Equations and Inequalities | 方程与不等式

📚 Equations and Inequalities | 方程与不等式

In A-Level Mathematics, equations and inequalities form the foundation of algebraic reasoning. You will solve linear and quadratic equations, work with simultaneous equations, interpret inequalities, and use both algebraic and graphical methods. This article covers the key Edexcel specification points and gives you clear examples to build confidence.

在 A-Level 数学中，方程与不等式是代数推理的基础。你将学习解线性方程和二次方程、处理联立方程、解释不等式，并运用代数与图像方法。本文涵盖 Edexcel 大纲的关键考点，并通过清晰的示例帮助你建立信心。

1. Solving Linear Equations | 解线性方程

A linear equation has the general form ax + b = 0, where a ≠ 0. To solve it, isolate x by applying inverse operations to both sides of the equation. For example, solve 3x – 7 = 5: add 7 to both sides to get 3x = 12, then divide by 3 to get x = 4. Always substitute your answer back into the original equation to check.

线性方程的一般形式为 ax + b = 0，其中 a ≠ 0。解方程时，通过对等式两边进行逆运算来分离 x。例如，解 3x – 7 = 5：两边同时加 7，得到 3x = 12，再除以 3，得到 x = 4。始终将答案代回原方程进行检验。

If the equation contains brackets or fractions, simplify first. Expand brackets, or multiply every term by the lowest common denominator to clear fractions. For instance, (x/2) + 3 = 5 becomes x/2 = 2, so x = 4.

如果方程含有括号或分数，请先化简。展开括号，或将每一项乘以最小公分母以消去分数。例如，(x/2) + 3 = 5 变为 x/2 = 2，因此 x = 4。

2. Quadratic Equations and Factorisation | 二次方程与因式分解

A quadratic equation has the form ax² + bx + c = 0, with a ≠ 0. When the quadratic expression can be factorised, write it as (px + q)(rx + s) = 0. Then set each factor equal to zero and solve separately. For example, x² – 5x + 6 = 0 factorises to (x – 2)(x – 3) = 0, giving x = 2 or x = 3.

二次方程的形式为 ax² + bx + c = 0，其中 a ≠ 0。当二次式可以因式分解时，写成 (px + q)(rx + s) = 0。然后令每个因子等于零并分别求解。例如，x² – 5x + 6 = 0 因式分解为 (x – 2)(x – 3) = 0，得到 x = 2 或 x = 3。

Always check that the factorised form expands back to the original quadratic. If the quadratic does not factorise easily, use completing the square or the quadratic formula from the next sections.

务必检查因式分解形式展开后能否还原为原二次式。如果二次式不易因式分解，请使用接下来两节中的配方法或求根公式。

3. Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. For a simple quadratic with a = 1, use the identity x² + bx = (x + b/2)² – (b/2)². This method is useful for solving equations and for identifying the vertex of a parabola.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。对于 a = 1 的简单二次式，使用恒等式 x² + bx = (x + b/2)² – (b/2)²。该方法可用于解方程以及确定抛物线的顶点。

Example: solve x² + 6x + 1 = 0. Write x² + 6x = (x + 3)² – 9, so the equation becomes (x + 3)² – 9 + 1 = 0, hence (x + 3)² = 8. Taking square roots gives x + 3 = ±√8, so x = -3 ± 2√2.

示例：解 x² + 6x + 1 = 0。写出 x² + 6x = (x + 3)² – 9，因此方程变为 (x + 3)² – 9 + 1 = 0，即 (x + 3)² = 8。开平方得到 x + 3 = ±√8，所以 x = -3 ± 2√2。

4. The Quadratic Formula and the Discriminant | 求根公式与判别式

For any quadratic ax² + bx + c = 0, the solutions are given by the quadratic formula:

对于任意二次方程 ax² + bx + c = 0，其解由求根公式给出：

x = (-b ± √(b² – 4ac)) / (2a)

The expression under the square root, Δ = b² – 4ac, is called the discriminant. It tells you the nature of the roots without solving the equation.

