Exercise 4D: Quadratic Equations, the Discriminant and Inequalities | 习题4D：二次方程、判别式与不等式

📚 Exercise 4D: Quadratic Equations, the Discriminant and Inequalities | 习题4D：二次方程、判别式与不等式

Exercise 4D in the AQA A-Level Mathematics course focuses on quadratic equations, the discriminant, and the inequalities that arise from them. This revision guide breaks down every key skill you need, from solving quadratics by three different methods to interpreting the discriminant and tackling ‘range of values’ problems. Master these ideas and you will build a solid foundation for later topics such as functions, graphs and modelling.

AQA A-Level 数学课程中的习题4D 重点围绕二次方程、判别式及其相关不等式展开。本复习指南将逐项拆解你所需的核心技能：从三种方法求解二次方程，到理解判别式，再到处理“求取值范围”类问题。掌握这些内容，你将为基础函数、图像和建模等后续主题打下坚实基础。

1. The Standard Form of a Quadratic Function | 二次函数的标准形式

A quadratic expression is written in the standard form ax² + bx + c, where a, b and c are real constants and a ≠ 0. The graph of y = ax² + bx + c is always a parabola, and its shape depends on the sign of a. If a > 0, the parabola has a minimum point; if a < 0, it has a maximum point.

二次表达式写成标准形式 ax² + bx + c，其中 a、b、c 为实常数，且 a ≠ 0。函数 y = ax² + bx + c 的图像始终是一条抛物线，其形状由 a 的符号决定：当 a > 0 时，抛物线有最小值点；当 a < 0 时，则有最大值点。

In Exercise 4D you will frequently meet three equivalent forms of the same quadratic. The expanded form ax² + bx + c is useful for reading off the coefficients. The factorised form a(x − p)(x − q) is useful for locating roots. The completed-square form a(x − h)² + k is useful for identifying the turning point. Being fluent in moving between these forms is an essential examination skill.

在习题4D 中，你会频繁遇到同一个二次函数的三种等价形式。展开形式 ax² + bx + c 便于读出系数；因式分解形式 a(x − p)(x − q) 便于确定根；配方形式 a(x − h)² + k 便于找出顶点。熟练地在这些形式之间转换是一项必备的考试技能。

Standard form: ax² + bx + c = 0, with a ≠ 0.

标准形式：ax² + bx + c = 0，其中 a ≠ 0。
Roots are the x-values where the parabola meets the x-axis.

根是抛物线与 x 轴交点处的 x 值。
The turning point has x-coordinate h = −b/(2a).

顶点的 x 坐标为 h = −b/(2a)。

2. Solving by Factorisation | 用因式分解法求解

Factorisation is the fastest method when the quadratic has simple integer coefficients. To solve ax² + bx + c = 0, you rewrite the left-hand side as the product of two linear factors, then set each factor equal to zero. Since a product is zero only when at least one factor is zero, each linear equation gives one root.

当二次方程的系数为简单整数时，因式分解是最快捷的方法。要解 ax² + bx + c = 0，将左边写成两个一次因式的乘积，然后令每个因式等于零。因为乘积为零的前提是至少有一个因式为零，所以每个一次方程给出一个根。

For example, to solve x² − 5x + 6 = 0, look for two numbers that multiply to give 6 and add to give −5. These numbers are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3). Setting each factor to zero gives x = 2 or x = 3. Always expand your factorisation as a quick mental check.

例如，解 x² − 5x + 6 = 0 时，需要找出两个数，它们相乘等于 6，相加等于 −5。这两个数是 −2 和 −3，因此 x² − 5x + 6 = (x − 2)(x − 3)。令每个因式为零，得到 x = 2 或 x = 3。快速展开因式进行心算检查总是值得的。

(x − 2)(x − 3) = 0, so x = 2 or x = 3

When the coefficient of x² is not 1, factorisation requires greater care. Try combinations of factors of a and c that produce the correct middle term. For instance, 2x² + 7x + 3 = (2x + 1)(x + 3), since 2x × 3 + 1 × x = 6x + x = 7x. With practice, this ‘trial and improvement’ approach becomes very quick.

当 x² 的系数不为 1 时，因式分解需要更加谨慎。尝试 a 与 c 的因子组合，使得得到正确的一次项系数。例如，2x² + 7x + 3 = (2x + 1)(x + 3)，因为 2x × 3 + 1 × x = 6x + x = 7x。多加练习后，这种“试凑”方法会变得非常快捷。

3. Completing the Square | 配方法

Completing the square rewrites x² + bx + c as (x + b/2)² − (b/2)² + c. This converts the quadratic into a form that contains a perfect square, from which the vertex and the exact roots can be read directly. It is especially valuable when factorisation is impossible because the roots are irrational.

