📚 IB Mathematics: Quadratic Equations with Negative Discriminant | 判别式小于零的实系数二次方程
For a quadratic equation ax² + bx + c = 0 with real coefficients a, b and c, the discriminant Δ = b² − 4ac is the fastest indicator of the nature of its roots. When Δ < 0, many students fear that the equation 'has no solutions', but in the IB Mathematics syllabus the correct statement is that the equation has no real solutions and exactly two non-real complex conjugate solutions. This idea is tested in both Analysis and Approaches (AA) and Applications and Interpretation (AI), usually as short-answer questions involving the quadratic formula or the complex roots of a real-coefficient polynomial.
对于实系数二次方程 ax² + bx + c = 0（其中 a、b、c 为实数），判别式 Δ = b² − 4ac 是判断根的性质最快的工具。当 Δ < 0 时，很多同学误以为方程“无解”，但在 IB 数学课程中，准确说法是：方程没有实数解，但恰有两个非实数的共轭复数解。这一考点同时出现在分析与方法（AA）和应用与解释（AI）两份考纲中，常以运用求根公式或讨论实系数多项式复数根的形式出现。
1. The Discriminant and the Nature of Roots | 判别式与根的类别
Let a ≠ 0. For the equation ax² + bx + c = 0, the three possible signs of the discriminant give three different types of root behaviour:
设 a ≠ 0。对于方程 ax² + bx + c = 0，判别式的正负号对应三类不同的根的性质：
Δ = b² − 4ac
| Sign of Δ | Number of roots | Type of roots |
| Δ > 0 | two distinct roots | both roots are real |
| Δ = 0 | one repeated root | a real double root |
| Δ < 0 | two distinct roots | non-real complex conjugate roots p ± qi, q ≠ 0 |
When Δ < 0, the two roots cannot be plotted on a real number line. They are written in the form p ± qi, where p and q are real and i² = −1. Because the original coefficients are real, the two roots are always perfect mirrors of each other in the complex plane.
当 Δ < 0 时，两个根无法画在实数轴上。它们必须写成 p ± qi 的形式，其中 p、q 为实数，且 i² = −1。由于原方程的系数均为实数，这两个根在复平面中必定互为镜像，即一对共轭复数。
2. What Does Δ < 0 Actually Tell Us? | Δ < 0 到底说明了什么？
Completing the square gives an excellent explanation. Rewrite the equation as:
配方可以给出很好的解释。将方程改写为：
(x + b / (2a))² = (b² − 4ac) / (4a²)
The right-hand side is Δ / (4a²). Since a² > 0, the denominator is positive; if Δ < 0, the whole right-hand side is a negative real number. For any real x, the left-hand side is a square and cannot be negative, so no real solution exists.
等号右边就是 Δ / (4a²)。由于 a² > 0，分母为正；若 Δ < 0，整个右边就是一个负数。对任意实数 x 而言，左边是一个完全平方，不可能为负，因此方程没有实数解。
But over the complex numbers every negative real number has two square roots: for a positive number k, the two square roots of −k are i√k and −i√k. Therefore the equation still has two distinct complex-number solutions. This agrees with the fundamental theorem of algebra, which guarantees that a quadratic has exactly two roots when complex roots are counted.
但在复数范围内，每个负实数都有两个平方根：对于正数 k，−k 的两个平方根是 i√k 与 −i√k。因此原方程仍然有两个不同的复数解。这正与代数基本定理一致：只要计入复数根，二次方程一定恰有两个根。
3. Solving Using the Quadratic Formula | 用求根公式求解
The ordinary quadratic formula remains valid for Δ < 0:
普通的求根公式在 Δ < 0 时依然成立：
x = (−b ± √b² − 4ac) / (2a)
When b² − 4ac < 0, it is helpful to write the square root as i √4ac − b², because 4ac − b² is positive. Hence:
当 b² − 4ac < 0 时，将根号写成 i√4ac − b² 会更方便，因为此时 4ac − b² 是一个正实数。因此：
x = (−b ± i √4ac − b²) / (2a)
If we separate real and imaginary parts, the two roots always share the same real part −b/(2a). The imaginary parts are opposites, so the pair is always z = m + ni and z' = m − ni.
