📚 Solving Quadratic Equations and Linear-Quadratic Simultaneous Equations | 二次方程与一次-二次联立方程求解
e constants and a ≠ 0. This article revises the main solution methods and extends them to simultaneous equations with one linear and one quadratic equation.
二次方程贯穿 IGCSE 数学大纲，从纯代数运算到图像、建模和解题都会出现。二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。本文复习主要求解方法，并延伸到由一个一次方程和一个二次方程组成的联立方程。
1. Standard Form and Key Vocabulary | 标准形式与核心术语
ultiplying x, and c is the constant term. If a = 0, the equation is linear, not quadratic.
在选择求解方法之前，一定要把二次方程整理成 ax² + bx + c = 0 的形式。系数 a 是 x² 前面的数，b 是 x 前面的数，c 是常数项。如果 a = 0，这个方程就是一次方程，而不是二次方程。
these numbers correctly is essential for factorising, using the formula, or completing the square.
例如，在 3x² − 5x + 2 = 0 中，a = 3，b = −5，c = 2。正确识别这些数对于因式分解、使用求根公式或配方法都至关重要。
2. Solving by Factorising | 因式分解法求解
to zero. For example, x² − 7x + 12 = 0 becomes (x − 3)(x − 4) = 0, giving x = 3 or x = 4.
当二次表达式可以分解为两个括号时，让每个括号等于 0。例如 x² − 7x + 12 = 0 可写成 (x − 3)(x − 4) = 0，得到 x = 3 或 x = 4。
Remember that the product of two factors is zero only if at least one factor is zero. This is the zero-product property that makes factorising such a powerful method. Always check your factorisation by expanding the brackets.
记住，两个因式的乘积为 0，只有至少一个因式为 0 时才成立。这个零乘积性质让因式分解法非常有效。一定要通过展开括号来检查因式分解是否正确。
x² + 5x + 6 = (x + 2)(x + 3), so x = −2 or x = −3
x² + 5x + 6 = (x + 2)(x + 3)，所以 x = −2 或 x = −3
3. The Quadratic Formula | 求根公式
If a quadratic cannot be factorised easily, use the quadratic formula. It works for every quadratic equation, even when the roots are irrational or negative.
如果二次方程不容易因式分解，就使用求根公式。它适用于所有二次方程，即使根是无理数或负数也能求解。
x = (−b ± √b² − 4ac) / (2a)
Substitute a, b and c carefully, paying attention to negative signs. The expression b² − 4ac is called the discriminant; it determines whether the roots are real and distinct, repeated, or not real.
代入 a、b、c 时要仔细，特别注意负号。表达式 b² − 4ac 叫作判别式，它决定根是实数且不同、重根还是非实数。
² + 3x − 2 = 0, a = 2, b = 3, c = −2, so x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √25) / 4, giving x = 1/2 or x = −2.
对于 2x² + 3x − 2 = 0，a = 2，b = 3，c = −2，所以 x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √25) / 4，得到 x = 1/2 或 x = −2。
4. Completing the Square | 配方法
Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This is useful for finding the vertex of a parabola and for solving when factorising is not obvious.
配方法把 ax² + bx + c 改写成 a(x + p)² + q 的形式。这在求抛物线顶点以及无法直接因式分解时非常有用。
example, x² + 6x + 5 = 0 becomes (x + 3)² − 4 = 0, so x + 3 = ±2, giving x = −1 or x = −5.
例如，x² + 6x + 5 = 0 可化为 (x + 3)² − 4 = 0，所以 x + 3 = ±2，得到 x = −1 或 x = −5。
To complete the square for x² + bx, add and subtract (b/2)². This keeps the expression balanced while creating a perfect square. The vertex is then (−b/2, q).
