例如，解 2x² - 4x - 3 = 0。这里 a = 2，b = -4，c = -3。代入公式得到 x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4，化简后 x ≈ 2.58 或 x ≈ -0.58。
Always simplify the square root if possible, and round only at the final step to avoid rounding errors.
如果可以，务必化简根号；只在最后一步四舍五入，以避免舍入误差。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This is useful for finding the turning point of a parabola and for solving equations without using the formula.
配方法将二次式改写为 a(x + p)² + q 的形式。这对于求抛物线的顶点以及在不使用公式的情况下解方程非常有用。
ubtract (b/2)². Since x² + bx + (b/2)² = (x + b/2)², we get x² + bx = (x + b/2)² - (b/2)².
对于 x² + bx，我们加上并减去 (b/2)²。因为 x² + bx + (b/2)² = (x + b/2)²，所以 x² + bx = (x + b/2)² - (b/2)²。
= (x + 3)² - 9 - 1 = (x + 3)² - 10. Setting this to zero gives (x + 3)² = 10, so x = -3 ± √10.
例如，x² + 6x - 1 = (x + 3)² - 9 - 1 = (x + 3)² - 10。令其等于零，得到 (x + 3)² = 10，所以 x = -3 ± √10。
a first. For 2x² + 8x + 5 = 0, write 2(x² + 4x) + 5 = 2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3.
当 a 不是 1 时，先提出 a。对于 2x² + 8x + 5 = 0，写成 2(x² + 4x) + 5 = 2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3。
5. The Discriminant | 判别式
Inside the quadratic formula, the expression b² - 4ac is called the discriminant,
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