配方法将 x² + bx + c 改写为 (x + b/2)² − (b/2)² + c。这样做可以将二次式转化为包含完全平方的形式，从而直接读出顶点和精确根。当因式分解不可行且根为无理数时，配方法尤其有价值。

Worked example: complete the square for x² − 6x + 5. Half of −6 is −3, so write x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4. To solve (x − 3)² − 4 = 0, rearrange to (x − 3)² = 4, then take square roots: x − 3 = ±2, giving x = 5 or x = 1.

示例：对 x² − 6x + 5 配方。−6 的一半是 −3，因此 x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4。解 (x − 3)² − 4 = 0，移项得 (x − 3)² = 4，再开平方：x − 3 = ±2，得到 x = 5 或 x = 1。

(x − 3)² = 4, so x − 3 = ±2, giving x = 5 or x = 1

If the leading coefficient is not 1, factor it out before completing the square. For 2x² + 8x − 5, write 2(x² + 4x) − 5 = 2[(x + 2)² − 4] − 5 = 2(x + 2)² − 8 − 5 = 2(x + 2)² − 13. This immediately reveals the minimum point (−2, −13).

如果首项系数不为 1，先将其提出来再进行配方。对于 2x² + 8x − 5，写成 2(x² + 4x) − 5 = 2[(x + 2)² − 4] − 5 = 2(x + 2)² − 8 − 5 = 2(x + 2)² − 13。这立即显示出最小值点 (−2, −13)。

4. The Quadratic Formula | 二次公式

The quadratic formula provides a universal method for solving any quadratic equation. For ax² + bx + c = 0, the solutions are given by the formula below. This formula is derived from completing the square and is guaranteed to work for every quadratic, including those with irrational or complex roots.

二次公式为求解任何二次方程提供了通用方法。对于 ax² + bx + c = 0，解由下方公式给出。该公式由配方法推导而来，对每个二次方程都一定适用，包括根为无理数或复数的情况。

x = (−b ± √(b² − 4ac)) / (2a)

To use the formula, first identfy a, b and c carefully. Substitute them into the expression and simplify. For example, solve 2x² − 3x − 2 = 0 by setting a = 2, b = −3, c = −2. Then b² − 4ac = 9 + 16 = 25, and x = (3 ± 5)/4, which gives x = 2 or x = −0.5.

使用公式时，首先仔细确定 a、b、c。将其代入表达式并进行化简。例如，解 2x² − 3x − 2 = 0，令 a = 2，b = −3，c = −2。则 b² − 4ac = 9 + 16 = 25，于是 x = (3 ± 5)/4，得到 x = 2 或 x = −0.5。

x = (3 ± √25)/4 = (3 ± 5)/4, so x = 2 or x = −0.5

Always check whether the discriminant b² − 4ac is a perfect square before applying the formula. If it is, factorisation would have worked more quickly. The formula should be your reliable fallback whenever factorisation fails or the coefficients are awkward decimals.

在使用公式前，先检查判别式 b² − 4ac 是否为完全平方数。如果是，因式分解会更快捷。当因式分解失败或系数是复杂的数时，公式应是你可靠的备用手段。

5. The Discriminant and the Nature of Roots | 判别式与根的性质

The quantity Δ = b² − 4ac is called the discriminant. It determines exactly how many real roots a quadratic equation has, without solving the equation fully. This makes it a powerful tool for classifying graphs and for solving parameter problems.

量 Δ = b² − 4ac 称为判别式。它可以在不求方程全部解的情况下，精确判断二次方程有多少个实根。这使得它成为图像分类和参数问题求解的强大工具。

If Δ > 0, the equation has two distinct real roots.

若 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, the equation has one repeated real root (a tangent to the x-axis).

若 Δ = 0，方程有一个重根（抛物线与 x 轴相切）。
If Δ < 0, the equation has no real roots; the curve lies entirely above or below the x-axis.

若 Δ < 0，方程没有实数根；抛物线完全位于 x 轴上方或下方。

For example, the equation x² + 2x + 5 = 0 has discriminant 2² − 4 × 1 × 5 = 4 − 20 = −16, which is negative, so it has no real roots. Geometrically, the parabola never touches the x-axis. You should be able to state this conclusion in one sentence, using the sign of the discriminant as justification.

例如，方程 x² + 2x + 5 = 0 的判别式为 2² − 4 × 1 × 5 = 4 − 20 = −16，为负数，因此方程没有实数根。从几何上看，抛物线永远不会触及 x 轴。你应该能用一句话陈述这一结论，并以判别式的符号作为依据。

6. Parametric ‘k’ Problems | 含参数 k 的问题

A typical AQA exam question gives a quadratic involving an unknown constant k, such as x² + kx + (k + 3) = 0, and asks for the set of values of k for which the equation has two distinct real roots, or equal roots, or no real roots. The strategy is to compute the discriminant in terms of k and then form an equation or inequality.

