📚 Example 1.6.2: Solving Quadratic Forms | 例 1.6.2：求解二次型方程
This worked example demonstrates how to solve an equation that looks like a quartic yet behaves as a quadratic. By making a clever substitution, we reduce the problem to familiar territory and find all four real roots.
本例题展示如何求解一个看似四次、实为二次型的方程。通过巧妙的换元，我们将问题转化为熟悉的二次方程，并求出四个实数根。
1. Problem Statement | 问题陈述
We are asked to solve the equation:
要求解方程：
x⁴ − 5x² + 4 = 0
f. This structure suggests that the equation is actually quadratic in the variable u = x².
最高次数为 4，因此这是一个四次方程。但注意，每一项都是 x² 的幂：x⁴ 是 (x²)²，x² 本身就是 x²。这种结构说明，若令 u = x²，则方程实际上是关于 u 的二次方程。
2. Substitution Technique | 换元法
Define the substitution:
我们作如下换元：
u = x²
Then x⁴ = (x²)² = u². Substitute these into the original equation:
于是 x⁴ = (x²)² = u²。将两者代入原方程：
u² − 5u + 4 = 0
This is now a simple quadratic. The substitution has reduced the degree of the equation from 4 to 2, making it far easier to solve.
现在得到一个简单的二次方程。换元将方程的次数从 4 降为 2，大大简化了求解过程。
3. Solving the Quadratic in u | 求解关于 u 的二次方程
5u + 4 = 0. We look for two numbers that multiply to 4 and add to −5. These are −1 and −4:
对 u² − 5u + 4 = 0 进行因式分解。寻找两个相乘得 4、相加得 −5 的数，即 −1 和 −4：
(u − 1)(u − 4) = 0
Hence, u = 1 or u = 4.
因此，u = 1 或 u = 4。
Alternatively, one could use the quadratic formula:
或者，也可以使用求根公式：
u = (5 ± √25 − 16) / 2 = (5 ± 3) / 2
This gives the same results: u = 1 and u = 4.
得到相同结果：u = 1 和 u = 4。
4. Recovering x | 回到变量 x
Remember that we set u = x². We must now solve two separate equations for x:
别忘了我们令 u = x²。现在需要分别求解两个关于 x 的方程：
For u = 1:
当 u = 1 时：
x² = 1 ⇒ x = ±1
For u = 4:
当 u = 4 时：
x² = 4 ⇒ x = ±2
Therefore, the full set of real solutions is:
因此，全部实数解为：
x = 1, −1, 2, −2
5. Checking the Solutions | 检验解
is always wise to check at least one or two solutions in the original equation. For x = 1:
在检验时至少代入一两个解是明智的做法。验证 x = 1：
1⁴ − 5(1²) + 4 = 1 − 5 + 4 = 0
For x = −2:
验证 x = −2：
(−2)⁴ − 5(−2)² + 4 = 16 − 20 + 4 = 0
All four values satisfy the equation, confirming our solution set.
四个值均满足原方程，说明解集正确无误。
6. Graphical Interpretation | 图像解释
The curve y = x⁴ − 5x² + 4 is a W‑shaped quartic. It crosses the x‑axis at exactly the four points we found: x = −2, −1, 1, 2.
曲线 y = x⁴ − 5x² + 4 是 W 形的四次曲线，它与 x 轴的交点恰好是我们求得的四个点：x = −2, −1, 1, 2。
u = 4 correspond to the turning points of the quartic in terms of its geometric structure.
换元 u = x² 将正负 x 映射到同一个 u。在这个映射下，图形的两支（x > 0 和 x < 0）折叠在一起，把四次曲线变成 u 域中的抛物线。根 u = 1 和 u = 4 在几何结构上对应四次的某种对称性。
7. Alternative Method: Direct Factorisation | 另法：直接因式分解
We can factorise the original quartic directly by treating x² as a single entity:
我们也可以把 x² 视为一个整体直接分解原四次式：
x⁴ − 5x² + 4 = (x² − 1)(x² − 4)
Then factorise each quadratic further:
再将每个二次式继续分解：
(x² − 1) = (x − 1)(x + 1)
(x² − 4) = (x − 2)(x + 2)
Hence the original equation becomes:
于是原方程变为：
(x − 1)(x + 1)(x − 2)(x + 2) = 0
Setting each factor to zero gives the same four roots.
令每个因式为零，同样得到四个根。
8. Common Pitfalls | 常见误区
Forgetting the ± sign. When solving x² = u for x, do not write x = √u only; remember x = ±√u.
忘记 ± 号。求解 x² = u 时，不要只写 x = √u，要记得 x = ±√u。
Stopping at u. Some students solve for u and then stop, forgetting to substitute back to find x.
算到 u 就停下。有些学生求出 u 后不再回代，忘了求 x。
Losing roots. When using the substitution method, you must consider both possible values of u and both signs for each.
丢根。使用换元法时，必须考虑 u 的两个值以及每个值对应的正负号。
Factorisation errors. In the quadratic in u, signs matter. Check that the product is correct.
因式分解错误。在关于 u 的二次方程中，符号很关键。检查乘积是否正确。
9. Extension: More General Quadratic Forms | 推广：更一般的二次型方程
he form ax²ⁿ + bxⁿ + c = 0. Choose u = xⁿ, then the equation becomes a standard quadratic.
同样的方法适用于一切形如 ax²ⁿ + bxⁿ + c = 0 的方程，令 u = xⁿ 即可化为标准二次方程。
For example, solve x⁶ − 7x³ − 8 = 0:
例如，求解 x⁶ − 7x³ − 8 = 0：
Let u = x³, then u² − 7u − 8 = 0 = (u − 8)(u + 1), giving u = 8 or u = −1.
令 u = x³，则 u² − 7u − 8 = 0 = (u − 8)(u + 1)，得 u = 8 或 u = −1。
If u = 8, then x³ = 8 so x = 2 (real). If u = −1, then x³ = −1 so x = −1 (real).
若 u = 8，则 x³ = 8，所以 x = 2（实数）；若 u = −1，则 x³ = −1，所以 x = −1（实数）。
Be careful: if n is even, each u leads to two real roots; if n is odd, each u leads to exactly one real root.
注意：若 n 为偶数，每个 u 对应两个实数根；若 n 为奇数，每个 u 恰对应一个实数根。
10. Practice Problems | 练习
Try these questions on your own, then check your answers.
请先独立尝试以下习题，再核对答案。
1. Solve x⁴ − 10x² + 9 = 0
求解 x⁴ − 10x² + 9 = 0
2. Solve x⁶ + 9x³ + 8 = 0
求解 x⁶ + 9x³ + 8 = 0
3. Solve x⁸ − 17x⁴ + 16 = 0
求解 x⁸ − 17x⁴ + 16 = 0
= ±1, ±3. 2. x = −1, x = −2 (real roots; the other solutions are complex). 3. x = ±1, ±2.
答案：1. x = ±1, ±3；2. 实数根为 x = −1, x = −2（其余为复数解）；3. x = ±1, ±2。
In this example, we saw that recognising hidden quadratic structure is a powerful tool. It reduces complicated equations to simple ones and ensures no roots are missed. Always substitute back and verify your final answers.
在本例中，我们看到识别隐藏的二次型结构是一种强力工具。它能把复杂的方程化繁为简，并确保不漏根。务必回代并检验最终答案。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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