Solving Quadratic Inequalities | 二次不等式求解

📚 Solving Quadratic Inequalities | 二次不等式求解

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

二次不等式是包含二次表达式的不等式，它通常可以写成以下几种一般形式：

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

where a, b, and c are constants and a ≠ 0. The first step in solving a quadratic inequality is to rearrange it into one of these standard forms, with zero on one side.

其中 a、b、c 为常数且 a ≠ 0。求解二次不等式的第一步是将其整理成上述标准形式之一，即让一侧为零。

For example, the expression x² – 4x + 3 < 0 is a quadratic inequality in standard form. Here a = 1, b = -4, and c = 3.

例如，x² – 4x + 3 < 0 就是一个标准形式的二次不等式，这里 a = 1，b = -4，c = 3。

To solve this inequality, we need to find the range of x values for which the quadratic expression is negative. Understanding the graph of y = x² – 4x + 3 will help us greatly.

要解这个不等式，我们需要找出使二次表达式为负的 x 取值范围。理解 y = x² – 4x + 3 的图像对我们非常有帮助。

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1. The Graph of y = ax² + bx + c | 二次函数 y = ax² + bx + c 的图像

The graph of a quadratic function is a parabola. Its shape depends on the sign of a:

二次函数的图像是抛物线，其形状取决于 a 的符号：

• If a > 0, the parabola opens upward and has a minimum point.
• 如果 a > 0，抛物线开口向上，并具有最小值点。
• If a < 0, the parabola opens downward and has a maximum point.
• 如果 a < 0，抛物线开口向下，并具有最大值点。

The x-intercepts of the parabola are the real roots of the equation ax² + bx + c = 0. If the quadratic has two distinct real roots α and β (with α < β), the parabola will cross the x-axis at these two points.

抛物线与 x 轴的交点就是方程 ax² + bx + c = 0 的实根。如果二次方程有两个不同的实根 α 和 β（假设 α < β），那么抛物线就会在这两点与 x 轴相交。

For a > 0, the part of the parabola below the x-axis lies between α and β. At these points, the quadratic expression is negative, i.e., ax² + bx + c < 0. Outside the interval [α, β], the expression is positive.

当 a > 0 时，抛物线位于 x 轴下方的部分在 α 和 β 之间。在这些点上，二次表达式为负，即 ax² + bx + c < 0；在区间 [α, β] 之外，表达式为正。

For a < 0, the situation is reversed: the expression is positive between the roots and negative outside.

当 a < 0 时，情况正好相反：在两个根之间表达式为正，而在两根之外为负。

Thus, solving a quadratic inequality often reduces to finding the roots and then determining the sign of the quadratic on the intervals defined by those roots.

因此，解二次不等式通常归结为先求出根，然后判断由这些根划分的各个区间上二次表达式的符号。

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2. Solving by Factorization | 通过因式分解求解

If the quadratic expression can be factorized easily, we can use a number line or a sign table to solve the inequality.

如果二次表达式可以轻松分解因式，我们可以使用数轴或符号表来解不等式。

Let us solve x² – 4x + 3 < 0 step by step.

让我们逐步求解 x² – 4x + 3 < 0。

First, factorize the expression:

首先，对表达式进行因式分解：

x² – 4x + 3 = (x – 1)(x – 3)

The product (x – 1)(x – 3) is less than zero if the two factors have opposite signs. There are two cases:

两个因式 (x – 1)(x – 3) 的乘积小于零，说明两个因式的符号相反。因此有两种情况：

• Case 1: x – 1 > 0 and x – 3 < 0, which gives x > 1 and x < 3, so 1 < x < 3.
• 情况 1：x – 1 > 0 且 x – 3 < 0，即 x > 1 且 x < 3，所以 1 < x < 3。
• Case 2: x – 1 < 0 and x - 3 > 0, which gives x < 1 and x > 3, impossible since no number can be both less than 1 and greater than 3.
• 情况 2：x – 1 < 0 且 x - 3 > 0，即 x < 1 且 x > 3，这是不可能的，因为没有任何数既小于 1 又大于 3。

Therefore, the solution set is {x : 1 < x < 3}.

因此，解集为 {x : 1 < x < 3}。

Alternatively, we can use a sign table. Mark the roots 1 and 3 on a number line. Then test a point in each interval to determine the sign of the product.

或者，我们可以使用符号表。在数轴上标出根 1 和 3，然后检验每个区间内的一个点，确定乘积的符号。

Interval	x < 1	1 < x < 3	x > 3
Test x	0	2	4
Sign of (x – 1)	–	+	+
Sign of (x – 3)	–	–	+
Sign of product	+	–	+

The product is negative only when 1 < x < 3, so that is the solution.

乘积仅在 1 < x < 3 时为负，因此这就是解。

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3. Solving by Completing the Square | 通过配方法求解

When the quadratic does not factorize easily, we can use the method of completing the square to solve the inequality.

当二次式不易因式分解时，我们可以使用配方法来解不等式。

Consider the inequality x² + 6x + 2 ≥ 0.

考虑不等式 x² + 6x + 2 ≥ 0。

First, complete the square on the left side:

首先，对左边进行配方：

x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7

So the inequality becomes:

因此不等式变为：

(x + 3)² – 7 ≥ 0

Add 7 to both sides:

两边同时加 7：

(x + 3)² ≥ 7

Taking square roots gives two cases:

开平方后得到两种情况：

x + 3 ≥ √7 or x + 3 ≤ -√7

Thus:

因此：

x ≥ √7 – 3 or x ≤ -√7 – 3

So the solution is {x : x ≤ -3 – √7 or x ≥ -3 + √7}.

所以解集为 {x : x ≤ -3 – √7 或 x ≥ -3 + √7}。

In this case, because the coefficient of x² is positive, the parabola opens upward and the expression is non-negative outside the interval between the roots.

在这种情况下，由于 x² 的系数为正，抛物线开口向上，表达式在两根之间的区间之外非负。

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4. Solving Using the Quadratic Formula | 使用求根公式求解

For any quadratic equation ax² + bx + c = 0, the roots can be found using the quadratic formula:

对于任何二次方程 ax² + bx + c = 0，可以使用求根公式求出根：

x = (-b ± √(b² – 4ac)) / (2a)

Once the roots α and β are found, we can determine the sign of the quadratic on each interval by testing a point.

一旦求出根 α 和 β，我们就可以通过检验每个区间上的一个点来确定二次表达式的符号。

Let us solve 2x² + 3x – 2 > 0.

让我们求解 2x² + 3x – 2 > 0。

First, find the roots of 2x² + 3x – 2 = 0. Using the quadratic formula with a = 2, b = 3, c = -2:

首先求方程 2x² + 3x – 2 = 0 的根。使用求根公式，其中 a = 2，b = 3，c = -2：

x = (-3 ± √(3² – 4 × 2 × (-2))) / (2 × 2) = (-3 ± √(9 + 16)) / 4 = (-3 ± 5) / 4

So x = ( -3 + 5 ) / 4 = 0.5 or x = ( -3 – 5 ) /

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