Solving Quadratic Inequalities: Number Line and Intervals | 二次不等式的解法：数轴与区间

📚 Solving Quadratic Inequalities: Number Line and Intervals | 二次不等式的解法：数轴与区间

A quadratic inequality is an inequality that can be written in the form ax² + bx + c > 0, ax² + bx + c < 0, ax² + bx + c ≥ 0, or ax² + bx + c ≤ 0, where a ≠ 0. The goal is to find the set of real numbers x that make the inequality true.

二次不等式是可以写成 ax² + bx + c > 0、ax² + bx + c < 0、ax² + bx + c ≥ 0 或 ax² + bx + c ≤ 0 形式的不等式，其中 a ≠ 0。解二次不等式的目标是求出使不等式成立的全体实数 x。

1. What Is a Quadratic Inequality | 什么是二次不等式

A quadratic expression is a polynomial of degree 2. When we compare such a polynomial to zero using >, <, ≥, or ≤, we obtain a quadratic inequality.

二次表达式是次数为 2 的多项式。当我们将这样的多项式与 0 用 >、<、≥ 或 ≤ 进行比较时，就得到了二次不等式。

Examples include x² − 3x + 2 > 0 and 2x² + 5x − 3 ≤ 0. The solution set is usually one or two intervals on the real number line.

例如 x² − 3x + 2 > 0 和 2x² + 5x − 3 ≤ 0。解集通常是实数数轴上的一个或两个区间。

2. The Standard Form and Basic Facts | 标准形式与基本事实

Before solving, move all terms to one side so that the right side is 0. The standard form is ax² + bx + c > 0, with a ≠ 0.

在求解之前，先把所有项移到一边，使右边为 0。标准形式是 ax² + bx + c > 0，其中 a ≠ 0。

If the quadratic polynomial has two distinct real roots r₁ and r₂, then it can be factored as a(x − r₁)(x − r₂). The sign of this product changes at each root.

若二次多项式有两个不同的实根 r₁ 和 r₂，则可以因式分解为 a(x − r₁)(x − r₂)。这个乘积的符号在每个根处会发生变化。

The number line method uses these roots as boundary points. Between consecutive roots, the sign of the quadratic expression is constant.

数轴法就是以这些根为分界点。在相邻根之间，二次表达式的符号保持不变。

3. Step 1: Bring to Standard Form | 第一步：化为标准形式

Rewrite the inequality so that all terms are on the left and 0 is on the right. For example, x² + 1 < 2x becomes x² − 2x + 1 < 0.

将不等式改写为所有项在左边、右边为 0 的形式。例如 x² + 1 < 2x 可化为 x² − 2x + 1 < 0。

If the coefficient of x² is negative, multiply both sides by −1 and reverse the inequality sign. This makes the leading coefficient positive and simplifies the number-line analysis.

如果 x² 的系数为负，则将两边同乘 −1 并改变不等号方向。这样能使首项系数为正，从而简化数轴分析。

For instance, −x² + 4x − 3 > 0 is equivalent to x² − 4x + 3 < 0.

例如 −x² + 4x − 3 > 0 等价于 x² − 4x + 3 < 0。

4. Step 2: Factor or Find Roots | 第二步：因式分解或求根

Factor the quadratic if possible. If x² − 5x + 6 > 0, factor as (x − 2)(x − 3) > 0. The roots are x = 2 and x = 3.

如果可能，先对二次式进行因式分解。若 x² − 5x + 6 > 0，可分解为 (x − 2)(x − 3) > 0，根为 x = 2 和 x = 3。

When factoring is not simple, use the quadratic formula:

当因式分解不容易时，使用求根公式：

x = (−b ± √(b² − 4ac)) / (2a)

The discriminant Δ = b² − 4ac determines the number of real roots. If Δ > 0, there are two distinct roots. If Δ = 0, there is one double root. If Δ < 0, there are no real roots.

判别式 Δ = b² − 4ac 决定实根的个数。若 Δ > 0，有两个不同实根；若 Δ = 0，有一个重根；若 Δ < 0，没有实根。

5. Step 3: Number Line and Marking Roots | 第三步：数轴与标根

Draw a horizontal number line and mark the roots. Use an open circle for strict inequalities > or <, and a closed circle for non-strict inequalities ≥ or ≤.

