📚 The Discriminant Δ: Determining the Nature of Roots of Quadratic Equations | 判别式Δ：判断二次方程根的情况
= 0, the value of Δ = b² − 4ac tells us everything about the number and type of its roots — without actually solving the equation.
判别式Δ是代数中最强大的工具之一。对于任意二次方程 ax² + bx + c = 0，Δ = b² − 4ac 的值可以告诉我们根的数量和类型——而无需真正解方程。
1. Definition of the Discriminant | 判别式的定义
where a ≠ 0, the discriminant is defined as Δ = b² − 4ac. The symbol Δ is the Greek letter 'Delta'.
对于标准形式为 ax² + bx + c = 0 的二次方程，其中 a ≠ 0，判别式定义为 Δ = b² − 4ac。符号Δ是希腊字母“德尔塔”。
Δ = b² − 4ac
x = (−b ± √Δ) / (2a), so its sign determines whether the square root is real or imaginary.
判别式实际上就是二次公式中根号下的表达式。它出现在公式 x = (−b ± √Δ) / (2a) 中，因此它的正负决定了根号内是实数还是虚数。
2. The Three Cases: Δ > 0 | 三种情况之一：Δ > 0
quare root √Δ is a positive real number, so −b + √Δ and −b − √Δ give two different values.
当 Δ > 0 时，二次方程有两个不相等的实数根。这意味着 √Δ 是一个正实数，因此 −b + √Δ 和 −b − √Δ 给出两个不同的值。
x₁ = (−b + √Δ) / (2a), x₂ = (−b − √Δ) / (2a)
+ c intersects the x-axis at exactly two distinct points. These two intersection points are the two real roots.
从图形上看，抛物线 y = ax² + bx + c 与 x 轴恰好有两个不同的交点。这两个交点就是两个实数根。
3. The Case: Δ = 0 | 情况之二：Δ = 0
th the plus and minus branches of the quadratic formula give the same value x = −b / (2a).
当 Δ = 0 时，二次方程恰好有一个实数根，称为重根或二重根。由于 √Δ = 0，二次公式中的加号和减号分支给出相同的值 x = −b / (2a)。
x = −b / (2a)
Graphically, the parabola touches the x-axis at exactly one point — lies on the x-axis. The equation can be written as a perfect square, such as (x − 3)² = 0.
从图形上看，抛物线与 x 轴恰好相切于一点——抛物线的顶点位于 x 轴上。该方程可以写成完全平方的形式，例如 (x − 3)² = 0。
4. The Case: Δ < 0 | 情况之三：Δ < 0
omplex conjugate roots. Since √Δ is imaginary, the roots are of the form p ± qi, where i = √−1.
当 Δ < 0 时，二次方程没有实数根。相反，它有两个共轭复数根。由于 √Δ 是虚数，根的形式为 p ± qi，其中 i = √−1。
Graphically, the parabola does not intersect the x-axis at all. If a > 0, the parabola lies entirely above the x-axis; if a < 0, it lies entirely below the x-axis.
从图形上看，抛物线完全不与 x 轴相交。如果 a > 0，抛物线完全位于 x 轴上方；如果 a < 0，则完全位于 x 轴下方。
5. Deriving the Discriminant from the Quadratic Formula | 从二次公式推导判别式
c formula is derived by completing the square on ax² + bx + c = 0. The result is x = (−b ± √b² − 4ac) / (2a). The expression under the square root is precisely the discriminant.
二次公式通过对 ax² + bx + c = 0 配方法推导而来。结果是 x = (−b ± √b² − 4ac) / (2a)。根号下的表达式正是判别式。
This process reveals that the equation is solvable in real numbers only when b² − 4ac ≥ 0.
从 ax² + bx + c = 0 出发，我们除以 a，配方，并解出 x。这个过程揭示了只有在 b² − 4ac ≥ 0 时，方程才有实数解。
6. Worked Example 1: Two Distinct Real Roots | 例题1：两个不相等的实数根
0. Here a = 2, b = −5, and c = 1. Compute Δ = b² − 4ac = (−5)² − 4 × 2 × 1 = 25 − 8 = 17.
考虑方程 2x² − 5x + 1 = 0。这里 a = 2，b = −5，c = 1。计算 Δ = b² − 4ac = (−5)² − 4 × 2 × 1 = 25 − 8 = 17。
Δ = 17 > 0
Since Δ > 0, the equation has two distinct real roots. They are x = (5 ± √17) / 4. Note that we do not need to simplify the roots to know their nature — the sign of Δ alone is sufficient.
因为 Δ > 0，方程有两个不相等的实数根。它们是 x = (5 ± √17) / 4。注意我们不需要化简根来判断根的性质——仅凭Δ的符号就足够了。
7. Worked Example 2: A Repeated Root | 例题2：重根
− 6x + 9 = 0. Here a = 1, b = −6, and c = 9. Compute Δ = (−6)² − 4 × 1 × 9 = 36 − 36 = 0.
