📚 Example 2.8.2: Using the Discriminant in Quadratics | 例 2.8.2：二次方程判别式应用
n which one coefficient contains an unknown constant, then asks you to decide for which values of that constant the equation has two distinct real roots, one repeated root, or no real roots.
在 AQA A-Level 数学中，例 2.8.2 是二次方程判别式的经典练习。题目通常给出含未知常数的方程 ax² + bx + c = 0，要求判断方程有实根、重根或无实根时参数的范围。
The method itself is short, but every mark comes from careful bookkeeping: identify a, b and c, form the discriminant, then solve the resulting inequality. This article unpacks the method, shows a full worked solution, and highlights the exam traps that cost the most marks.
方法本身很短，但每一分都来自严谨的记账：确定 a、b、c，写出判别式，再解出相应的不等式。本文将拆解方法，展示完整解题过程，并强调最容易丢分的考试陷阱。
1. Why This Example Matters | 为什么这个例题重要
Example 2.8.2 sits in the Quadratics topic, but the idea extends far beyond it: circles, tangents, curve sketching and optimisation all rely on the same test. AQA examiners often use the discriminant to decide whether a line meets a curve, so mastering this example improves your algebraic reasoning across the whole paper.
例 2.8.2 属于二次函数主题，但它的思想远不止于此：圆、切线、曲线作图与最优化都依赖同一个判别。AQA 考官经常用判别式来判断直线与曲线是否相交，因此掌握这道例题能提升你在整份试卷中的代数推理能力。
At first glance, the example looks like an abstract algebra exercise. But AQA often embeds the same task in real-world contexts, such as finding when a projectile reaches a given height, or when a cost model breaks even. The discriminant is the fastest tool for all of these.
乍一看，这道例题像是一道抽象代数练习。但 AQA 常将同样的任务嵌入实际情境，例如求抛射体何时达到给定高度，或成本模型何时收支平衡。判别式是解决这类问题最快捷的工具。
It is also a reliable source of marks. Because the working is short and the reasoning is visible, a clean solution to Example 2.8.2 rarely fails to pick up the method marks at the start of a question.
它同时是稳定的得分点。因为过程简短、推理可见，一份整洁的例 2.8.2 解答几乎总能拿到题目开头的方法分。
2. The Discriminant Formula | 判别式公式
It is exactly the expression that appears under the square root in the quadratic formula:
对于二次方程 ax² + bx + c = 0，其中 a ≠ 0，判别式定义为 Δ = b² − 4ac。它正是求根公式中位于平方根号内的表达式：
Δ = b² − 4ac
x = (−b ± √Δ) / (2a)
It is called the discriminant because it discriminates between the three possible types of roots. Once Δ is known, the equation can be classified without being solved.
它被称为判别式，因为它能"判别"三种可能的根的情形。只要知道 Δ，无需真正解方程就能对方程分类。
ient of x, so if b = k − 3, then b² means (k − 3)², and the whole bracket must be squared.
在动手运算之前，必须先正确读出各系数。特别注意符号：b 是 x 的系数，因此如果 b = k − 3，那么 b² 表示 (k − 3)²，必须对整个括号平方。
- a = the coefficient of x², and a ≠ 0 | a = x² 的系数，且 a ≠ 0
- b = the coefficient of x, including its sign | b = x 的系数，需连同符号一起取
- c = the constant term, including its sign | c = 常数项，需连同符号一起取
If a itself contains the unknown constant, the question is no longer automatically quadratic. That special case is discussed in Section 12.
如果 a 本身也含未知常数，题目就不再自动是二次方程。这一特殊情形将在第 12 节讨论。
3. Interpreting the Sign of Δ | 判别式符号的意义
The sign of Δ tells you the nature of the roots without solving the equation. There are exactly three cases, and each one has an algebraic meaning and a graphical meaning.
Δ 的符号无需解方程即可判断根的性质。总共只有三种情形，每一种都同时具有代数意义和图像意义。
| Discriminant (判别式) | Roots (根) | Graph (图像) |
| Δ > 0 | Two distinct real roots (两个不相等实根) | Curve cuts the x-axis twice (曲线与 x 轴交于两点) |
| Δ = 0 | One repeated real root (一个重根) | Curve touches the x-axis (曲线与 x 轴相切) |
| Δ < 0 | No real roots (无实根) | Curve does not meet the x-axis (曲线不与 x 轴相交) |
Notice that the table says nothing about the size of the roots. The discriminant never tells you what the roots are, only how many real roots exist and whether they coincide.
注意：这张表完全没有涉及根的大小。判别式永远不告诉你根具体是多少，只告诉你实根的个数以及它们是否重合。
Two words carry the marks here. "Distinct" means strictly Δ > 0, while "real roots" on its own means Δ ≥ 0, because a repeated root is still real.
这里有两个词直接决定得分。"不相等"意味着严格地 Δ > 0；而只说"有实根"则是 Δ ≥ 0，因为重根也是实根。
4. Reading the Graph | 图像解读
xis at exactly one point, the vertex; if Δ < 0 the curve never meets the x-axis at all.
判别式是抛物线 y = ax² + bx + c 在代数上的投影。若 Δ > 0，抛物线与 x 轴交于两点；若 Δ = 0，它与 x 轴恰好相切于一点，即顶点；若 Δ < 0，曲线与 x 轴完全不相交。
When Δ < 0, the curve lies entirely above or entirely below the x-axis, and which one depends on the sign of a. If a > 0 the parabola opens upwards, so it stays positive for every real x. If a < 0 it opens downwards and stays negative for every real x.