画一条水平数轴并标出根。对于严格不等式 > 或 < 用空心圈，对于非严格不等式 ≥ 或 ≤ 用实心圈。

These roots divide the number line into three regions. Choose one test value from each region and substitute it into the quadratic expression.

这些根把数轴分成三个区域。从每个区域中选取一个检验值，代入二次表达式中。

For (x − 2)(x − 3) > 0, pick 0, 2.5, and 4. The results are positive, negative, and positive respectively.

对于 (x − 2)(x − 3) > 0，选取 0、2.5 和 4，结果分别为正、负、正。

6. Step 4: Test Intervals | 第四步：检验区间

A table helps organize the sign in each interval:

用表格可以清晰地整理每个区间内的符号：

Interval	Test value	Sign of (x − 2)(x − 3)
(−∞, 2)	0	(+)(−) = −
(2, 3)	2.5	(+)(+) = +
(3, ∞)	4	(+)(+) = +

Wait, the table above is not correct for (2,3): when x = 2.5, (x − 2) = 0.5 positive and (x − 3) = −0.5 negative, so the product is negative. The correct signs are −, +, − after correcting.

注意，上表在 (2,3) 中并不正确：当 x = 2.5 时，(x − 2) = 0.5 为正，(x − 3) = −0.5 为负，因此乘积为负。修正后的符号依次为 −、+、−。

Let us correct the table carefully:

让我们仔细修正表格：

Interval \| 区间	Test value \| 检验值	Sign \| 符号
(−∞, 2)	0	(+)(−) = −
(2, 3)	2.5	(+)(−) = −
(3, ∞)	4	(+)(+) = +

Therefore (x − 2)(x − 3) > 0 is true in (−∞, 2) and (3, ∞).

因此 (x − 2)(x − 3) > 0 在 (−∞, 2) 和 (3, ∞) 上成立。

7. The “Outside-Inside” Rule | “外小内大”规则

For a quadratic with a > 0 and two distinct roots r₁ < r₂, the expression is positive outside the roots and negative inside the roots.

对于 a > 0 且有两个不同实根 r₁ < r₂ 的二次函数，表达式在两根之外为正，在两根之内为负。

So for ax² + bx + c > 0 with a > 0, the solution is x < r₁ or x > r₂.

因此对于 a > 0 的 ax² + bx + c > 0，解为 x < r₁ 或 x > r₂。

For ax² + bx + c < 0 with a > 0, the solution is r₁ < x < r₂.

对于 a > 0 的 ax² + bx + c < 0，解为 r₁ < x < r₂。

If a < 0, the signs are reversed. It is usually safer to multiply by −1 first so that a > 0, then apply the rule.

若 a < 0，则符号相反。通常更安全的做法是先将不等式两边乘 −1 使 a > 0，再应用该规则。

8. Special Cases: Double Root | 特殊情形：重根

If the discriminant is zero, the quadratic has one double root, say x = r. The expression is zero at r and takes one sign everywhere else.

若判别式为零，二次式有一个重根，记为 x = r。该表达式在 r 处为零，在其余各处符号相同。

For example, x² − 6x + 9 = (x − 3)². This is always non-negative, and zero only at x = 3.

例如 x² − 6x + 9 = (x − 3)²。它总是非负的，仅在 x = 3 处为零。

Thus x² − 6x + 9 > 0 has solution x ≠ 3, which is (−∞, 3) ∪ (3, ∞). The inequality x² − 6x + 9 ≥ 0 has solution all real numbers.

因此 x² − 6x + 9 > 0 的解为 x ≠ 3，即 (−∞, 3) ∪ (3, ∞)。不等式 x² − 6x + 9 ≥ 0 的解为全体实数。

For x² − 6x + 9 < 0, there is no solution, because a square is never negative.

对于 x² − 6x + 9 < 0，由于完全平方不可能为负，所以无解。

9. Special Cases: No Real Roots | 特殊情形：无实根

If the discriminant is negative, the quadratic has no real roots. Its sign is the same as the sign of a for every x.

若判别式为负，则二次式没有实根。它对任意 x 的符号都与 a 的符号相同。

For example, x² + x + 1 has Δ = 1 − 4 = −3 < 0 and a = 1 > 0, so x² + x + 1 > 0 for all real x.