考虑方程 x² − 6x + 9 = 0。这里 a = 1，b = −6，c = 9。计算 Δ = (−6)² − 4 × 1 × 9 = 36 − 36 = 0。
Δ = 0
actly one repeated real root. Solving gives x = 6 / 2 = 3. Indeed, x² − 6x + 9 = (x − 3)².
因为 Δ = 0，方程恰好有一个重实数根。解出 x = 6 / 2 = 3。事实上，x² − 6x + 9 = (x − 3)²。
8. Worked Example 3: No Real Roots | 例题3：无实数根
x² + 2x + 5 = 0. Here a = 1, b = 2, and c = 5. Compute Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16.
考虑方程 x² + 2x + 5 = 0。这里 a = 1，b = 2，c = 5。计算 Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16。
Δ = −16 < 0
ts. The complex roots are x = −1 ± 2i, which can be found using the quadratic formula with √−16 = 4i.
因为 Δ < 0，方程没有实数根。复数根为 x = −1 ± 2i，可以用二次公式并利用 √−16 = 4i 求得。
9. Special Case: b Is Even | 特殊情况：b 为偶数
can be simplified. Substituting b = 2k into Δ = b² − 4ac gives Δ = 4k² − 4ac = 4(k² − ac).
如果系数 b 是偶数，设 b = 2k，则判别式可以简化。将 b = 2k 代入 Δ = b² − 4ac 得到 Δ = 4k² − 4ac = 4(k² − ac)。
Sometimes the reduced discriminant Δ' determine the nature of roots more quickly. For example, in x² + 4x + 1 = 0, b/2 = 2, so Δ' = 4 − 1 = 3 > 0, giving two real roots.
有时使用简化判别式 Δ' = (b/2)² − ac = k² − ac。它与Δ同号，因此可以更快地判断根的性质。例如，在 x² + 4x + 1 = 0 中，b/2 = 2，所以 Δ' = 4 − 1 = 3 > 0，有两个实数根。
10. Connection to the Sum and Product of Roots | 与根的和与积的联系
For a quadratic equation ax² + bx + c = 0 with roots r₁ and r₂, Vieta' /a and r₁ × r₂ = c/a. The discriminant relates to these roots through (r₁ − r₂)² = Δ / a².
对于根为 r₁ 和 r₂ 的二次方程 ax² + bx + c = 0，韦达定理给出 r₁ + r₂ = −b/a 和 r₁ × r₂ = c/a。判别式与这些根的关系为 (r₁ − r₂)² = Δ / a²。
the roots coincide; if Δ > 0, they are separated by a real distance; if Δ < 0, their difference is imaginary.
这个恒等式表明Δ度量了两个根之间的距离。如果 Δ = 0，根重合；如果 Δ > 0，它们之间有实距离；如果 Δ < 0，它们的差是虚数。
11. Summary Table of the Discriminant | 判别式汇总表
following table summarizes all possible cases for the quadratic equation ax² + bx + c = 0:
下表总结了二次方程 ax² + bx + c = 0 的所有可能情况：
| Discriminant Δ | Nature of Roots | Graphical Meaning |
| Δ > 0 | Two distinct real roots | Parabola crosses x-axis twice |
| Δ = 0 | One repeated real root | Parabola touches x-axis once |
| Δ < 0 | Two complex conjugate roots | Parabola does not meet x-axis |
ndard form with a ≠ 0. Always rewrite the equation as ax² + bx + c = 0 before computing Δ.
请记住：只有当方程是标准形式且 a ≠ 0 时，判别式才有意义。在计算Δ之前，一定要先将方程改写为 ax² + bx + c = 0。
12. Common Pitfalls and Exam Tips | 常见错误与考试技巧
in 2x² + 3 = 5x, you must rewrite it as 2x² − 5x + 3 = 0 before applying the discriminant.
一个常见错误是在确定 a、b、c 之前忘记将方程化为零。例如，在 2x² + 3 = 5x 中，必须先改写为 2x² − 5x + 3 = 0，然后再应用判别式。
- Always check a ≠ 0 — otherwise the equation is not quadratic. | 始终检查 a ≠ 0——否则方程不是二次方程。
- Be careful with negative signs: b² is always positive, but −4ac can be negative. | 小心负号：b² 总是正的，但 −4ac 可能是负的。
- If the question asks for 'real roots', include the case Δ ≥ 0. | 如果题目问“实数根”，要包含 Δ ≥ 0 的情况。
- When asked to 'show that the equation has no real roots', it is enough to compute Δ and state Δ < 0. | 当要求“证明方程没有实数根”时，计算Δ并说明Δ < 0 即可。
In many exam problems, you are given a parameter k and asked to find the range of k for which the equation has two distinct real roots. This becomes an inequality Δ > 0, which you solve to find k.
在许多考试题中，你会得到一个参数 k，并被要求找出使方程有两个不同实数根的 k 的取值范围。这转化为不等式 Δ > 0，解这个不等式即可求出 k。
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