一类典型的 AQA 考题给出含有未知常数 k 的二次方程，如 x² + kx + (k + 3) = 0，然后要求求出使方程有两个不等实根、等根或无实根时 k 的取值范围。解题策略是计算含 k 的判别式，再列出方程或不等式。

Worked example: find the values of k for which x² + kx + (k + 3) = 0 has equal roots. Here a = 1, b = k, c = k + 3, so Δ = k² − 4(1)(k + 3) = k² − 4k − 12. For equal roots we require Δ = 0, hence k² − 4k − 12 = 0, which factorises as (k − 6)(k + 2) = 0. Therefore k = 6 or k = −2.

示例：求使得 x² + kx + (k + 3) = 0 有等根时 k 的值。此处 a = 1，b = k，c = k + 3，所以 Δ = k² − 4(1)(k + 3) = k² − 4k − 12。等根要求 Δ = 0，因此 k² − 4k − 12 = 0，因式分解为 (k − 6)(k + 2) = 0。故 k = 6 或 k = −2。

Δ = k² − 4k − 12 = 0, so k = 6 or k = −2

If the question asks for two distinct real roots, set Δ > 0 and solve a quadratic inequality. If it asks for no real roots, set Δ < 0. Remember that the sign of a determines whether the quadratic in k opens upwards or downwards, which affects your final inequality answer.

若题目要求两个不等实根，则令 Δ > 0 并解二次不等式。若要求无实根，则令 Δ < 0。注意 k 的二次式开口方向由 a 的符号决定，这会影响最终的不等式答案。

7. Solving Quadratic Inequalities | 解二次不等式

To solve a quadratic inequality such as x² − 2x − 8 < 0, rewrite it as (x − 4)(x + 2) < 0. The roots are x = −2 and x = 4, and these divide the number line into three regions. Test a value from each region, or use the shape of the graph, to determine where the product is negative.

解二次不等式如 x² − 2x − 8 < 0 时，先改写为 (x − 4)(x + 2) < 0。根为 x = −2 和 x = 4，这两个值将数轴分成三个区域。在每个区域取一个测试值，或利用图像形状，来判断乘积在何处为负。

For x < −2, try x = −3: (−7)(−1) = 7 > 0, so the inequality is not satisfied.

当 x < −2 时，取 x = −3：(−7)(−1) = 7 > 0，不满足不等式。
For −2 < x < 4, try x = 0: (−4)(2) = −8 < 0, so this region works.

当 −2 < x < 4 时，取 x = 0：(−4)(2) = −8 < 0，该区域成立。
For x > 4, try x = 5: (1)(7) = 7 > 0, so the inequality is not satisfied.

当 x > 4 时，取 x = 5：(1)(7) = 7 > 0，不满足不等式。

(x − 4)(x + 2) < 0, so −2 < x < 4

Notice that the strict inequality < produces an open interval; the non-strict inequality ≤ would include the endpoints. For a > 0, the expression ax² + bx + c is negative only between the two roots and positive outside them. For a < 0, this pattern reverses. Memorising this 'inside negative, outside positive' rule saves time in the non-calculator paper.

注意，严格不等号 < 给出开区间；非严格不等号 ≤ 则包含端点。当 a > 0 时，ax² + bx + c 仅在两根之间为负，在两根之外为正。当 a < 0 时，规律恰好相反。记住“中间为负、两边为正”的规则，能在不使用计算器的试卷中节省时间。

8. Sketching the Graph to Solve Inequalities | 借助图像解不等式

A clear sketch is the fastest and most reliable way to solve a quadratic inequality. Draw the parabola with the correct orientation, mark the x-intercepts, and shade the region where the curve lies below the x-axis for ‘<' or above it for '>‘. The inequality sign then translates directly into a statement about x.

清晰的图像是解二次不等式最快捷且最可靠的方法。画出方向正确的抛物线，标出 x 轴交点，并在曲线位于 x 轴下方的区域（对应 “<”）或上方的区域（对应 “>”）进行标记。不等号即可直接转化为关于 x 的表述。

For example, solve x² − 3x − 4 ≥ 0. Since (x − 4)(x + 1) ≥ 0, the roots are x = −1 and x = 4. The parabola opens upwards, so the curve is above the x-axis when x ≤ −1 or x ≥ 4. Including the endpoints because the inequality is non-strict, the answer is x ≤ −1 or x ≥ 4.

