📚 The Discriminant: Determining the Nature of Roots of Quadratic Equations | 判别式：判断二次方程根的情况
In A-Level mathematics, the discriminant is one of the most elegant and powerful tools for analysing quadratic equations without solving them. It tells us exactly how many real roots a quadratic equation has, and whether those roots are distinct or repeated. This article explores the discriminant in depth, covering its definition, derivation, and applications across a range of problem types.
在A-Level数学中，判别式是无需解方程即可分析二次方程的最优雅、最强大的工具之一。它能准确告诉我们一个二次方程有多少实数根，以及这些根是相异根还是重根。本文将深入探讨判别式的定义、推导过程及其在各类问题中的应用。
1. The Quadratic Formula and the Discriminant | 二次公式与判别式
the form ax² + bx + c = 0, where a ≠ 0, the solutions are given by the quadratic formula:
x = (−b ± √b² − 4ac) / 2a
The expression under the square root, b² − 4ac, is called the discriminant and is conventionally denoted by the Greek letter Δ (Delta). It is the key to understanding the nature of the roots without performing any further calculation.
对于形式为 ax² + bx + c = 0（其中 a ≠ 0）的二次方程，其解由二次公式给出：
x = (−b ± √b² − 4ac) / 2a
根号下的表达式 b² − 4ac 被称为判别式，通常用希腊字母 Δ（德尔塔）表示。它是无需进一步计算就能理解根的性质的关键。
2. The Three Cases of the Discriminant | 判别式的三种情况
The value of the discriminant Δ = b² − 4ac determines the nature of the roots as follows:
判别式 Δ = b² − 4ac 的值决定根的性质如下：
| Value of Δ | Δ的值 | Nature of Roots | 根的性质 | Graphical Meaning | 图像意义 |
|---|---|---|
| Δ > 0 | Two distinct real roots | Curve crosses the x-axis at two points |
| Δ > 0（正数） | 两个相异的实数根 | 曲线与x轴相交于两点 |
| Δ = 0 | One repeated real root (a double root) | Curve touches the x-axis at exactly one point |
| Δ = 0（零） | 一个重根（二重根） | 曲线与x轴恰好相切于一点 |
| Δ < 0 | No real roots (two complex conjugate roots) | Curve does not intersect the x-axis |
| Δ < 0（负数） | 无实数根（两个共轭复根） | 曲线与x轴无交点 |
It is worth memorising these three cases, as they form the foundation for many exam questions.
这三种情况值得牢记，因为它们是许多考题的基础。
3. Derivation: Why the Discriminant Works | 推导：判别式为何有效
scriminant determines the nature of roots, we can complete the square on ax² + bx + c = 0:
为了理解判别式为何能决定根的性质，我们可以对 ax² + bx + c = 0 进行配方法：
a(x + b/2a)² = (b² − 4ac) / 4a
±√Δ); if Δ = 0, the square root is 0, giving exactly one value; if Δ < 0, the square root of a negative number is not real, so no real roots exist.
由于 a ≠ 0，左边是一个完全平方乘以 a，而完全平方总是非负的。因此，若要存在实数解，右边必须是非负的。
4. Worked Example: Two Distinct Real Roots | 例题：两个相异的实数根
nsider the equation 2x² − 5x + 1 = 0. Here, a = 2, b = −5, and c = 1. The discriminant is:
考虑方程 2x² − 5x + 1 = 0。这里 a = 2，b = −5，c = 1。判别式为：
Δ = (−5)² − 4 × 2 × 1 = 25 − 8 = 17
− 5x + 1 crosses the x-axis at two points. We can state this conclusion immediately, without solving the equation.
由于 Δ = 17 > 0，该方程有两个相异的实数根。函数 y = 2x² − 5x + 1 的图像与x轴相交于两点。我们无需解方程即可立即得出这个结论。
5. Worked Example: A Repeated Root | 例题：重根
onsider the equation x² − 6x + 9 = 0. Here, a = 1, b = −6, and c = 9. The discriminant is:
考虑方程 x² − 6x + 9 = 0。这里 a = 1，b = −6，c = 9。判别式为：
Δ = (−6)² − 4 × 1 × 9 = 36 − 36 = 0
t x = 3, which is the double root. Indeed, the equation can be factorised as (x − 3)² = 0.