当 Δ < 0 时，曲线完全位于 x 轴上方或下方，具体取决于 a 的符号。若 a > 0，抛物线开口向上，对所有实数 x 都取正值；若 a < 0，开口向下，对所有实数 x 都取负值。
This graph view is what makes the discriminant useful in tangency problems. "The line is a tangent to the curve" and "the resulting quadratic has a repeated root" are two ways of saying exactly the same thing, and the bridge between them is Δ = 0.
正是这种图像视角，让判别式在相切问题中格外有用。"直线与曲线相切"和"所得二次方程有重根"是同一件事的两种说法，而两者之间的桥梁就是 Δ = 0。
5. Setting Up the Worked Example | 建立例题：题目与读题
Consider the following version of the example, which is typical of the AQA style:
下面这道题是例 2.8.2 的典型版本，风格与 AQA 真题一致：
is a real constant, has (a) two distinct real roots, (b) equal roots, (c) no real roots. Find the possible values of k in each case.
方程 x² + (k − 3)x + 4 = 0，其中 k 为实常数。当方程 (a) 有两个不相等实根，(b) 有重根，(c) 无实根时，分别求 k 的取值范围。
them down explicitly, in a line of their own, is worth doing even when the answer feels obvious.
第一步是读出系数。这里 a = 1，b = k − 3，c = 4。即便答案看起来很明显，也值得把这三个系数单独写在一行上。
The second job is to keep the bracket. The expression (k − 3)² is a perfect square of a binomial, not k² − 9. Expanding it correctly at the next stage is where most lost marks are created.
第二步是保留括号。(k − 3)² 是二项式的完全平方，并不等于 k² − 9。下一阶段能否正确展开，正是多数失分的来源。
6. Step-by-Step Solution | 分步解答
Substitute the coefficients into the discriminant:
把各系数代入判别式：
Δ = b² − 4ac = (k − 3)² − 4 × 1 × 4
Δ = (k − 3)² − 16
+ 9 − 16, so Δ = k² − 6k − 7. Factorising produces the form that is easiest to work with:
展开括号得 k² − 6k + 9 − 16，因此 Δ = k² − 6k − 7。因式分解后得到最便于处理的形式：
Δ = (k − 7)(k + 1)
changes sign, and the critical values k = 7 and k = −1 are visible only after factorising.
因式形式不是可有可无的装饰。解 (k − 7)(k + 1) > 0 需要知道表达式在哪里变号，而临界值 k = 7 与 k = −1 只有在分解之后才一目了然。
0 gives Δ = −7 < 0, and indeed (0 − 7)(0 + 1) = −7, so the factorisation is consistent.
做一次数值检验可以避免疏漏：取 k = 0 得 Δ = −7 < 0，而 (0 − 7)(0 + 1) = −7，说明分解结果自洽。
7. Solving the Inequality | 解不等式与区间表达
. Because the leading coefficient of the quadratic in k is positive, the product is positive outside the roots and negative between them.
三种情形都转化为关于 Δ = (k − 7)(k + 1) 符号的陈述。由于关于 k 的二次式首项系数为正，乘积在两根之外为正、在两根之间为负。
(a) Δ > 0 ⇒ k < −1 or k > 7
(b) Δ = 0 ⇒ k = −1 or k = 7
(c) Δ < 0 ⇒ −1 < k < 7
Write the answers as intervals or as inequalities in k, and keep the strictness of each sign. Part (b) is not an interval at all but a pair of isolated values, which is exactly why "equal roots" questions usually carry a smaller mark than the other two.
答案要用区间或关于 k 的不等式写出，并保持各符号的严格性。(b) 部分根本不是区间，而是两个孤立的值，这也是"重根"小题通常分值较低的原因。
) makes the three answers visible at once and is often accepted as justification in an exam, provided the critical values are marked.
画出抛物线 y = (k − 7)(k + 1) 的草图，可以同时看清三个答案；只要标出临界值，考试中通常认可这种作图作为理由。
8. The Repeated Root Case in Detail | 重根情形详解
the quadratic formula collapses to a single value, because the square root term vanishes:
当 Δ = 0 时，求根公式中的平方根项消失，只剩一个值：
x = −b / (2a)
ly, k = −1 gives b = −4, so x = 4 / 2 = 2 and the equation becomes x² − 4x + 4 = (x − 2)².
在本题中，k = 7 时 b = 7 − 3 = 4，于是 x = −4 / 2 = −2，方程化为 x² + 4x + 4 = (x + 2)²。同样地，k = −1 时 b = −4，于是 x = 4 / 2 = 2，方程化为 x² − 4x + 4 = (x − 2)²。
the condition for a perfect square. A quadratic with a repeated root always factorises into a squared linear bracket, and the repeated root is the x-coordinate of the vertex.
这正是 Δ = 0 被称为"完全平方条件"的原因。有重根的二次式总能分解为一个一次式的平方，而重根就是顶点的横坐标。
The repeated-root case is the one that answers tangency questions, because a tangent touches a curve at exactly one point and therefore produces a quadratic with exactly one solution.
重根情形正是回答相切问题的关键：切线与曲线恰好只交于一点，因此对应的二次方程恰好只有一个解。
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