Example 1.6.2: Solving Quadratic Forms | 例 1.6.2：求解二次型方程

📚 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

The highest power is 4, so this is a quartic equation. However, note that every term is a power of x²: x⁴ is (x²)² and x² is itself. 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

Factorise u² − 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

It 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。

The substitution u = x² maps both positive and negative x to a single u. Under this map, the two branches of the graph (for x > 0 and x < 0) are folded together, converting the quartic into a parabola in u. The roots u = 1 and 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

The same idea works for any equation of the 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

Answers: 1. x = ±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。

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Example 1.6.2: Solving Quadratic Forms | 例 1.6.2：求解二次型方程