根号下的表达式 Δ = b² – 4ac 称为判别式。它无需解方程即可判断根的性质。

Discriminant	Nature of roots
Δ > 0	Two distinct real roots
Δ = 0	One repeated real root
Δ < 0	No real roots

For example, x² + 4x + 5 = 0 has Δ = 16 – 20 = -4, so it has no real roots. Edexcel questions often ask you to state the number of roots or find the range of a parameter for which roots are real.

例如，x² + 4x + 5 = 0 的判别式 Δ = 16 – 20 = -4，因此它没有实数根。Edexcel 考题常要求你说明根的个数，或求参数使方程有实数根的范围。

5. Simultaneous Equations | 联立方程

Simultaneous equations involve two or more unknowns that satisfy two or more equations. For two linear equations, use elimination or substitution. Example: solve 2x + y = 7 and x – y = 2. Adding the equations gives 3x = 9, so x = 3, then y = 1.

联立方程涉及两个或更多未知数同时满足两个或更多方程。对于两个线性方程，可使用消元法或代入法。示例：解 2x + y = 7 和 x – y = 2。两式相加得到 3x = 9，因此 x = 3，然后 y = 1。

When one equation is linear and the other is quadratic, substitute the linear expression into the quadratic. You will often obtain a quadratic equation, giving up to two solution pairs. Always substitute back to verify both equations are satisfied.

当一个方程是线性、另一个是二次时，将线性表达式代入二次方程。你通常会得到一个二次方程，最多产生两对解。始终代回检验两个方程是否同时满足。

6. Linear Inequalities | 线性不等式

A linear inequality has the form ax + b < c, ax + b ≤ c, ax + b > c, or ax + b ≥ c. Solving is similar to solving equations, but if you multiply or divide both sides by a negative number, you must reverse the inequality sign. For example, -2x < 6 becomes x > -3.

线性不等式的形式为 ax + b < c、ax + b ≤ c、ax + b > c 或 ax + b ≥ c。解法与解方程类似，但如果两边乘以或除以负数，必须反转不等号。例如，-2x < 6 变为 x > -3。

Solutions can be written using set notation or interval notation. For instance, x > 3 is written as (3, ∞). On a number line, use an open circle for strict inequalities and a closed circle for inclusive inequalities.

解可以用集合记号或区间记号表示。例如，x > 3 可写成 (3, ∞)。在数轴上，严格不等式用空心圆表示，包含等号的不等式用实心圆表示。

7. Quadratic Inequalities | 二次不等式

To solve a quadratic inequality such as ax² + bx + c > 0 or ax² + bx + c < 0, first find the critical values by solving the corresponding quadratic equation ax² + bx + c = 0. Then use a sign table or sketch the graph to determine the intervals where the expression is positive or negative.

解二次不等式（如 ax² + bx + c > 0 或 ax² + bx + c < 0）时，先通过解相应的二次方程 ax² + bx + c = 0 求出临界值。然后使用符号表或画出草图，确定表达式为正或负的区间。

Example: solve x² – x – 6 > 0. Factorise to (x – 3)(x + 2) > 0, so critical values are x = 3 and x = -2. The parabola opens upwards, so the expression is positive when x < -2 or x > 3. The solution is x < -2 or x > 3.

示例：解 x² – x – 6 > 0。因式分解为 (x – 3)(x + 2) > 0，因此临界值为 x = 3 和 x = -2。抛物线开口向上，所以当 x < -2 或 x > 3 时表达式为正。解为 x < -2 或 x > 3。

8. Inequalities Involving Absolute Values | 含绝对值的不等式

Absolute value inequalities use the modulus function. The inequality |ax + b| < c, where c > 0, is equivalent to the double inequality -c < ax + b < c. The inequality |ax + b| > c is equivalent to ax + b < -c or ax + b > c.

绝对值不等式使用模函数。不等式 |ax + b| < c（其中 c > 0）等价于双重不等式 -c < ax + b < c。不等式 |ax + b| > c 等价于 ax + b < -c 或 ax + b > c。

Example: solve |2x – 1| ≤ 5. This gives -5 ≤ 2x – 1 ≤ 5. Add 1 to all parts to get -4 ≤ 2x ≤ 6, then divide by 2 to obtain -2 ≤ x ≤ 3.

