Quadratic Functions
and Equations二次函数与方程

Start from $f(x) = a x^{2} + b x + c$ and apply the algebraic identity for completing the square:从 $f(x) = a x^{2} + b x + c$ 出发，套用配方法的代数恒等式：

The first term $a (x + b/(2 a))^{2}$ is non-negative when $a > 0$ and equals zero only at $x = -b/(2 a)$. The remaining constant $c - b^{2}/(4 a)$ is the value of $f$ at that point. Symmetry of the squared term about $x = -b/(2 a)$ is what makes that vertical line the axis of symmetry of the whole parabola: $f(-b/(2 a) + h) = f(-b/(2 a) - h)$ for every $h$. The argument runs in reverse for $a < 0$ with "non-negative" replaced by "non-positive," and the same line is the axis of symmetry.第一项 $a (x + b/(2 a))^{2}$ 当 $a > 0$ 时非负，仅在 $x = -b/(2 a)$ 处为零；其余常数 $c - b^{2}/(4 a)$ 为该点的 $f$ 值。平方项关于 $x = -b/(2 a)$ 对称，使该竖直线成为整条抛物线的对称轴：对任意 $h$，$f(-b/(2 a) + h) = f(-b/(2 a) - h)$。$a < 0$ 时把"非负"换成"非正"，结论一致。

Expand $a (x - r_{1})(x - r_{2})$ step by step:逐步展开 $a (x - r_{1})(x - r_{2})$：

Compare with standard form $a x^{2} + b x + c$. Matching coefficients gives与一般式 $a x^{2} + b x + c$ 对照，比较系数得

These are Vieta's formulas for a quadratic. They are a fast sanity check when factoring: if you propose roots $r_{1}, r_{2}$, the sum should equal $-b/a$ and the product should equal $c/a$. If either fails, the proposed factoring is wrong.这就是二次的 Vieta 公式（韦达定理）。因式分解时是速效核验：若提出根 $r_{1}, r_{2}$，则根之和应等于 $-b/a$，根之积应等于 $c/a$。任一条不满足，因式分解就错。

Assume $a x^{2} + b x + c$ factors over the integers as $(p x + r)(q x + s)$ with $p q = a$, $r s = c$, and $p s + q r = b$. Multiply out:设 $a x^{2} + b x + c$ 在整数范围内可分解为 $(p x + r)(q x + s)$，满足 $p q = a$、$r s = c$、$p s + q r = b$。展开：

Set $u = p s$ and $v = q r$. Then $u + v = b$ and令 $u = p s$、$v = q r$。则 $u + v = b$ 且

So we just need to find two integers $u, v$ whose product is $a c$ and whose sum is $b$ , that is the AC method's search step. Once you have $u, v$, splitting $b x = u x + v x$ and grouping recovers the factorisation. If no such integer pair exists, the quadratic does not factor over the integers and you should switch to the quadratic formula or completing the square.所以只需找乘积为 $a c$、和为 $b$ 的两个整数 $u, v$ , 这正是 AC 法的搜索步骤。得到 $u, v$ 后，把 $b x = u x + v x$ 拆开再分组，即恢复因式分解。若无此整数对，则二次在整数范围内不可分解，应改用求根公式或配方法。

Start from $a x^{2} + b x + c = 0$ with $a \ne 0$. Divide by $a$:从 $a x^{2} + b x + c = 0$（$a \ne 0$）出发，两边除以 $a$：

Move the constant and complete the square (see §5):移项并配方（见 §5）：

Take square roots and isolate $x$:开平方并解出 $x$：

The radicand $b^{2} - 4 a c$ is the discriminant. If it is positive, $\sqrt{\Delta}$ is a positive real number and the $\pm$ gives two distinct roots. If it is zero, the $\pm$ collapses and we get one repeated root. If it is negative, $\sqrt{\Delta}$ is not real and the roots are complex conjugates.根号内 $b^{2} - 4 a c$ 即判别式。正时 $\sqrt{\Delta}$ 为正实数，$\pm$ 给出两个不同实根；为零时 $\pm$ 合并，得一重根；为负时 $\sqrt{\Delta}$ 非实，根为共轭复根。