例如，解 x² − 3x − 4 ≥ 0。因为 (x − 4)(x + 1) ≥ 0，根为 x = −1 和 x = 4。抛物线开口向上，所以当 x ≤ −1 或 x ≥ 4 时曲线位于 x 轴上方。由于是非严格不等式，需包含端点，答案为 x ≤ −1 或 x ≥ 4。

x ≤ −1 or x ≥ 4

In an exam, always label the roots and the y-intercept on your sketch. An unlabelled sketch earns no credit, but a well-labelled one can earn method marks even if a minor arithmetic slip later occurs. The graph is also the clearest way to check whether a ‘k’ inequality was solved correctly.

考试中，务必在草图上标出根和 y 轴截距。未标注的草图得不到分数，而标注清晰的草图即使后续出现轻微计算失误，也能获得方法分。图像同样是检查含 k 的不等式是否解对的最直观方式。

9. Common Mistakes in Exercise 4D | 习题4D的常见错误

Many students lose marks on Exercise 4D for avoidable reasons. One recurring error is forgetting to rearrange the equation into standard form before applying the formula. Another is misreading the signs of a, b and c when b or c is negative, which leads to a wrong discriminant and two wrong roots.

许多学生在习题4D 中因可避免的原因丢分。一个反复出现的错误是忘记在应用公式前将方程整理为标准形式。另一个错误是在 b 或 c 为负数时误读 a、b、c 的符号，导致判别式错误，进而得到两个错误根。

Incorrect sign of Δ: always write b² − 4ac, not b² + 4ac.

判别式符号错误：始终写 b² − 4ac，而不是 b² + 4ac。
Forgetting that a = 0 is not allowed in a quadratic.

忘记 a = 0 时不是二次方程。
Dropping the ± sign when taking square roots in completing the square.

配方开平方时遗漏 ± 号。
Writing ‘x < −2 and x > 4′ instead of ‘x < −2 or x > 4′ for a ‘greater than’ inequality.

对“大于”型不等式，把 x < −2 或 x > 4 误写成 x < −2 且 x > 4。

To avoid these pitfalls, adopt a consistent written structure: state the equation in standard form, write down a, b and c explicitly, compute Δ, and then give the roots in a clear boxed form. This discipline is especially important in the AQA non-calculator paper, where every method mark matters.

为避免这些陷阱，请采用一致的书写结构：写出标准形式方程，明确列出 a、b、c，计算 Δ，然后以清晰的方框形式给出根。这种规范性在 AQA 非计算器试卷中尤为重要，因为每一分方法分都举足轻重。

10. Exam-Style Practice Questions | 真题风格练习

The following three questions mirror the style of Exercise 4D in the AQA specification. Attempt each one fully, using the methods described in this guide, then compare your working with the concise solutions provided.

以下三道题模仿 AQA 大纲中习题4D 的风格。请使用本指南所讲的方法完整作答，然后对照简要解答检查你的过程。

Question 1: Solve 3x² + 5x − 2 = 0 by factorisation.

题目1：用因式分解法解 3x² + 5x − 2 = 0。

Solution 1: 3x² + 5x − 2 = (3x − 1)(x + 2) = 0, so x = 1/3 or x = −2.

解答1：3x² + 5x − 2 = (3x − 1)(x + 2) = 0，所以 x = 1/3 或 x = −2。

Question 2: Find the range of values of m for which mx² + 2mx + 4 = 0 has no real roots, given m ≠ 0.

题目2：已知 m ≠ 0，求使 mx² + 2mx + 4 = 0 无实数根时 m 的取值范围。

Solution 2: Δ = (2m)² − 16m = 4m² − 16m = 4m(m − 4). Since m ≠ 0, we require m(m − 4) < 0, giving 0 < m < 4.

解答2：Δ = (2m)² − 16m = 4m² − 16m = 4m(m − 4)。由于 m ≠ 0，要求 m(m − 4) < 0，得到 0 < m < 4。

Question 3: Solve x² − 4x + 1 = 0 by completing the square, giving your answers in exact surd form.

题目3：用配方法解 x² − 4x + 1 = 0，以精确根式形式写出答案。

Solution 3: (x − 2)² − 4 + 1 = 0, so (x − 2)² = 3. Hence x = 2 ± √3.

解答3：(x − 2)² − 4 + 1 = 0，因此 (x − 2)² = 3。故 x = 2 ± √3。

x = 2 ± √3

After completing these exercises, review any errors carefully. In the AQA examination, quadratic skills appear not only within pure mathematics but also as tools in mechanics and statistics. A secure understanding of Exercise 4D will serve you throughout the whole A-Level course.

完成这些练习后，请仔细检查每一个错误。在 AQA 考试中，二次函数技能不仅出现在纯数学模块中，还作为工具用于力学和统计学。扎实掌握习题4D 的内容，将在整个 A-Level 课程中为你提供持续助力。

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Exercise 4D: Quadratic Equations, the Discriminant and Inequalities | 习题4D：二次方程、判别式与不等式