📚 Solving Quadratic Equations | 解二次方程
In this revision guide, we will explore the key techniques for solving quadratic equations, a fundamental topic in the Edexcel IGCSE Mathematics syllabus. You will learn how to factorise, apply the quadratic formula, complete the square, and interpret the discriminant, with worked examples to reinforce your understanding.
本复习指南将带你探索解二次方程的关键技巧，这是 Edexcel IGCSE 数学考纲中的一个基础主题。你将学习如何因式分解、使用二次公式、配方、理解判别式，并通过例题巩固理解。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its standard form is written as:
二次方程是最高次数为2的多项式方程，即变量的最高幂为2。其标准形式写作：
ax² + bx + c = 0, where a ≠ 0
Here, a, b, and c are constants, and x is the unknown variable. The coefficient a must not be zero; otherwise the equation becomes linear.
其中，a、b、c 是常数，x 是未知变量。系数 a 不能为零，否则方程就变成一次方程。
For example, 2x² − 3x + 1 = 0 is a quadratic equation, while x + 5 = 0 is not.
例如，2x² − 3x + 1 = 0 是二次方程，而 x + 5 = 0 不是。
2. Solving by Factorising | 因式分解法
If a quadratic expression can be written as the product of two linear factors, then the equation can be solved by setting each factor equal to zero. This method is often the quickest when factorisation is straightforward.
如果一个二次表达式可以写成两个一次因式的乘积，那么可以通过令每个因式等于零来求解。当因式分解较为简单时，这通常是最快的方法。
x + 6 = 0, we look for two numbers that multiply to 6 and add to 5. These are 2 and 3, so:
对于方程 x² + 5x + 6 = 0，我们需要找到两个数，它们相乘得6，相加得5。这两个数是2和3，所以：
x² + 5x + 6 = (x + 2)(x + 3) = 0
x + 2 = 0, so x = −2, and x + 3 = 0, so x = −3. Hence the solutions are x = −2 and x = −3.
令每个因式等于零，得到 x + 2 = 0，即 x = −2；以及 x + 3 = 0，即 x = −3。因此解为 x = −2 和 x = −3。
on is in standard form first. If not, rearrange it to ax² + bx + c = 0 before factorising.
在因式分解前，务必先检查方程是否处于标准形式。如果不是，先将其整理为 ax² + bx + c = 0 再进行分解。
3. The Quadratic Formula | 二次公式
those that cannot be factorised easily. For ax² + bx + c = 0, the solutions are given by:
二次公式可以求解任何二次方程，即使那些不易因式分解的方程。对于 ax² + bx + c = 0，其解为：
x = (−b ± √b² − 4ac) / 2a
and c into the expression. For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2:
使用公式时，将 a、b、c 的值代入表达式。例如，解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2：
x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4
Since √25 /4 = 2/4 = 0.5, and x = (−3 − 5)/4 = −8/4 = −2. Thus the solutions are x = 0.5 and x = −2.
因为 √25 = 5，所以 x = (−3 + 5)/4 = 2/4 = 0.5，x = (−3 − 5)/4 = −8/4 = −2。因此解为 x = 0.5 和 x = −2。
Be careful with negative signs and use brackets when substituting into the formula.
代入公式时要注意负号，并使用括号以避免符号错误。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This technique is useful for solving equations and also for finding the vertex of a parabola.
配方法是将二次表达式改写为 a(x + p)² + q 的形式。这种技巧对于解方程以及求抛物线顶点非常有用。
x + c, complete the square by adding and subtracting (b/2)². For example, x² + 6x + 1 = 0:
对于简单的二次项 x² + bx + c，通过加上和减去 (b/2)² 来配方。例如，x² + 6x + 1 = 0：
x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8 = 0
Then (x + 3)² = 8, so x + 3 = ±√8, giving x = −3 ± 2√2.
于是 (x + 3)² = 8，所以 x + 3 = ±√8，得到 x = −3 ± 2√2。
When the coefficient of x² is not 1, factor it out first before completing the square.
