Solving Quadratic Equations | 二次方程求解

📚 Solving Quadratic Equations | 二次方程求解

Quadratic equations appear throughout A-Level mathematics, from pure algebra to applied mechanics. Mastering their solution methods is essential for success in the AQA specification.

二次方程贯穿 A-Level 数学始终，从纯代数到应用力学无不涉及。掌握其求解方法是 AQA 考纲中取得高分的关键。

1. Standard Form and Factorisation | 标准形式与因式分解

A quadratic equation is any expression of the form ax² + bx + c = 0, where a, b, c are real constants and a ≠ 0. The first method you should attempt is factorisation, which works when the expression can be written as a product of two linear factors.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为实数常数且 a ≠ 0。首先应尝试的方法是因式分解，当表达式可写成两个一次因式的乘积时适用。

For example, to solve x² − 5x + 6 = 0, look for two numbers that multiply to 6 and add to −5. These are −2 and −3, so:

例如，解 x² − 5x + 6 = 0 时，寻找两个数相乘为 6 且相加为 −5。它们是 −2 和 −3，因此：

(x − 2)(x − 3) = 0

Then x = 2 or x = 3. Always check whether the coefficient of x² is not 1, such as 2x² + 7x + 3 = (2x + 1)(x + 3).

于是 x = 2 或 x = 3。若 x² 的系数不为 1，如 2x² + 7x + 3 = (2x + 1)(x + 3)，仍需检查。

2. Quadratic Formula | 二次公式

When factorisation is difficult or impossible, use the quadratic formula. For ax² + bx + c = 0, the solutions are given by:

当因式分解困难甚至无法进行时，使用二次公式。对于 ax² + bx + c = 0，解由下式给出：

x = (−b ± √(b² − 4ac)) / (2a)

This formula works for all real quadratic equations. For example, 2x² − 4x − 1 = 0 gives a = 2, b = −4, c = −1:

此公式适用于所有实数二次方程。例如 2x² − 4x − 1 = 0 中 a = 2，b = −4，c = −1：

x = (4 ± √(16 + 8)) / 4 = (4 ± √24) / 4 = 1 ± √6 / 2

Remember to simplify surds where possible, and leave answers in exact form unless a decimal is requested.

记住尽可能化简根式，除非题目要求小数，否则保留精确形式。

3. Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. Start by factoring out a, then halve the coefficient of x.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。先提出 a，再将 x 的系数取半。

For example, x² + 6x + 1 becomes (x + 3)² − 9 + 1 = (x + 3)² − 8. Setting this equal to zero gives:

例如，x² + 6x + 1 变为 (x + 3)² − 9 + 1 = (x + 3)² − 8。令其等于零可得：

(x + 3)² = 8 → x = −3 ± 2√2

This method is crucial for finding turning points and solving equations where the quadratic formula is less elegant.

这一方法在求顶点坐标以及处理某些公式不够简洁的方程时至关重要。

4. The Discriminant | 判别式

The discriminant, Δ = b² − 4ac, determines the nature of the roots without solving the equation.

判别式 Δ = b² − 4ac 无需解方程即可判断根的性质。

If Δ > 0, there are two distinct real roots.

若 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, there is one repeated real root.

若 Δ = 0，方程有一个重根。
If Δ < 0, there are no real roots (two complex conjugate roots).

若 Δ < 0，方程没有实数根（有一对共轭复根）。

This is especially useful in coordinate geometry when determining whether a line intersects a curve, and in mechanics for projectile motion problems.

在解析几何中判断直线与曲线是否相交，以及力学中的抛体运动问题中，这一概念特别有用。

5. Sketching Quadratics | 绘制二次函数图像

A quadratic graph is a parabola. To sketch it accurately you need the roots, the y-intercept, and the vertex (turning point).

二次函数图像是抛物线。要准确绘制，需要知道根、y 轴截距以及顶点（驻点）。

The vertex can be found by completing the square: y = a(x + p)² + q gives vertex at (−p, q). Alternatively, the x-coordinate of the vertex is −b/(2a).

顶点可通过配方法得到：y = a(x + p)² + q 的顶点为 (−p, q)。另一种方法是用 x 坐标公式 −b/(2a)。

If a > 0 the parabola opens upwards; if a < 0 it opens downwards. Always mark the axis of symmetry x = −b/(2a).

