Solving Quadratic Inequalities: Methods and Interval Determination | 二次不等式：解法与区间判断

📚 Solving Quadratic Inequalities: Methods and Interval Determination | 二次不等式：解法与区间判断

key idea is not simply to solve an equation, but to determine all real values of x for which the inequality is true.

二次不等式是指含有二次表达式的不等式。在 A-level 数学中，你需要会求解形如 ax² + bx + c > 0、ax² + bx + c ≥ 0、ax² + bx + c < 0 或 ax² + bx + c ≤ 0 的不等式，其中 a ≠ 0。核心思想不是简单地解方程，而是确定使得不等式成立的所有实数 x 的集合。

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1. Standard Form and Rearranging | 标准形式与移项整理

A quadratic inequality must first be written in standard form, with all terms on one side and zero on the other side. For example, x² + 2x > 8 is not in standard form because the right-hand side is not zero.

二次不等式首先应写成标准形式，即把所有项移到一边，另一边为零。例如 x² + 2x > 8 就不在标准形式，因为右边不是零。

Subtract 8 from both sides to obtain x² + 2x - 8 > 0. Always keep the inequality sign pointing the same way unless you multiply or divide by a negative number.

两边同时减去 8，得到 x² + 2x - 8 > 0。除非乘以或除以负数，否则始终保持不等号方向不变。

x ≥ 0 becomes 2x(x - 4) ≥ 0. This simplification makes the critical values easier to find.

如果存在公因式，先把它提出来。例如 2x² - 8x ≥ 0 可化为 2x(x - 4) ≥ 0。这种化简会使临界值更容易求出。

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2. Critical Values by Factorisation | 因式分解求临界值

xpression. For x² + 2x - 8, we find (x + 4)(x - 2) because 4 × (-2) = -8 and 4 + (-2) = 2.

对二次表达式进行因式分解。对于 x² + 2x - 8，可得 (x + 4)(x - 2)，因为 4 × (-2) = -8，4 + (-2) = 2。

o zero: x = -4 and x = 2. These values split the real number line into separate intervals.

令每个因式为零，就得到临界值（表达式等于零的点）：x = -4 和 x = 2。这些值把实数轴分成若干独立的区间。

f a. Remember that the value of a only affects whether the parabola opens upwards or downwards.

对于一般因式分解形式 a(x - p)(x - q)，临界值始终是 x = p 和 x = q，与 a 的正负无关。请记住，a 的取值只影响抛物线开口向上还是向下。

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3. Sign Diagram and Sign Table | 符号图与符号表

Plot the critical values on a number line. Then test a point from each interval to determine whether each factor is positive or negative. Multiply the signs to get the sign of the quadratic.

在数轴上标出临界值，然后从每个区间取一个测试点，判断每个因式的正负。将符号相乘即可得到二次表达式的正负。

Interval	x + 4	x - 2	(x + 4)(x - 2)
x < -4	negative	negative	positive
-4 < x < 2	positive	negative	negative
x > 2	positive	positive	positive

In the table, the final column gives the sign of x² + 2x - 8 on each interval. For the inequality x² + 2x - 8 > 0, choose the intervals marked positive.

在表中，最后一列给出 x² + 2x - 8 在每个区间上的符号。对于不等式 x² + 2x - 8 > 0，选择标记为正的区间。

Therefore the solution is x < -4 or x > 2. Notice that the critical values themselves are not included because the inequality is strict.

因此解为 x < -4 或 x > 2。注意临界值本身不包含在解集中，因为这是严格不等式。

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4. Interval Notation and Set Notation | 区间记号与集合记号

The solution x < -4 or x > 2 is written in interval notation as (-∞, -4) ∪ (2, ∞). The symbol ∪ means the union of the two sets.

解 x < -4 或 x > 2 用区间记号写作 (-∞, -4) ∪ (2, ∞)。符号 ∪ 表示两个集合的并集。

brackets for weak inequalities ≤ or ≥. For example, x² + 2x - 8 ≥ 0 has solution (-∞, -4] ∪ [2, ∞).

严格不等式 < 或 > 使用圆括号，非严格不等式 ≤ 或 ≥ 使用方括号。例如 x² + 2x - 8 ≥ 0 的解为 (-∞, -4] ∪ [2, ∞)。

For the opposite inequality x² + 2x - 8 < 0, the solution is the interval between the critical values: -4 < x < 2, which is written as (-4, 2).

对于相反的不等式 x² + 2x - 8 < 0，解恰为两个临界值之间的区间：-4 < x < 2，写作 (-4, 2)。

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5. Graphical Interpretation | 图像解释

bola because the coefficient of x² is positive. It crosses the x-axis at x = -4 and x = 2.

y = x² + 2x - 8 的图像是开口向上的抛物线，因为 x² 的系数为正。它与 x 轴相交于 x = -4 和 x = 2。

The inequality > 0 asks where the graph lies above the x-axis. Looking at the graph, this occurs to the left of -4 and to the right of 2. The inequality < 0 asks where the graph lies below the x-axis, which occurs between -4 and 2.

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