对于 x² + bx，要添加并减去 (b/2)² 来配方。这样可以在保持表达式平衡的同时构造完全平方。顶点坐标为 (−b/2, q)。
5. Discriminant and Nature of Roots | 判别式与根的性质
epeated real root. If Δ < 0, there are no real roots; the graph does not cross the x-axis.
对于 ax² + bx + c = 0，判别式为 Δ = b² − 4ac。如果 Δ > 0，有两个不同的实根；如果 Δ = 0，有一个重根；如果 Δ < 0，没有实根，图像与 x 轴不相交。
- Δ > 0: two real roots, graph cuts the x-axis twice | Δ > 0：两个实根，图像与 x 轴相交两次
- Δ = 0: one repeated root, graph touches the x-axis | Δ = 0：一个重根，图像与 x 轴相切
- Δ < 0: no real roots, graph is entirely above or below the x-axis | Δ < 0：没有实根，图像完全在 x 轴上方或下方
IGCSE questions often ask you to find the value of k for which a quadratic has equal roots. Set the discriminant equal to zero and solve for k.
IGCSE 题目经常要求求出使二次方程有重根的 k 值。令判别式等于 0，然后解出 k。
6. Linear-Quadratic Simultaneous Equations | 一次-二次联立方程
qual to each other: 2x + 1 = x² + x − 1, then rearrange and solve the resulting quadratic.
IGCSE 常见题型是求解形如 y = 2x + 1 和 y = x² + x − 1 的方程组。因为两个式子都等于 y，就让它们相等：2x + 1 = x² + x − 1，然后整理并解所得的二次方程。
After finding x, substitute back into the linear equation to find the corresponding y-values. Always give solutions as coordinate pairs, not just x-values.
求出 x 后，代回一次方程求出对应的 y 值。答案要写成坐标对，不能只写 x 值。
Sometimes the quadratic equation gives no real roots. In that case, the line and parabola do not meet, and the simultaneous equations have no real solutions.
有时二次方程没有实根。在这种情况下，直线与抛物线不相交，联立方程没有实数解。
7. Worked Example: Linear-Quadratic System | 例题：一次-二次方程组
Solve y = x + 3 and y = x² + x − 2.
解方程组 y = x + 3 和 y = x² + x − 2。
ions equal: x + 3 = x² + x − 2, which simplifies to x² − 5 = 0. Therefore x² = 5, so x = ±√5.
让两个表达式相等：x + 3 = x² + x − 2，化简得 x² − 5 = 0。因此 x² = 5，所以 x = ±√5。
Substitute into y = x + 3: when x = √5, y = 3 + √5; when x = −√5, y = 3 − √5. The solutions are (√5, 3 + √5) and (−√5, 3 − √5).
代入 y = x + 3：当 x = √5 时，y = 3 + √5；当 x = −√5 时，y = 3 − √5。解为 (√5, 3 + √5) 和 (−√5, 3 − √5)。
Always verify each solution in both original equations. This catches sign errors and confirms that the coordinate pair lies on both graphs.
一定要把每个解代回原方程组中的两个方程进行验证。这样可以发现符号错误，并确认坐标对同时落在两个图像上。
8. Graphical Interpretation | 图像解释
tions of a linear-quadratic system are the intersection points of the line and the parabola.
方程 ax² + bx + c = 0 的解是函数 y = ax² + bx + c 的图像与 x 轴的交点。一次-二次方程组的解就是直线与抛物线的交点。
If the discriminant is negative, the parabola does not meet the x-axis. Similarly, a line may cut a parabola twice, touch it once, or miss it completely, depending on the discriminant of the resulting quadratic.
如果判别式为负，抛物线与 x 轴没有交点。类似地，直线可能与抛物线相交于两点、相切于一点，或者完全不相交，这取决于消元后所得二次方程的判别式。
Sketching the graphs can help you check whether your algebra produces a reasonable number of solutions. One intersection point often means a repeated root.
画出图像草图有助于检查代数求解得到的解的数量是否合理。一个交点通常意味着重根。
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