若把实部与虚部分开，两个根总共有相同的实部 −b/(2a)。虚部互为相反数，所以两个根总是成对出现：z = m + ni 与 z' = m − ni。
4. The Conjugate Root Theorem | 共轭复根定理
Suppose a polynomial has real coefficients. If a complex number z is a root, then its complex conjugate z̄ is also a root. For a quadratic this is easy to verify with Vieta's formulas.
设一个多项式的系数全为实数。若复数 z 是它的根，那么 z 的共轭复数 z̄ 也一定是它的根。对于二次方程，用韦达定理即可方便地验证这一点。
Let the roots be p + qi and r + si. Their sum is real because −b/a is real, and their product is real because c/a is real. If p + qi is one root and q ≠ 0, the other root must have imaginary part −q to cancel the imaginary part of the sum. Substituting r = p and s = −q also makes the product p² + q², which is real. Therefore the only possible partner of p + qi is p − qi.
设两根为 p + qi 与 r + si。由于 −b/a 为实数，两根之和必为实数；由于 c/a 为实数，两根之积也必为实数。若 p + qi 是其中一个根且 q ≠ 0，为了让“和”的虚部抵消，另一个根的虚部必须是 −q。再取 r = p、s = −q，则乘积为 p² + q²，确实是实数。因此 p + qi 唯一的搭档只能是 p − qi。
The conjugate root theorem is the core reason why complex roots of quadratic equations with real coefficients always appear as a ± bi.
共轭复根定理正是“实系数二次方程的复数根一定以 a ± bi 形式成对出现”的根本原因。
5. Geometric Interpretation | 几何解释
Graphically, the roots of ax² + bx + c = 0 are the x-intercepts of the parabola y = ax² + bx + c. If Δ < 0, the parabola has no intersection with the x-axis.
从图像上看，方程 ax² + bx + c = 0 的根就是抛物线 y = ax² + bx + c 与 x 轴的交点。若 Δ < 0，抛物线不与 x 轴相交。
The vertex of the parabola has coordinates (−b/(2a), −Δ/(4a)). When Δ < 0, the quantity −Δ is positive, so the sign of the y-coordinate of the vertex is exactly the sign of a. If a > 0, the vertex lies above the x-axis and the whole parabola stays above the x-axis. If a < 0, the vertex lies below the x-axis and the whole parabola stays below it.
抛物线的顶点坐标为 (−b/(2a), −Δ/(4a))。当 Δ < 0 时，−Δ 为正，因此顶点纵坐标的正负号恰好与 a 相同。若 a > 0，顶点在 x 轴上方，整条抛物线也都在 x 轴上方；若 a < 0，顶点在 x 轴下方，整条抛物线也都在 x 轴下方。
This explains why a quadratic with negative discriminant never crosses the horizontal axis, yet the algebraic equation still has valid answers inside the complex number system.
这就解释了为什么判别式为负的二次方程图像永远不穿过 x 轴，但从代数角度看，方程在复数系中仍然有确定的解。
6. Sum and Product of Complex Roots | 复数根的和与积
Vieta's formulas continue to hold when the roots are complex. For ax² + bx + c = 0 with roots z₁ and z₂:
当根为复数时，韦达定理仍然成立。对于 ax² + bx + c = 0，设两根为 z₁ 与 z₂：
z₁ + z₂ = −b/a, z₁ × z₂ = c/a
For a conjugate pair z = p + qi and z̄ = p − qi, the sum is 2p and the product is p² + q². Both are real numbers. In fact, z₁ × z₂ = p² + q² = |z|², the squared modulus of either root.