示例：解 |2x – 1| ≤ 5。得到 -5 ≤ 2x – 1 ≤ 5。所有部分同时加 1，得 -4 ≤ 2x ≤ 6，再除以 2，得到 -2 ≤ x ≤ 3。

9. Graphical Methods for Equations | 方程的图解法

Graphical methods allow you to estimate solutions to equations, especially when algebraic methods are difficult. To solve f(x) = 0, draw the graph of y = f(x) and find the x-coordinates of the points where the curve crosses the x-axis.

图像方法可以估算方程的解，尤其是在代数方法较困难时。要解 f(x) = 0，画出 y = f(x) 的图像，并找出曲线与 x 轴交点的 x 坐标。

You can also solve f(x) = g(x) by drawing y = f(x) and y = g(x) on the same axes. The x-coordinates of the intersection points are the solutions. This is particularly useful when one equation is quadratic or cubic and the other is linear.

你也可以通过在同一坐标系中画出 y = f(x) 和 y = g(x)，解 f(x) = g(x)。交点处的 x 坐标就是解。当一个方程是二次或三次、另一个是线性时，这种方法尤其有用。

10. Graphical Methods for Inequalities | 不等式的图解法

For inequalities in two variables, such as y < 2x + 1 or y ≥ x², shade the region that satisfies the inequality. Draw the boundary line or curve: use a dashed line for strict inequalities and a solid line for inequalities that include equality.

对于两个变量的不等式，例如 y < 2x + 1 或 y ≥ x²，要画出满足不等式的区域。绘制边界线或曲线：严格不等式使用虚线，包含等号的不等式使用实线。

Choose a test point not on the boundary, such as (0, 0), and substitute it into the inequality. If it satisfies the inequality, shade the region containing the test point; otherwise shade the opposite side. Edexcel questions often ask you to identify or describe a region bounded by several inequalities.

选择一个不在边界上的测试点，如 (0, 0)，并代入不等式。如果满足不等式，则着色包含测试点的区域；否则着色另一侧。Edexcel 考题常要求你识别或描述由多个不等式围成的区域。

11. Algebraic Fractions in Equations | 方程中的代数分式

Equations involving algebraic fractions require careful handling of denominators. Multiply every term by the common denominator to eliminate fractions, or cross-multiply when there is a single fraction on each side. For example, solve 1/(x – 2) = 3/(x + 1). Cross-multiplying gives x + 1 = 3x – 6, so x = 7/2.

含有代数分式的方程需要仔细处理分母。将每一项乘以公分母以消去分式，或当两边各只有一个分式时交叉相乘。例如，解 1/(x – 2) = 3/(x + 1)。交叉相乘得到 x + 1 = 3x – 6，所以 x = 7/2。

Always check for extraneous solutions by ensuring no denominator equals zero in the original equation. If a solution makes a denominator zero, it must be rejected.

务必通过确认原方程中分母不为零来检查无效解。如果某个解使分母为零，则必须舍去。

12. Problem Solving and Exam Tips | 解题策略与考试技巧

When solving equations and inequalities, watch out for common mistakes: forgetting to reverse the inequality sign when dividing by a negative number, losing a root when cancelling a factor, or ignoring excluded values in rational equations. Always show clear working and check your final answers.

解方程和不等式时，注意常见错误：除以负数时忘记反转不等号，约去因式时丢根，或在有理方程中忽略排除值。始终展示清晰的步骤并检查最终答案。

In Edexcel exams, questions may combine topics, such as using the discriminant to find the range of k for which a quadratic has two distinct real roots, or solving an inequality represented graphically. Practice drawing diagrams, using interval notation, and interpreting questions precisely.

在 Edexcel 考试中，题目可能综合多个知识点，例如利用判别式求参数 k 的范围，使二次方程有两个不同实根，或解图像表示的不等式。练习画图、使用区间记号并准确解读题意。

Use your calculator wisely for checking numerical answers, but always write down the algebraic method. Marks are awarded for reasoning, not just the final answer. Revise the standard forms in this article regularly to build speed and accuracy.

合理使用计算器检查数值答案，但始终写出代数方法。分数根据推理过程给分，而不仅仅是最终答案。定期复习本文中的标准形式，以提高速度和准确性。

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Equations and Inequalities | 方程与不等式