例如 x² + x + 1 的 Δ = 1 − 4 = −3 < 0，且 a = 1 > 0，所以 x² + x + 1 > 0 对所有实数 x 成立。

Therefore the solution of x² + x + 1 > 0 is (−∞, ∞). Meanwhile x² + x + 1 ≤ 0 has no real solution.

因此 x² + x + 1 > 0 的解为 (−∞, ∞)。而 x² + x + 1 ≤ 0 无实数解。

A table summarizes the three cases for a > 0:

下表总结了 a > 0 时的三种情形：

Discriminant \| 判别式	Roots \| 根	ax² + bx + c > 0	ax² + bx + c < 0
Δ > 0	r₁ < r₂	x < r₁ or x > r₂	r₁ < x < r₂
Δ = 0	r	x ≠ r	no solution
Δ < 0	none	all real x	no solution

If a < 0, the inequality signs in the second and third columns are swapped.

若 a < 0，则第二列和第三列的不等号要互换。

10. Interval Notation | 区间表示法

Interval notation is a compact way to write solution sets. A round bracket ( ) excludes an endpoint, and a square bracket [ ] includes it.

区间表示法是书写解集的紧凑方式。圆括号 ( ) 表示不包含端点，方括号 [ ] 表示包含端点。

Inequality \| 不等式	Interval \| 区间
x > 3	(3, ∞)
x ≥ 3	[3, ∞)
x < −2	(−∞, −2)
−1 ≤ x ≤ 4	[−1, 4]
x < 1 or x > 5	(−∞, 1) ∪ (5, ∞)

The symbol ∪ means “union” and combines separate intervals. Always use ∞ or −∞ with round brackets because infinity is not a number.

符号 ∪ 表示“并集”，用于合并不相连的区间。∞ 或 −∞ 前面永远用圆括号，因为无穷大不是确定的数。

11. Worked Example | 综合例题

Solve 2x² − 5x − 3 ≤ 0 step by step.

逐步求解 2x² − 5x − 3 ≤ 0。

Factor the quadratic: 2x² − 5x − 3 = (2x + 1)(x − 3).

因式分解：2x² − 5x − 3 = (2x + 1)(x − 3)。

Set each factor to zero: 2x + 1 = 0 gives x = −1/2; x − 3 = 0 gives x = 3.

令每个因式为 0：2x + 1 = 0 得 x = −1/2；x − 3 = 0 得 x = 3。

The roots divide the number line into three intervals:

这两个根将数轴分成三个区间：

For x < −1/2, choose x = −1: (2(−1)+1)(−1−3) = (−1)(−4) = 4 > 0.

当 x < −1/2 时，取 x = −1：(2(−1)+1)(−1−3) = (−1)(−4) = 4 > 0。
For −1/2 < x < 3, choose x = 0: (0+1)(0−3) = −3 < 0.

当 −1/2 < x < 3 时，取 x = 0：(0+1)(0−3) = −3 < 0。
For x > 3, choose x = 4: (8+1)(4−3) = 9 > 0.

当 x > 3 时，取 x = 4：(8+1)(4−3) = 9 > 0。

Since the inequality is ≤ 0, we include the endpoints. The solution is −1/2 ≤ x ≤ 3, or in interval notation [−1/2, 3].

因为不等式是 ≤ 0，所以包含端点。解为 −1/2 ≤ x ≤ 3，即区间 [−1/2, 3]。

12. Common Mistakes and Tips | 常见错误与提示

One common mistake is forgetting to reverse the inequality sign when multiplying or dividing by a negative number. Always check your work.

一个常见错误是当乘以或除以负数时忘记改变不等号方向。一定要检查每一步。

Another mistake is applying the “outside-inside” rule when a < 0 without adjusting. Convert to a > 0 first to avoid sign errors.

另一个错误是在 a < 0 时未经调整就套用“外小内大”规则。先化为 a > 0 可以避免符号错误。

When an inequality includes equality, such as ≥ or ≤, the roots must be included in the solution set. Use closed circles on the number line and square brackets in interval notation.

当不等式包含等号，如 ≥ 或 ≤ 时，根必须包含在解集中。数轴上用实心圈，区间表示法中用方括号。

Finally, always test one value in each interval after writing your answer. This quick verification catches most mistakes.

最后，写出答案后一定要在每个区间中选一个值进行检验。这种快速验证能发现大多数错误。

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Solving Quadratic Inequalities: Number Line and Intervals | 二次不等式的解法：数轴与区间