对于共轭对 z = p + qi 与 z̄ = p − qi，它们的和为 2p，乘积为 p² + q²。两者都是实数。实际上 z₁ × z₂ = p² + q² = |z|²，即任意一个根的模长的平方。
Because Δ < 0 implies 4ac > b² ≥ 0, the product c/a is positive. Consequently the two complex roots have equal modulus √c/a. This fact is particularly useful when you are asked to recover a quadratic from a given complex root.
因为 Δ < 0 意味着 4ac > b² ≥ 0，所以 c/a 必为正数。于是两个复数根具有相同的模长 √c/a。当题目给出一个复数根并要求还原原二次方程时，这个结论尤其好用。
7. Complex Plane Representation | 复平面上的表示
In the Argand diagram, draw the two roots z = p + qi and z̄ = p − qi. They are symmetric about the horizontal real axis.
在阿甘图中画出两个根 z = p + qi 与 z̄ = p − qi。它们关于水平的实轴对称。
English: Their common real part p = −b/(2a) is the x-coordinate of both points.
中文：它们的公共实部 p = −b/(2a) 是两个点的横坐标。
English: Their imaginary parts are opposite, so the segment joining them is vertical and is bisected by the real axis.
中文：它们的虚部互为相反数，因此连接两点的线段是竖直的，并且被实轴垂直平分。
English: Both points lie on a circle centred at the origin with radius √c/a.
中文：两个点都位于以原点为圆心、半径为 √c/a 的同一个圆上。
This symmetric picture is a quick visual check: whenever the answer to a real-coefficient quadratic is one complex number, you must immediately write its conjugate as the second solution.
这个对称的图像是一个快速自检方法：只要实系数二次方程解出一个复数，就必须立刻写出它的共轭复数作为另一个解。
8. Worked Example: Solving a Quadratic | 例题一：解具体方程
Question: Solve x² − 6x + 25 = 0.
题目：解方程 x² − 6x + 25 = 0。
Solution: Here a = 1, b = −6 and c = 25. The discriminant is:
解答：这里 a = 1，b = −6，c = 25。判别式为：
Δ = (−6)² − 4 × 1 × 25 = 36 − 100 = −64
Because Δ < 0, the roots are complex conjugates. Using the formula x = (−b ± i √4ac − b²)/(2a):
因为 Δ < 0，两根为共轭复数。代入公式 x = (−b ± i√4ac − b²)/(2a)：
x = (6 ± i √64) / 2 = (6 ± 8i) / 2 = 3 ± 4i
So the solution set is {3 + 4i, 3 − 4i}. Check the product: (3 + 4i)(3 − 4i) = 9 + 16 = 25 = c/a, and the sum is 6 = −b/a.
所以解集为 {3 + 4i, 3 − 4i}。检验：乘积 (3 + 4i)(3 − 4i) = 9 + 16 = 25 = c/a，和为 6 = −b/a，完全吻合。
9. Worked Example: Rebuilding the Equation | 例题二：由根反求方程
Question: A quadratic equation with real coefficients has one root 2 − 3i. Find the equation in the form x² + px + q = 0.
题目：已知一个实系数二次方程的一个根为 2 − 3i，求该方程形如 x² + px + q = 0 的表达式。
Solution: Since the coefficients are real, the conjugate root 2 + 3i must also be a root. The sum of the two roots is:
解答：由于系数为实数，共轭根 2 + 3i 必为另一个根。两根之和为：
(2 − 3i) + (2 + 3i) = 4
and their product is:
两根之积为：
(2 − 3i)(2 + 3i) = 4 + 9 = 13
For a monic quadratic x² + px + q = 0, the sum of roots is −p and the product is q. Hence p = −4 and q = 13. The required equation is x² − 4x + 13 = 0.
对于首项系数为 1 的二次方程 x² + px + q = 0，两根和为 −p，两根积为 q。因此 p = −4，q = 13，所求方程为 x² − 4x + 13 = 0。
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