Picture $x^{2} + b x$ as a literal square of side $x$ plus a rectangle of dimensions $b \times x$. Split the rectangle into two strips of dimensions $(b/2) \times x$ and attach one strip to each of two adjacent sides of the square. You now have an L-shape with a missing corner. The missing corner is a $(b/2) \times (b/2) = (b/2)^{2}$ square. Adding it fills out the L into a perfect $(x + b/2) \times (x + b/2)$ square , hence "completing the square."把 $x^{2} + b x$ 看成边长为 $x$ 的正方形加上 $b \times x$ 的长方形。将长方形分成两条 $(b/2) \times x$ 的条，分别贴在正方形相邻两边上，得到一个缺角的 L 形。缺的角是 $(b/2) \times (b/2) = (b/2)^{2}$ 的方块。补上它就把 L 形补成边长 $(x + b/2)$ 的完整正方形 , 这就是"配方法"得名的由来。

Algebraically, the same move says: $x^{2} + b x + (b/2)^{2} = (x + b/2)^{2}$. Subtracting $(b/2)^{2}$ to undo the addition gives $x^{2} + b x = (x + b/2)^{2} - (b/2)^{2}$, which is the substitution that turns standard form into vertex form.代数上同等于：$x^{2} + b x + (b/2)^{2} = (x + b/2)^{2}$。再减去 $(b/2)^{2}$ 抵消所加，得 $x^{2} + b x = (x + b/2)^{2} - (b/2)^{2}$ , 这就是把一般式变成顶点式的代换。

A streaming service models monthly demand as $x = 1000 - 50 p$ subscribers when the price is $p$ dollars per month. Find the price that maximises revenue, and the resulting revenue.某流媒体把月需求建模为 $x = 1000 - 50 p$（订阅人数），月费 $p$ 美元。求最大化营收的价格与对应营收。

Revenue is $R(p) = p \cdot x = p (1000 - 50 p) = 1000 p - 50 p^{2}$, which is a downward-opening parabola in $p$. The maximum is at the vertex:营收 $R(p) = p \cdot x = 1000 p - 50 p^{2}$，是关于 $p$ 的向下开口抛物线。最大值在顶点：

At $p^{*} = 10$, demand is $x^{*} = 1000 - 50 \cdot 10 = 500$ subscribers, and revenue is$p^{*} = 10$ 时需求 $x^{*} = 500$ 人，营收

The same calculation appears in AP Macroeconomics (monopoly revenue maximisation) and in IB Math HL E4 (problem-solving using calculus), where the calculus version uses $\frac{dR}{dp} = 0$ instead of the vertex formula. Both methods give the same answer; the calculus approach generalises beyond quadratics.同一计算在 AP 宏观经济（垄断营收最大化）与 IB Math HL E4（用微积分求解问题）中再现，那里用 $\frac{dR}{dp} = 0$ 取代顶点公式。结果一致；微积分方法可推广到非二次情形。

Solve $x^{2} + 2 x + 5 \le 0$.求解 $x^{2} + 2 x + 5 \le 0$。

Compute the discriminant: $\Delta = 4 - 20 = -16 < 0$. The parabola has no real roots and never crosses the $x$-axis. Since $a = 1 > 0$, the parabola opens upward and lies entirely above the $x$-axis , the expression $x^{2} + 2 x + 5$ is always strictly positive.算判别式：$\Delta = 4 - 20 = -16 < 0$。抛物线无实根，不与 $x$ 轴相交。$a = 1 > 0$，开口向上，整条曲线在 $x$ 轴上方 , 表达式 $x^{2} + 2 x + 5$ 恒为正。

The inequality $\le 0$ asks where the expression is non-positive. It never is. So the solution set is empty: no real $x$ satisfies it.不等式 $\le 0$ 问表达式何时非正，但它从不非正，故解集为空。

Compare with the opposite direction: $x^{2} + 2 x + 5 > 0$ is true for all real $x$. The contrast is a useful sanity check whenever $\Delta < 0$: depending on the direction of the inequality and the sign of $a$, the solution set is either everything or nothing.反向对照：$x^{2} + 2 x + 5 > 0$ 对所有实 $x$ 成立。每当 $\Delta < 0$，根据不等式方向与 $a$ 的符号，解集要么"全部实数"，要么"空集"。