由于 Δ = 0，该方程恰好有一个重实数根。函数 y = x² − 6x + 9 的图像在 x = 3 处与x轴相切，3 就是重根。事实上，该方程可以因式分解为 (x − 3)² = 0。
6. Worked Example: No Real Roots | 例题：无实数根
onsider the equation 3x² + 2x + 5 = 0. Here, a = 3, b = 2, and c = 5. The discriminant is:
考虑方程 3x² + 2x + 5 = 0。这里 a = 3，b = 2，c = 5。判别式为：
Δ = 2² − 4 × 3 × 5 = 4 − 60 = −56
s entirely above the x-axis (since a > 0) and never intersects it. In the context of complex numbers, the equation has two conjugate complex roots, given by:
由于 Δ = −56 < 0，该方程没有实数根。函数 y = 3x² + 2x + 5 的图像完全位于x轴上方（因为 a > 0），与x轴永不相交。在复数的背景下，该方程有两个共轭复根：
x = (−2 ± i√56) / 6 = (−1 ± i√14) / 3
7. Finding Unknown Coefficients Using the Discriminant | 利用判别式求未知系数
find the values of k for which the equation kx² + 4x + 2 = 0 has two distinct real roots.
一种非常常见的考题类型涉及求未知参数 k 的值，使得方程具有特定数量的根。例如，求 k 的值使方程 kx² + 4x + 2 = 0 有两个相异的实数根。
For two distinct real roots, we require Δ > 0:
要存在两个相异的实数根，我们需要 Δ > 0：
Δ = 4² − 4 × k × 2 = 16 − 8k > 0
16 > 8k → k < 2
However, we must also remember that the equation is quadratic, so k ≠ 0. The final answer is k < 2 and k ≠ 0.
然而，我们必须记住该方程是二次的，所以 k ≠ 0。最终答案是 k < 2 且 k ≠ 0。
8. The Discriminant and Tangent Lines | 判别式与切线
c, the resulting quadratic equation in x has a discriminant that reveals the relationship:
判别式还可用于确定一条直线与一条曲线的相交、相切或相离关系。当直线 y = mx + c 代入二次曲线（例如 y = ax² + bx + c）时，得到的关于 x 的二次方程的判别式揭示了它们之间的关系：
Δ > 0: The line intersects the curve at two distinct points.
Δ > 0：直线与曲线相交于两个不同的点。
Δ = 0: The line is tangent to the curve (touches at exactly one point).
Δ = 0：直线与曲线相切（恰好接触于一点）。
Δ < 0: The line does not meet the curve.
Δ < 0：直线与曲线不相交。
s. For example, determine whether the line y = 3x − 1 intersects the curve y = x² + x + 2.
这种技巧在坐标几何题目中经常考查。例如，判断直线 y = 3x − 1 是否与曲线 y = x² + x + 2 相交。
Substituting gives:
代入得到：
x² + x + 2 = 3x − 1 → x² − 2x + 3 = 0
Δ = (−2)² − 4 × 1 × 3 = 4 − 12 = −8 < 0
Since Δ < 0, the line does not intersect the curve.
由于 Δ < 0，直线与曲线不相交。
9. Connection to the Vertex (Turning Point) | 与顶点（转向点）的联系
The discriminant is closely related to the vertex of a parabola. After completing the square, the vertex is at (−b/2a, −Δ/4a). The y-coordinate of the vertex, −Δ/4a, directly involves the discriminant. If a > 0:
判别式与抛物线的顶点密切相关。配方后，顶点位于 (−b/2a, −Δ/4a)。顶点的 y 坐标 −Δ/4a 直接包含判别式。若 a > 0：
Δ > 0 means the vertex lies below the x-axis (since −Δ/4a < 0), so the curve crosses the axis twice.
Δ > 0 意味着顶点位于x轴下方（因为 −Δ/4a < 0），所以曲线与x轴相交于两点。
Δ = 0 means the vertex lies exactly on the x-axis, giving a tangent point.
Δ = 0 意味着顶点恰好位于x轴上，产生一个切点。
Δ < 0 means the vertex lies above the x-axis, so the curve never crosses it.
Δ < 0 意味着顶点位于x轴上方，所以曲线永远不会穿过x轴。
This geometric interpretation can help you visualise problems and verify your algebraic answers.
这种几何解释有助于你形象化地理解问题并验证代数答案。
10. The Discriminant with Inequalities | 判别式与不等式
For inequalities involving quadratics, the discriminant tells us whether the quadratic expression is always positive or always negative. For instance, if a > 0 and Δ < 0, then ax² + bx + c > 0 for all real values of x. This is because the parabola opens upward and never crosses the x-axis, so it is entirely above the x-axis.
