📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, problem solving, and even in real-life contexts such as area and projectile motion. A solid understanding of quadratics will help you solve equations, sketch graphs, and analyse the nature of roots with confidence.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、坐标几何、问题解决中，甚至在面积和抛体运动等实际情境中也会遇到。扎实掌握二次方程将帮助你自信地解方程、画图像并分析根的性质。
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 unknown variable x is 2, so the equation can have up to two real solutions, called roots.
二次方程是一个二次多项式方程。未知变量 x 的最高次数是 2，因此方程最多可以有两个实数解，这些解称为根。
In IGCSE Mathematics, you must be able to solve quadratic equations by factorising, completing the square, and using the quadratic formula. You also need to link equations to their parabolic graphs.
在 IGCSE 数学中，你必须掌握用因式分解法、配方法和求根公式解二次方程。你还需要把方程与抛物线图像联系起来。
2. Standard Form and Coefficients | 标准形式与系数
a, b, and c are real numbers and a cannot be zero. If a = 0, the equation becomes linear.
二次方程的标准形式是 ax² + bx + c = 0，其中 a、b、c 是实数，且 a 不能为零。如果 a = 0，方程就变成线性方程。
ax² + bx + c = 0, a ≠ 0
The coefficient a controls the width and direction of the parabola, b affects the position of the axis of symmetry, and c is the y-intercept.
系数 a 控制抛物线的宽度和开口方向，b 影响对称轴的位置，c 是 y 轴截距。
3. Solving by Factorising | 因式分解法
product of two binomials. If (px + q)(rx + s) = 0, then at least one factor must be zero.
当二次方程可以表示为两个一次因式的乘积时，因式分解通常是最快的方法。如果 (px + q)(rx + s) = 0，那么至少有一个因式必须为零。
ations: px + q = 0 or rx + s = 0. Solving these gives the roots of the original quadratic.
这样就得到两个独立的一次方程：px + q = 0 或 rx + s = 0。解出它们就得到原二次方程的根。
Always check that the factorised form expands back to the original expression, especially when a ≠ 1.
一定要检查因式分解后展开是否等于原表达式，尤其是当 a ≠ 1 时。
4. Worked Example: Factorising | 因式分解例题
tiply to +6 and add to −5. The numbers are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0.
解方程 x² − 5x + 6 = 0。我们寻找两个数，使它们的乘积为 +6、和为 −5。这两个数是 −2 和 −3，所以 x² − 5x + 6 = (x − 2)(x − 3) = 0。
Setting each factor to zero gives x − 2 = 0 or x − 3 = 0, hence x = 2 or x = 3.
令每个因式为零，得到 x − 2 = 0 或 x − 3 = 0，因此 x = 2 或 x = 3。
plit the middle term or test factor pairs: (2x + 1)(x + 3) = 0, giving x = −1/2 or x = −3.
对于较难的情况，例如 2x² + 7x + 3 = 0，我们拆分中间项或尝试因式对：(2x + 1)(x + 3) = 0，得到 x = −1/2 或 x = −3。
5. Completing the Square | 配方法
Completing the square is a method that always works, even when the quadratic cannot be factorised neatly. It rewrites ax² + bx + c in the form a(x + p)² + q.
配方法是一种始终有效的方法，即使二次方程不能很好地因式分解。它把 ax² + bx + c 改写成 a(x + p)² + q 的形式。
b/2)² to create a perfect square: x² + bx + (b/2)² − (b/2)² + c = (x + b/2)² + c − (b/2)².
对于首项系数为 1 的二次式 x² + bx + c，加减 (b/2)² 来构造完全平方：x² + bx + (b/2)² − (b/2)² + c = (x + b/2)² + c − (b/2)²。
This form is useful for solving equations and for finding the vertex of a parabola.
这种形式对于解方程和求抛物线的顶点都很有用。
6. Worked Example: Completing the Square | 配方法例题
Solve x² + 6x + 2 = 0 by completing the square. First, move the constant: x² + 6x = −2.
用配方法解 x² + 6x + 2 = 0。首先移项：x² + 6x = −2。
Add (6/2)² = 9 to both sides: x² + 6x + 9 = −2 + 9, so (x + 3)² = 7.
两边加上 (6/2)² = 9：x² + 6x + 9 = −2 + 9，所以 (x + 3)² = 7。
Taking square roots gives x + 3 = ±√7, hence x = −3 ± √7. The two roots are x = −3 + √7 and x = −3 − √7.
开平方得到 x + 3 = ±√7，因此 x = −3 ± √7。两个根分别是 x = −3 + √7 和 x = −3 − √7。
7. The Quadratic Formula | 求根公式
oots of any quadratic equation directly. For ax² + bx + c = 0, the solutions are x = (−b ± √b² − 4ac) / (2a).
求根公式直接给出任何二次方程的根。对于 ax² + bx + c = 0，解为 x = (−b ± √b² − 4ac) / (2a)。
x = (−b ± √b² − 4ac) / (2a)
This formula is derived from completing the square and is provided in most IGCSE formula sheets. It is especially reliable when factorising is difficult.
这个公式由配方法推导而来，通常在 IGCSE 公式表中给出。当因式分解困难时，它尤其可靠。
Example: Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3, so x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (2 ± √10) / 2.
例题：解 2x² − 4x − 3 = 0。这里 a = 2、b = −4、c = −3，所以 x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (2 ± √10) / 2。
8. Discriminant and Nature of Roots | 判别式与根的性质
Δ = b² − 4ac. It tells us the nature of the roots without solving the equation completely.
判别式是求根公式中平方根下的表达式：Δ = b² − 4ac。它可以在不完全解方程的情况下告诉我们根的性质。
Δ = b² − 4ac
ne real root, sometimes called a repeated root. If Δ < 0, there are no real roots, but two complex roots.
如果 Δ > 0，方程有两个不同的实数根。如果 Δ = 0，则恰好有一个实数根，有时称为重根。如果 Δ < 0，则没有实数根，但有两个复数根。
+ 2x + 5 = 0, Δ = 2² − 4(3)(5) = 4 − 60 = −56 < 0, so the equation has no real solutions.
例题：对于 3x² + 2x + 5 = 0，Δ
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图