Solving Minimum Value and Quadratic Inequality: Example 3.5.1 | 求最小值与二次不等式：例 3.5.1

📚 Solving Minimum Value and Quadratic Inequality: Example 3.5.1 | 求最小值与二次不等式：例 3.5.1

This worked example demonstrates how to find the minimum value of a quadratic function using the method of completing the square, and then applies the result to solve a quadratic inequality. The problem is taken from a typical AQA A-Level Mathematics exercise.

本例题演示如何通过配方法（completing the square）求二次函数的最小值，并利用所得结果解二次不等式。本题选自 AQA A-Level 数学的典型练习。

The function under consideration is y = 2x² − 8x + 5. We will first rewrite it in vertex form, identify the minimum point, and then solve the inequality 2x² − 8x + 5 ≥ 3.

所研究的函数为 y = 2x² − 8x + 5。我们首先将其改写为顶点式，找出最小值点，然后解不等式 2x² − 8x + 5 ≥ 3。

1. Completing the Square | 配方法

To complete the square for a quadratic in the form ax² + bx + c, we first factor out the coefficient of x² from the first two terms. Here, a = 2, b = −8, c = 5.

对于形如 ax² + bx + c 的二次式，我们先从前两项中提出 x² 的系数。此处 a = 2，b = −8，c = 5。

We factor out 2 from the terms involving x:

我们从含 x 的项中提出 2：

y = 2(x² − 4x) + 5

Next, inside the bracket, we complete the square on x² − 4x. The formula is x² − 4x = (x − 2)² − 4. This works because (x − 2)² = x² − 4x + 4, so we subtract 4 to compensate.

接下来，在括号内对 x² − 4x 配方。利用公式 x² − 4x = (x − 2)² − 4。这是因为 (x − 2)² = x² − 4x + 4，因此减去 4 以保持相等。

Substituting this back gives:

将其代回得到：

y = 2[(x − 2)² − 4] + 5

2. Simplifying to Vertex Form | 化简为顶点式

We now expand the bracket and simplify the constant terms:

现在我们展开括号并合并常数项：

y = 2(x − 2)² − 8 + 5

y = 2(x − 2)² − 3

This is the vertex form of the quadratic, written as y = a(x − h)² + k, where a = 2, h = 2, and k = −3.

这就是该二次函数的顶点式，写作 y = a(x − h)² + k，其中 a = 2，h = 2，k = −3。

Since a = 2 > 0, the parabola opens upwards, and the vertex (h, k) = (2, −3) is the minimum point of the curve.

由于 a = 2 > 0，抛物线开口向上，因此顶点 (h, k) = (2, −3) 是曲线的最小值点。

3. Minimum Value and Its x-coordinate | 最小值及对应的 x 值

From the vertex form, the minimum value of the function is simply the value of k, because the squared term (x − 2)² is always non-negative, and is zero only when x = 2.

由顶点式可知，函数的最小值就是 k 的值，因为平方项 (x − 2)² 始终非负，且仅当 x = 2 时为零。

Answer: The minimum value is y = −3, attained at x = 2.

答案： 最小值为 y = −3，在 x = 2 时取得。

Equivalently, the minimum point on the curve is (2, −3).

等价地，曲线上的最小点为 (2, −3)。

4. Setting Up the Inequality | 建立不等式

We are asked to solve 2x² − 8x + 5 ≥ 3. Using the vertex form we derived, we substitute y = 2(x − 2)² − 3 into the inequality.

题目要求解 2x² − 8x + 5 ≥ 3。利用已得到的顶点式，我们将 y = 2(x − 2)² − 3 代入该不等式。

2(x − 2)² − 3 ≥ 3

Adding 3 to both sides:

两边同时加上 3：

2(x − 2)² ≥ 6

Dividing both sides by 2:

两边同时除以 2：

(x − 2)² ≥ 3

5. Solving the Reduced Inequality | 解化简后的不等式

The inequality (x − 2)² ≥ 3 involves a square. We must consider both the positive and negative square roots, because x − 2 could be either large enough or negative enough.

不等式 (x − 2)² ≥ 3 涉及平方项。我们必须同时考虑正负平方根，因为 x − 2 可能足够大，也可能足够负。

Taking square roots gives two cases:

开平方得到两种情况：

x − 2 ≤ −√3 or x − 2 ≥ √3

Adding 2 to both sides of each inequality:

每个不等式两边同时加上 2：

x ≤ 2 − √3 or x ≥ 2 + √3

6. Writing the Solution Set | 写出解集

The solution to the original inequality is therefore the union of two intervals:

因此，原不等式的解集是两个区间的并集：

x ∈ (−∞, 2 − √3] ∪ [2 + √3, ∞)

Using approximate values, √3 ≈ 1.732, so 2 − √3 ≈ 0.268 and 2 + √3 ≈ 3.732. Thus the solution is approximately x ≤ 0.268 or x ≥ 3.732.