当 x² 的系数不为 1 时，先将其提出，再进行配方。
5. The Discriminant | 判别式
c. It tells us how many real roots a quadratic equation has without solving it completely.
判别式是二次公式中根号下的表达式：Δ = b² − 4ac。它无需完全求解即可告诉我们二次方程有多少个实数根。
If Δ > 0, the equation has two distinct real roots.
如果 Δ > 0，方程有两个不同的实数根。
If Δ = 0, the equation has one repeated real root (a double root).
如果 Δ = 0，方程有一个重根（二重实数根）。
If Δ < 0, the equation has no real roots; the roots are complex.
如果 Δ < 0，方程没有实数根；根为复数根。
Then Δ = (−4)² − 4 × 3 × 2 = 16 − 24 = −8, which is negative, so there are no real roots.
例如，对于 3x² − 4x + 2 = 0，a = 3，b = −4，c = 2。则 Δ = (−4)² − 4 × 3 × 2 = 16 − 24 = −8，为负数，因此没有实数根。
6. Graphs and Quadratic Functions | 图形与二次函数
The roots of the equation ax² + bx + c = 0 correspond to the x-intercepts of the parabola.
二次函数 y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的根对应抛物线与 x 轴的交点。
nwards and has a maximum point. The turning point (vertex) can be found using x = −b / 2a.
如果 a > 0，抛物线开口向上，有最小值点；如果 a < 0，抛物线开口向下，有最大值点。顶点（转向点）可用 x = −b / 2a 求得。
te −(−2)/(2 × 1) = 1, and substituting gives y = 1 − 2 − 3 = −4, so the vertex is (1, −4).
例如，y = x² − 2x − 3 的根为 x = −1 和 x = 3（可通过因式分解验证）。其顶点的 x 坐标为 −(−2)/(2 × 1) = 1，代入得 y = 1 − 2 − 3 = −4，所以顶点为 (1, −4)。
7. Roots and Coefficients | 根与系数的关系
f its roots can be expressed in terms of the coefficients. If the roots are α and β, then:
对于二次方程 ax² + bx + c = 0，其根的和与积可以用系数表示。设根为 α 和 β，则：
α + β = −b / a, αβ = c / a
= 0. It is useful when checking solutions or when forming a new equation from known roots.
这个关系由展开 (x − α)(x − β) = 0 推导而来。它可用于核对解，或根据已知根构造新的方程。
or example, if the roots of 2x² + 5x − 3 = 0 are α and β, then α + β = −5/2 and αβ = −3/2.
例如，若 2x² + 5x − 3 = 0 的根为 α 和 β，则 α + β = −5/2，αβ = −3/2。
8. Solving Quadratic Equations by Completing the Square in Application | 配方法在应用题中的技巧
Some problems involving areas, projectile motion, or optimisation naturally lead to quadratic equations. When the equation cannot be factorised, the quadratic formula or completing the square are reliable methods.
一些涉及面积、抛体运动或优化的问题会自然产生二次方程。当方程无法因式分解时，二次公式或配方法是可靠的方法。
is 10 cm², then (x + 3)(x − 2) = 10. Expanding gives x² + x − 6 = 10, so x² + x − 16 = 0.
考虑一个长方形，长为 (x + 3)，宽为 (x − 2)。若其面积为 10 cm²，则 (x + 3)(x − 2) = 10。展开得到 x² + x − 6 = 10，即 x² + x − 16 = 0。
Using the quadratic formula with a = 1, b = 1, c = −16:
使用二次公式，a = 1，b = 1，c = −16：
x = (−1 ± √1 + 64) / 2 = (−1 ± √65) / 2
(−1 + 8.0622)/2 ≈ 3.531. The negative value is not physically meaningful in this context.
由于 x 表示长度，我们取正值：x ≈ (−1 + 8.0622)/2 ≈ 3.531。负值在此情境下没有物理意义。
Always consider whether both solutions are valid in the context of the problem.
在应用问题中，务必检查两个解是否都有实际意义。
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