📚 Quadratic Equations for IGCSE Mathematics | IGCSE 数学：二次方程
Quadratic equations appear throughout the IGCSE syllabus, from algebra and graphs to measurement and word problems. A strong understanding of solving quadratics by factorisation, completing the square, and the quadratic formula will help you tackle both Core and Extended questions with confidence.
二次方程贯穿 IGCSE 数学大纲，从代数、图像到测量和应用题都有涉及。熟练掌握因式分解法、配方法和求根公式，能帮助你自信应对核心卷和扩展卷中的相关题目。
1. What is a Quadratic Equation? | 什么是二次方程？
re a, b and c are constants and a ≠ 0. If a were zero, the equation would become linear, not quadratic.
二次方程是次数为 2 的多项式方程。它的一般形式为 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。如果 a 等于零，方程就变成一次方程，而不是二次方程。
9 = 0. The highest power of x is 2 in each case, which is the key feature of a quadratic.
常见的例子包括 x² − 5x + 6 = 0、2x² + 3x − 2 = 0 和 4x² − 9 = 0。每个方程中 x 的最高次数都是 2，这是二次方程的关键特征。
2. Standard Form and Key Terms | 标准形式与关键术语
ct all terms on one side and simplify. The coefficient a is the leading coefficient, b is the coefficient of x, and c is the constant term.
解题前先把方程化为标准形式 ax² + bx + c = 0。将所有项移到一边并化简。系数 a 是二次项系数，b 是 x 的系数，c 是常数项。
- Leading coefficient a controls the width and direction of the parabola. 二次项系数 a 控制抛物线的开口方向和宽窄。
- Constant c gives the y-intercept of the graph y = ax² + bx + c. 常数项 c 对应图像 y = ax² + bx + c 与 y 轴交点的纵坐标。
- Roots or solutions are the x-values that make the equation true. 根或解 是使方程成立的 x 值。
c = 3. Identifying these values correctly is essential before using the quadratic formula.
例如在 2x² − 7x + 3 = 0 中，a = 2，b = −7，c = 3。在使用求根公式前，正确识别这些数值非常重要。
3. Solving by Factorisation | 因式分解法求解
for x. This method relies on the zero product property: if p × q = 0, then p = 0 or q = 0.
当二次式可以写成两个一次因式的乘积时，因式分解法非常快捷。令每个括号等于零并解出 x。该方法依据零积性质：若 p × q = 0，则 p = 0 或 q = 0。
For x² − 5x + 6 = 0, factorise to (x − 2)(x − 3) = 0. The solutions are x = 2 or x = 3.
例如 x² − 5x + 6 = 0，因式分解为 (x − 2)(x − 3) = 0。解为 x = 2 或 x = 3。
tra care. For 2x² + 3x − 2 = 0, write it as (2x − 1)(x + 2) = 0, giving x = 1/2 or x = −2.
当 a ≠ 1 时，因式分解需要更加仔细。例如 2x² + 3x − 2 = 0，可写成 (2x − 1)(x + 2) = 0，得到 x = 1/2 或 x = −2。
Always check by expanding the brackets. Remember that the signs must multiply to give c and add to give b.
务必通过展开括号来检验。记住两个因式的常数项相乘必须得到 c，相加必须得到 b。
4. Solving by Completing the Square | 配方法求解
od is especially useful when the quadratic does not factorise neatly and when you need to find the vertex of a parabola.
配方法将二次式转化为 (x + p)² = q 的形式。当二次式不容易因式分解或需要求抛物线顶点时，该方法尤其有用。
+ 3)² − 9. The equation becomes (x + 3)² − 9 + 2 = 0, so (x + 3)² = 7. Therefore x = −3 ± √7.
例如 x² + 6x + 2 = 0，将 x² + 6x 写成 (x + 3)² − 9。方程变为 (x + 3)² − 9 + 2 = 0，即 (x + 3)² = 7。因此 x = −3 ± √7。
x² + bx = (x + b ÷ 2)² − (b ÷ 2)²
The identity above is the core step of completing the square. Divide the coefficient of x by 2, square the result, and keep the equation balanced.
上述恒等式是配方法的核心步骤。先把 x 的系数除以 2，再平方，最后在等式另一边保持平衡。
If the leading coefficient is not 1, first factor it out from the x² and x terms before completing the square.
如果二次项系数不是 1，通常先将其从 x² 项和 x 项中提出，再进行配方。
5. The Quadratic Formula | 求根公式
ard form. It is derived by completing the square on the general equation ax² + bx + c = 0.
求根公式可以求解任何标准形式的二次方程。它由对一般方程 ax² + bx + c = 0 进行配方法推导而来。
x = (−b ± √b² − 4ac) ÷ (2a)
Substitute the values of a, b and c carefully, paying attention to negative signs. Simplify the square root and the fraction fully.
代入 a、b、c 时要特别注意负号。最后将根号和分式化简到最简形式。
For 2x² + 3x − 2 = 0, a = 2, b = 3, c = −2. The formula gives x = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4, so x = 1/2 or x = −2.
例如 2x² + 3x − 2 = 0，a = 2，b = 3，c = −2。代入公式得 x = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4，所以 x = 1/2 或 x = −2。
This method works even when factorisation is difficult, such as when the roots involve square roots or fractions.
当因式分解较困难时，例如根含有根号或分数时，求根公式仍然有效。
6. Discriminant and Nature of Roots | 判别式与根的性质
4ac. It tells you how many real solutions the quadratic has without solving it completely.
判别式是根号内的表达式：Δ = b² − 4ac。它能在不完整求解的情况下判断二次方程有多少个实数解。
| Discriminant Δ = b² − 4ac | Nature of roots 根的性质 |
|---|---|
| Δ > 0 | Two distinct real roots 两个不相等的实数根 |
| Δ = 0
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