对于涉及二次函数的不等式，判别式告诉我们二次表达式是否恒正或恒负。例如，若 a > 0 且 Δ < 0，则对所有实数 x 都有 ax² + bx + c > 0。这是因为抛物线开口向上且永不与x轴相交，完全位于x轴上方。
Similarly, if a < 0 and Δ < 0, then ax² + bx + c < 0 for all real x. This property is useful in proving inequalities and in optimisation problems.
类似地，若 a < 0 且 Δ < 0，则对所有实数 x 都有 ax² + bx + c < 0。这个性质在证明不等式和优化问题中非常有用。
11. Common Mistakes and Pitfalls | 常见错误与陷阱
Students often make specific errors when working with the discriminant. Being aware of these will help you avoid them:
学生在处理判别式时经常会犯一些特定错误。了解这些错误有助于你避免它们：
Forgetting that a ≠ 0: When a parameter appears as the coefficient of x², you must exclude values that make the equation linear.
忘记 a ≠ 0：当参数作为 x² 的系数时，必须排除使方程变为一次方程的值。
Sign errors in b: If b is negative, remember to square it: (−5)² = 25, not −25.
b 的符号错误：如果 b 是负数，记得要平方：(−5)² = 25，而不是 −25。
Misinterpreting Δ = 0: Some students think the equation has no roots; in fact, it has exactly one root (a repeated root).
误解 Δ = 0：有些学生认为方程没有根；事实上，它恰好有一个根（重根）。
Ignoring the direction of the inequality: When multiplying or dividing an inequality by a negative number, the direction must be reversed. This often arises when solving for k with a negative coefficient.
忽略不等式方向：当不等式两边乘以或除以一个负数时，不等号方向必须反转。这在求解含负系数的 k 时经常出现。
12. Exam-Style Practice Questions | 考试风格练习题
To consolidate your understanding, try the following questions:
为了巩固理解，请尝试以下问题：
Question 1: Determine the nature of the roots of the equation 4x² − 12x + 9 = 0.
问题1：判断方程 4x² − 12x + 9 = 0 的根的性质。
Solution: Δ = (−12)² − 4 × 4 × 9 = 144 − 144 = 0, so the equation has one repeated real root.
解答：Δ = (−12)² − 4 × 4 × 9 = 144 − 144 = 0，因此方程有一个重实数根。
Question 2: Find the range of values of k for which the equation x² + kx + 4 = 0 has no real roots.
问题2：求使方程 x² + kx + 4 = 0 无实数根的 k 的取值范围。
Solution: We require Δ < 0: k² − 16 < 0, so k² < 16, giving −4 < k < 4.
解答：需要 Δ < 0：k² − 16 < 0，即 k² < 16，得到 −4 < k < 4。
Question 3: he line y = 2x + 3 is tangent to the curve y = x² + px + 1. Find the possible values of p.
问题3：直线 y = 2x + 3 与曲线 y = x² + px + 1 相切。求 p 的可能取值。
Solution: Substituting gives x² + px + 1 = 2x + 3, so x² + (p − 2)x − 2 = 0. For tangency, Δ = 0:
解答：代入得 x² + px + 1 = 2x + 3，即 x² + (p − 2)x − 2 = 0。相切要求 Δ = 0：
(p − 2)² − 4 × 1 × (−2) = 0 → (p − 2)² + 8 = 0
n, there is no real value of p for which the line is tangent. Indeed, the discriminant of the resulting equation is always positive, meaning the line always intersects the curve in two points.
由于 (p − 2)² + 8 = 0 没有实数解，所以不存在实数 p 使直线与该曲线相切。事实上，所得方程的判别式恒为正，意味着直线总是与曲线相交于两点。
In summary, the discriminant b² − 4ac is a simple yet powerful tool. It provides immediate information about the number and nature of roots of any quadratic equation, connects algebraic and geometric perspectives, and appears across a wide range of problem types in Edexcel A-Level Mathematics. Mastery of the discriminant is not just about memorising three cases — it is about understanding how algebra and geometry complement each other, a skill that will serve you well throughout your A-Level studies and beyond.
总之，判别式 b² − 4ac 是一个简单而强大的工具。它即时提供任何二次方程的根的数量和性质的信息，连接了代数与几何的视角，并广泛出现在埃德克斯A-Level数学的各种题型中。掌握判别式不仅仅是记住三种情况——更在于理解代数与几何如何相互补充，这一技能将在你的A-Level学习乃至更远的未来中大有裨益。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图