How to Solve Quadratic Inequalities Step by Step | 一元二次不等式分步解法

📚 How to Solve Quadratic Inequalities Step by Step | 一元二次不等式分步解法

A quadratic inequality is an inequality that involves a quadratic expression. It can be written in one of the following forms:

一元二次不等式是包含二次表达式的不等式，通常可以写成以下四种形式之一：

   ax² + bx + c < 0    ax² + bx + c ≥ 0    ax² + bx + c ≤ 0

where a, b and c are constants and a ≠ 0. Unlike a linear inequality, the solution set is usually an interval or a union of intervals, not a single half-line.

其中 a、b、c 为常数，且 a ≠ 0。与一元一次不等式不同，一元二次不等式的解集通常是一个区间，或者是两个区间的并集，而不是一条简单的射线。

Example: Solve x² − 5x + 6 > 0.

例如：解不等式 x² − 5x + 6 > 0。

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1. Step 1: Rewrite in Standard Form | 第一步：化为标准形式

Make sure the inequality is written as ax² + bx + c compared with 0. If it is not, move all terms to one side.

首先确保不等式的一侧为 0，即写成 ax² + bx + c 与 0 比较的形式。如果不是，把所有项移到同一边。

Example: Solve 2x² ≤ 3x + 5.

例：解不等式 2x² ≤ 3x + 5。

Rewrite as:

改写为：

2x² − 3x − 5 ≤ 0

Now the quadratic expression is on the left and 0 is on the right.

现在二次表达式在左边，右边为 0。

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2. Step 2: Find the Roots | 第二步：求根

on ax² + bx + c = 0. You can factorise, use the quadratic formula, or complete the square.

解对应的二次方程 ax² + bx + c = 0。可以使用因式分解、求根公式或配方法。

For 2x² − 3x − 5 = 0:

对于 2x² − 3x − 5 = 0：

(2x − 5)(x + 1) = 0

x = 5/2   or   x = −1

These roots divide the number line into intervals. For this example, the intervals are:

这两个根把数轴分成几个区间。本例中区间为：

• x < −1
• −1 < x < 5/2
• x > 5/2

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3. Step 3: Test Each Interval | 第三步：在每个区间内取点测试

Pick a test point from each interval and substitute it into the original quadratic expression. Only the sign (positive or negative) matters.

在每个区间内选取一个测试点，代入原二次表达式，判断结果的正负。只需关注符号。

For 2x² − 3x − 5:

对于 2x² − 3x − 5：

• In x < −1, choose x = −2: 2(4) −3(−2) −5 = 8 + 6 − 5 = 9 > 0
• 在 x < −1 区间，取 x = −2：2(4) −3(−2) −5 = 9 > 0
• In −1 < x < 5/2, choose x = 0: 0 − 0 − 5 = −5 < 0
• 在 −1 < x < 5/2 区间，取 x = 0：0 − 0 − 5 = −5 < 0
• In x > 5/2, choose x = 3: 18 − 9 − 5 = 4 > 0
• 在 x > 5/2 区间，取 x = 3：18 − 9 − 5 = 4 > 0

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4. Step 4: Write the Solution Set | 第四步：写出解集

Use the inequality symbol to decide which intervals are included.

根据原不等式的不等号确定哪些区间属于解集。

5 ≤ 0. The expression is negative in the middle interval and zero at the roots. Therefore:

原不等式为 2x² − 3x − 5 ≤ 0。表达式在中间区间为负，在根处为 0。因此：

−1 ≤ x ≤ 5/2

If the inequality had been strict (<), the roots would be excluded:

如果原不等式是严格小于 (<)，则根处要排除：

−1 < x < 5/2

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5. Using a Sign Table | 使用符号表

A sign table is a compact way to determine the sign of the quadratic expression without testing every interval with arithmetic.

符号表是一种简洁的方法，不需要逐一代数运算即可判断二次表达式在各区间的符号。

For (2x − 5)(x + 1):

以 (2x − 5)(x + 1) 为例：

Interval	x < −1	−1 < x < 5/2	x > 5/2
(2x − 5)	−	−	+
(x + 1)	−	+	+
Product	+	−	+

The table confirms that the product is negative on the interval −1 < x < 5/2.

符号表确认了乘积在 −1 < x < 5/2 区间内为负。

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6. Special Case: No Real Roots | 特殊情况：无实数根

If the quadratic equation has no real roots (discriminant Δ < 0), the quadratic expression does not change sign for any real x.

