📚 Example 2.7.1: Hidden Quadratics | 例题2.7.1：隐藏的二次方程
In this worked example, we will explore how to solve an equation that looks like a mixture of algebraic terms but can be rewritten as a quadratic equation. This is a classic AQA A-Level Mathematics technique known as solving a hidden quadratic.
在这个例题中，我们将探讨如何求解一个看似包含不同代数项、但可以重新写成二次方程的方程。这是 AQA 高中进阶数学中的经典技巧，称为“隐藏的二次方程”。
1. The Problem | 问题
Solve the equation for real values of x:
求解实数 x：
x − 5√x + 6 = 0
At first glance, this is not a quadratic equation because it contains both x and √x. We need to find a clever substitution.
乍一看，这并不是二次方程，因为它同时含有 x 和 √x。我们需要找到一个巧妙的换元方法。
2. Recognising the Quadratic Pattern | 识别二次模式
Notice that x is actually the square of √x. In symbols, x = (√x)². Therefore the equation can be viewed as a quadratic in the variable √x.
注意到 x 实际上是 √x 的平方，即 x = (√x)²。因此，该方程可以被看作关于变量 √x 的二次方程。
This type of appearance is called a hidden quadratic because the underlying structure is a quadratic expression in a smaller “base” expression.
这种形式被称为“隐藏的二次方程”，因为其底层结构是关于某个更小“基础”表达式的二次式。
3. Substitution | 换元
Let u = √x. Since u² = x, we can replace every term in the original equation.
令 u = √x。因为 u² = x，我们可以替换原方程中的每一项。
u² − 5u + 6 = 0
This is now a standard quadratic equation in u, which we can factorise or solve using the quadratic formula.
现在这是一个关于 u 的标准二次方程，我们可以通过因式分解或使用求根公式来求解。
4. Solving the Quadratic in u | 求解关于 u 的二次方程
Factorise the left-hand side:
对左边进行因式分解：
(u − 2)(u − 3) = 0
Therefore, u = 2 or u = 3. Both values are positive, which is important because u represents a square root.
因此，u = 2 或 u = 3。这两个值都是正的，这很重要，因为 u 代表的是一个平方根。
5. Back-Substitution | 反代
Recall that u = √x. We now substitute each solution back to find x.
回忆 u = √x。现在我们将每个解代回以求出 x。
When u = 2:
当 u = 2 时：
√x = 2 → x = 4
When u = 3:
当 u = 3 时：
√x = 3 → x = 9
So the proposed solutions are x = 4 and x = 9.
因此初步解为 x = 4 和 x = 9。
6. Verification | 验证解
Always check solutions by substituting them back into the original equation.
务必通过将解代回原方程来检验。
For x = 4:
对于 x = 4：
4 − 5√4 + 6 = 4 − 10 + 6 = 0 ✓
For x = 9:
对于 x = 9：
9 − 5√9 + 6 = 9 − 15 + 6 = 0 ✓
Both values satisfy the original equation, so they are both valid solutions.
两个值都满足原方程，因此它们都是有效解。
7. Handling Domain Restrictions | 定义域限制的处理
The expression √x is only defined for x ≥ 0. In our substitution we used u = √x, so u must also be non-negative.
表达式 √x 仅对 x ≥ 0 有定义。在我们的换元中使用了 u = √x，因此 u 也必须是非负的。
In this example, both solutions for u were positive, so no solutions were lost. If we had found a negative u, it would have been rejected because √x cannot be negative.
在这个例子中，u 的两个解均为正，因此没有丢失解。如果我们找到一个负的 u，它将被舍去，因为 √x 不可能为负。
8. Graphical Interpretation | 图形解释
Consider the graph of y = x − 5√x + 6. The x-intercepts of this curve are exactly the solutions to the equation.
考虑函数 y = x − 5√x + 6 的图像。该曲线与 x 轴的交点正是方程的解。
At x = 4 and x = 9, the curve crosses the x-axis, confirming the algebraic answers. The curve is not a usual parabola because of the square root term, but it still intersects the x-axis at two points.
在 x = 4 和 x = 9 处，曲线与 x 轴相交，验证了代数答案。由于包含平方根项，该曲线不是通常的抛物线，但它仍然与 x 轴相交于两点。
9. Common Pitfalls | 常见陷阱
One common mistake is to square the whole equation immediately, which usually introduces extra terms and makes the problem harder.
一个常见错误是立即对整个方程两边平方，这通常会产生额外的项，使问题更加困难。
Another mistake is forgetting to reject negative substitutions or forgetting to check the domain of the original variable.
另一个错误是忘记舍去负的换元值，或忘记检查原变量的定义域。
Always write down the substitution clearly and state that u = √x, so that the domain restriction is visible.
始终清晰写出换元 u = √x，以便定义域限制变得可见。
10. Exam Technique | 考试技巧
In the AQA exam, hidden quadratic questions often appear as “solve” or “find the exact values of x”. You should look for terms that are powers of each other, such as x and √x, or x² and x⁴.
在 AQA 考试中，隐藏二次方程的问题通常以“求解”或“求 x 的精确值”的形式出现。你应该寻找互为幂次关系的项，例如 x 和 √x，或 x² 和 x⁴。
Factorising the quadratic in u is usually faster than using the quadratic formula. However, if factorisation is not obvious, use the formula with confidence.
对 u 的二次式进行因式分解通常比使用求根公式更快。然而，如果因式分解不明显，可以放心使用求根公式。
11. Try It Yourself | 自己练习
Solve the following hidden quadratic equation:
求解以下隐藏二次方程：
x − 3√x + 2 = 0
Use the substitution u = √x, then solve the resulting quadratic. Check your answers by substitution.
使用换元 u = √x，然后求解所得的二次方程。通过代入检验你的答案。
Answer: x = 1 or x = 4.
答案：x = 1 或 x = 4。
12. Conclusion | 总结
Hidden quadratics are a powerful tool in A-Level Mathematics. By choosing a suitable substitution, an equation that looks unfamiliar can be transformed into a standard quadratic that is easy to solve.
隐藏二次方程是高中进阶数学中的一个有力工具。通过选择适当的换元，一个看起来陌生的方程可以被转化为容易求解的标准二次方程。
The key steps are: recognise the pattern, substitute, solve the quadratic, substitute back, and check all solutions within the domain.
关键步骤是：识别模式、换元、求解二次方程、代回，并在定义域内检验所有解。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导