若 a > 0，抛物线开口向上；若 a < 0，开口向下。务必标出对称轴 x = −b/(2a)。

6. Quadratic Inequalities | 二次不等式

Solving a quadratic inequality involves finding the roots first, then determining the sign of the quadratic on each interval.

解二次不等式要先求出根，再判断二次函数在各区间的符号。

For x² − x − 6 < 0, factorise to (x − 3)(x + 2) < 0. The critical points are −2 and 3. A quick sign table shows the solution is −2 < x < 3.

对于 x² − x − 6 < 0，分解为 (x − 3)(x + 2) < 0。临界点为 −2 和 3。快速符号表表明解为 −2 < x < 3。

For ≥ 0 or ≤ 0, include the roots. Remember to deal with a negative leading coefficient by multiplying through by −1 and reversing the inequality sign.

对于 ≥ 0 或 ≤ 0，需要包含根。若首项系数为负，可两边乘以 −1 并反转不等号。

7. Roots and Coefficients | 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, the sum and product of the roots are:

对于二次方程 ax² + bx + c = 0，设其根为 α 和 β，则根的和与积分别为：

α + β = −b/a, αβ = c/a

These relationships allow you to construct quadratic equations with given roots, or to find symmetric functions of roots without solving the original equation.

这些关系可用于构造具有给定根的二次方程，或在不解原方程的情况下求根的对称函数值。

For example, if α and β are roots of 2x² + 3x − 1 = 0, then α + β = −3/2 and αβ = −1/2. You can then find α² + β² = (α + β)² − 2αβ.

例如，若 α 和 β 是 2x² + 3x − 1 = 0 的根，则 α + β = −3/2，αβ = −1/2。于是可求得 α² + β² = (α + β)² − 2αβ。

8. Applications in Problem Solving | 应用问题求解

Quadratic equations model many real-world situations, including projectile motion, area optimisation, and geometric problems.

二次方程能模拟许多现实情境，包括抛体运动、面积优化和几何问题。

Suppose a rectangle has perimeter 20 m and area 24 m². Let length be x and width be 10 − x. Then x(10 − x) = 24, leading to x² − 10x + 24 = 0. Factorising gives x = 4 or 6.

假设一个矩形周长为 20 m，面积为 24 m²。设长为 x，宽为 10 − x。则 x(10 − x) = 24，得到 x² − 10x + 24 = 0。因式分解得 x = 4 或 6。

In mechanics, the height of a projectile is often given by h = ut − ½gt². Setting h to a target value creates a quadratic equation in time t, whose roots reveal when the projectile reaches that height.

在力学中，抛射体高度常由 h = ut − ½gt² 给出。将 h 设为某一目标值，就得到关于时间 t 的二次方程，其根揭示了抛射体何时达到该高度。

9. Common Mistakes and Tips | 常见错误与提示

One frequent error is forgetting to rearrange the equation into standard form before applying the formula or factorisation.

一个常见错误是在应用公式或因式分解前忘记将方程整理为标准形式。

Another is losing solutions when dividing both sides by a variable. Never divide by x unless you are certain x ≠ 0, because x = 0 may be a valid root.

另一个错误是两边除以变量而丢失解。切勿除以 x，除非你确定 x ≠ 0，因为 x = 0 可能是一个有效的根。

When using the quadratic formula, check that you have correctly substituted the signs of b and c. For instance, if b = −5, then −b = 5, not −5.

使用二次公式时，要检查 b 和 c 的符号是否正确代入。例如若 b = −5，则 −b = 5，而不是 −5。

Finally, always simplify surds such as √12 to 2√3. This is required for exact answers in AQA examinations.

最后，务必化简根式，如 √12 应写成 2√3。这是 AQA 考试中精确答案的要求。

10. Summary of Solution Methods | 方法总结

Choose the most efficient method according to the problem:

根据问题选择最有效的方法：

Method	Best Used When	Example
Factorisation	Simple integer roots	x² − 3x − 10 = 0
Completing the square	Finding vertex or dealing with inequalities	x² + 4x − 3 = 0
Quadratic formula	Any equation, especially non-factorisable	3x² + 2x − 5 = 0

Learn all three techniques, because AQA questions often specify the method, and later topics such as series and integration rely on your fluency with quadratics.

三种方法都需要掌握，因为 AQA 题目常常指定方法，而后续的级数与积分等内容都依赖于你对二次方程的熟练程度。

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Solving Quadratic Equations | 二次方程求解