如果二次方程没有实数根（判别式 Δ < 0），则二次表达式在整个实数范围内符号保持不变。

Example: Solve x² + 2x + 3 > 0.

例：解不等式 x² + 2x + 3 > 0。

The discriminant is:

判别式为：

Δ = 2² − 4(1)(3) = 4 − 12 = −8 < 0

Since the coefficient of x² is positive, the expression is always positive. Therefore the solution is all real numbers.

因为 x² 的系数为正，该表达式恒为正。因此解集为全体实数。

x ∈ ℝ

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7. Special Case: Perfect Square | 特殊情况：完全平方

When the quadratic is a perfect square, the discriminant is zero and the expression equals zero at exactly one point.

当二次式为完全平方时，判别式为零，表达式恰在一个点处等于零。

Example: Solve (x − 3)² > 0.

例：解不等式 (x − 3)² > 0。

tive, and it is zero only when x = 3. Hence the solution is all real numbers except x = 3.

平方总是非负的，且只有在 x = 3 时为零。因此解集为除 x = 3 外的全体实数。

x ≠ 3

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8. Common Mistakes to Avoid | 常见错误提醒

• Do not multiply or divide an inequality by a negative number without reversing the inequality sign.
• 不要忘记当乘以或除以负数时，不等号要反向。
• Do not forget that the roots are included when the inequality is ≥ or ≤.
• 当不等式为 ≥ 或 ≤ 时，不要忘记根处包含在解集中。
• Do not test only one point; the sign can change at each root.
• 不要只测试一个点；在每个根处符号都可能改变。
• Do not confuse the solution of the equation with the solution of the inequality.
• 不要把方程的解与不等式的解混为一谈。

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9. Worked Example (IGCSE style) | 典型考题（IGCSE 风格）

Question: Solve x² − x − 12 < 0.

题目：解不等式 x² − x − 12 < 0。

Step 1: The inequality is already in standard form.

第一步：该不等式已经是标准形式。

Step 2: Factorise:

第二步：因式分解：

x² − x − 12 = (x − 4)(x + 3)

Roots: x = 4 and x = −3.

根为 x = 4 和 x = −3。

Step 3: Test intervals:

第三步：测试区间：

• x < −3: choose x = −4 → (−8)(−1) = 8 > 0
• x < −3：取 x = −4 → (−8)(−1) = 8 > 0
• −3 < x < 4: choose x = 0 → (−4)(3) = −12 < 0
• −3 < x < 4：取 x = 0 → (−4)(3) = −12 < 0
• x > 4: choose x = 5 → (1)(8) = 8 > 0
• x > 4：取 x = 5 → (1)(8) = 8 > 0

Step 4: The original inequality is < 0, so we choose the interval where the product is negative:

第四步：原不等式为 < 0，因此选择乘积为负的区间：

−3 < x < 4

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10. Practice Questions | 练习题目

Try these on your own before checking the answers.

请先独立完成，再核对答案。

1. Solve x² + 3x − 10 > 0.
2. 解不等式 x² + 3x − 10 > 0。
3. Solve 2x² + x − 6 ≤ 0.
4. 解不等式 2x² + x − 6 ≤ 0。
5. Solve x² − 4x + 4 ≥ 0.
6. 解不等式 x² − 4x + 4 ≥ 0。
7. Solve −x² + 5x − 6 < 0.
8. 解不等式 −x² + 5x − 6 < 0。

x < −5 or x > 2   2. −2 ≤ x ≤ 3/2   3. all real numbers   4. x < 2 or x > 3

答案：1. x < −5 或 x > 2   2. −2 ≤ x ≤ 3/2   3. 全体实数   4. x < 2 或 x > 3

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11. Summary | 总结

To solve any quadratic inequality:

解一元二次不等式的步骤总结：

1. Rewrite in standard form with 0 on one side.
2. 将不等式化为一边为 0 的标准形式。
3. Find the roots of the corresponding quadratic equation.
4. 求出对应二次方程的根。
5. Use a sign table or test points to determine the sign on each interval.
6. 使用符号表或测试点确定每个区间内的符号。
7. t the intervals that satisfy the inequality, and remember to include the roots for ≥ or ≤.
8. 选择满足不等式的区间，注意 ≥ 或 ≤ 时包含根。

With practice, you will be able to solve these questions quickly and accurately in the IGCSE exam.

多加练习后，你一定能在 IGCSE 考试中快速准确地解答这类题目。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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