取近似值 √3 ≈ 1.732，则 2 − √3 ≈ 0.268，2 + √3 ≈ 3.732。因此解约为 x ≤ 0.268 或 x ≥ 3.732。

These values can be confirmed graphically: the parabola y = 2x² − 8x + 5 is above the horizontal line y = 3 on the two outer intervals.

这些值可以通过图像验证：抛物线 y = 2x² − 8x + 5 在两条外侧区间上位于水平线 y = 3 之上。

7. Graphical Interpretation | 图像解读

The vertex (2, −3) is the lowest point of the parabola. The horizontal line y = 3 intersects the curve at two points, which are exactly the boundary values x = 2 − √3 and x = 2 + √3.

顶点 (2, −3) 是抛物线的最低点。水平线 y = 3 与曲线相交于两点，恰好是边界值 x = 2 − √3 和 x = 2 + √3。

On the interval between these two x-values, the curve lies below the line y = 3; outside this interval, the curve lies above the line. This matches the algebraic solution perfectly.

在这两个 x 值之间的区间上，曲线位于直线 y = 3 的下方；在这个区间之外，曲线位于直线上方。这与代数解完全吻合。

Interval	Test value x	y = 2x² − 8x + 5	≥ 3?
(−∞, 0.268)	0	5	Yes
(0.268, 3.732)	2	−3	No
(3.732, ∞)	5	15	Yes

8. Common Mistakes to Avoid | 常见错误提醒

Mistake 1: Forgetting to divide by the coefficient a when completing the square. Always factor it out first.

错误一： 配方时忘记除以系数 a。应当先把 a 提出来。

Mistake 2: When solving x² ≥ a, writing only x ≥ √a and missing the negative branch. Remember to include x ≤ −√a.

错误二： 解 x² ≥ a 时只写 x ≥ √a，遗漏了负分支。记得要包含 x ≤ −√a。

Mistake 3: Adding 2 incorrectly to 2 − √3. The left boundary is exactly 2 − √3, not √3 − 2.

错误三： 将 2 − √3 错误地写成 √3 − 2。左边界就是 2 − √3。

9. General Method Summary | 一般方法总结

For any quadratic function y = ax² + bx + c with a > 0, the minimum value is found by completing the square to obtain y = a(x − h)² + k, where h = −b/(2a) and k = c − b²/(4a).

对任意 a > 0 的二次函数 y = ax² + bx + c，通过配方化为 y = a(x − h)² + k 即可求出最小值，其中 h = −b/(2a)，k = c − b²/(4a)。

To solve a quadratic inequality, rewrite it so that one side is zero, factor or complete the square, find the critical points, and test intervals. For ≥, include the critical points in the solution set.

解二次不等式时，先将一边化为零，再因式分解或配方，找出关键点，并用区间测试法确定解集。对于 ≥ 型不等式，关键点应包含在解集中。

10. Extension and Practice | 拓展与练习

Try these related problems to solidify your understanding:

尝试以下相关练习以巩固理解：

Find the minimum value of y = 3x² + 12x − 7 by completing the square.
求出 y = 3x² + 12x − 7 通过配方法得到的最小值。

Solve 2x² − 8x + 5 ≤ 3, the complementary inequality to the example above.
解 2x² − 8x + 5 ≤ 3，即上例的互补不等式。

Determine the range of values of x for which 2x² − 8x + 5 > 0.
确定使 2x² − 8x + 5 > 0 的 x 的取值范围。

For the first problem, y = 3(x + 2)² − 19, so the minimum is −19 at x = −2. For the second, the solution is 2 − √3 ≤ x ≤ 2 + √3. The third problem gives x < 2 − √(3/2) or x > 2 + √(3/2).

第一题中，y = 3(x + 2)² − 19，因此最小值为 −19，在 x = −2 时取得。第二题的解为 2 − √3 ≤ x ≤ 2 + √3。第三题的解为 x < 2 − √(3/2) 或 x > 2 + √(3/2)。

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Solving Minimum Value and Quadratic Inequality: Example 3.5.1 | 求最小值与二次不等式：例 3.5.1