📚 A-Level Maths: Efficient Methods for Solving Quadratic Simultaneous Equations | A-Level 数学：二次联立方程的高效解法
Quadratic simultaneous equations appear frequently in A-Level Mathematics, especially in coordinate geometry, curve sketching and exam problems involving intersections. They consist of one linear equation and one quadratic equation, or occasionally two quadratic equations, and require a systematic method to solve efficiently.
二次联立方程在 A-Level 数学中非常常见，尤其是在坐标几何、曲线作图以及涉及交点问题的考试题中。它们通常由一个线性方程和一个二次方程组成，偶尔也是两个二次方程，需要系统化的方法才能高效求解。
1. Understanding Quadratic Simultaneous Equations | 理解二次联立方程
A linear equation has degree 1, such as x + y = 5, while a quadratic equation has degree 2, such as y = x² − 3x + 1. When they are solved together, we are finding the coordinates of the point or points where the straight line and the curve intersect.
线性方程的次数为 1，例如 x + y = 5；二次方程的次数为 2，例如 y = x² − 3x + 1。当它们联立求解时，我们实际上是在求直线与曲线的交点坐标。
For two quadratic equations, the intersection points usually correspond to two curves crossing each other. The general approach is always to reduce the system to a single equation in one variable.
对于两个二次方程，交点通常对应两条曲线的相交位置。通用的方法始终是将方程组化简为只含一个变量的一元方程。
2. The Substitution Method | 代入法
The most reliable method is substitution. Solve the linear equation for one variable, then substitute this expression into the quadratic equation. This creates a quadratic equation in one variable, which can be solved by factorising, completing the square, or using the quadratic formula.
最可靠的方法是代入法。先从线性方程中解出一个变量，然后将该表达式代入二次方程。这样就得到一个一元二次方程，可以通过因式分解、配方法或求根公式来求解。
Consider the system:
考虑以下方程组：
y = x² − 2x + 3, y = x + 1
Since both equations are written as y = …, we can directly set the right-hand sides equal to each other:
由于两个方程都已写成 y = … 的形式，我们可以直接将等号右边的表达式相等：
x² − 2x + 3 = x + 1
This gives x² − 3x + 2 = 0, which factorises as (x − 1)(x − 2) = 0. Hence x = 1 or x = 2. Substituting back into y = x + 1 gives y = 2 or y = 3.
由此得到 x² − 3x + 2 = 0，因式分解为 (x − 1)(x − 2) = 0。因此 x = 1 或 x = 2。代回 y = x + 1 可得 y = 2 或 y = 3。
3. Rearranging the Linear Equation | 整理线性方程的技巧
When the linear equation is not already solved, the key is to choose the variable that gives the simplest substitution. For example, with 2x + y = 7, solving for y is usually easier because y = 7 − 2x, and no fractions appear.
当线性方程尚未变形时，关键在于选择使代入最简的变量。例如对于 2x + y = 7，通常解出 y 更容易，因为 y = 7 − 2x，不会出现分数。
If the linear equation is 3x − 2y = 5, solving for y gives y = (3x − 5)/2, which introduces fractions. In such cases, solving for x might sometimes be just as complicated, so it is often better to multiply through by a common denominator later, or clear fractions immediately.
如果线性方程是 3x − 2y = 5，则解出 y 得到 y = (3x − 5)/2，这会引入分数。在这种情况下，解出 x 有时也一样复杂，因此通常最好在后续步骤中乘以公分母，或者立即清除分数。
A useful trick is to keep the substitution expression in bracket form, for example y = (3x − 5)/2, and then substitute it into the quadratic equation before simplifying. This reduces the chance of sign errors.
一个实用的技巧是让代入表达式保持括号形式，例如 y = (3x − 5)/2，然后先代入二次方程再化简。这样可以减少符号错误。
4. Solving the Quadratic in One Variable | 解一元二次方程
After substitution, the resulting quadratic will have the form ax² + bx + c = 0. You should first attempt to factorise it, because this is the quickest method. If factorisation is not obvious, use the quadratic formula:
代入后，所得的一元二次方程具有 ax² + bx + c = 0 的形式。你应首先尝试因式分解，因为这是最快的方法。如果因式分解不明显，则使用求根公式：
x = (−b ± √(b² − 4ac)) / (2a)
For example, solving x² − 5x + 6 = 0 gives (x − 2)(x − 3) = 0, so x = 2 or x = 3. Then substitute each x-value into the linear equation to find the corresponding y-values.
例如，解 x² − 5x + 6 = 0 得到 (x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。然后将每个 x 值代入线性方程，求出对应的 y 值。
Remember that a quadratic equation can have two distinct real roots, one repeated real root, or no real roots. This directly tells you how many intersection points exist.
请记住，二次方程可以有两个不同的实根、一个重根，或者没有实根。这直接说明交点有多少个。
5. Using the Discriminant to Predict the Number of Solutions | 用判别式预判解的数量
Before solving fully, the discriminant Δ = b² − 4ac of the resulting quadratic tells you the nature of the roots without computation of the roots themselves. If Δ > 0, the line and curve intersect at two points; if Δ = 0, they touch at exactly one point; if Δ < 0, they do not intersect.
在完全求解之前，所得二次方程的判别式 Δ = b² − 4ac 可以告诉你根的性质，而无需实际计算根。如果 Δ > 0，则直线与曲线有两个交点；如果 Δ = 0，则刚好相切于一点；如果 Δ < 0，则没有交点。
For example, if the equation formed is 2x² − 4x + 5 = 0, then Δ = (−4)² − 4 × 2 × 5 = 16 − 40 = −24. Since Δ < 0, there are no real intersection points.
例如，如果形成的方程是 2x² − 4x + 5 = 0，则 Δ = (−4)² − 4 × 2 × 5 = 16 − 40 = −24。因为 Δ < 0，所以没有实数交点。
This is especially efficient in exam questions that only ask for the number of intersection points or whether a line is tangent to a curve.
这在考试题中尤其高效，例如只问交点个数，或判断直线是否为曲线的切线。
6. The Elimination Method for Special Cases | 特殊情况下的消元法
When both equations are quadratic and both have y², x², and xy terms with similar coefficients, elimination may be faster than substitution. For instance:
当两个方程都是二次方程，并且都具有 y²、x² 和 xy 项，且系数相似时，消元法可能比代入法更快。例如：
x² + y² = 25, x² − y² = 7
Adding the two equations eliminates y²: 2x² = 32, so x² = 16, hence x = ±4. Then substituting into x² + y² = 25 gives y² = 9, so y = ±3. This is far more efficient than substitution.
将两个方程相加可消去 y²：2x² = 32，因此 x² = 16，所以 x = ±4。代入 x² + y² = 25 得 y² = 9，所以 y = ±3。这比代入法高效得多。
Be careful to pair the x and y values correctly. In such symmetric problems, each x can pair with each y, but in general you must check each pair against the original equations.
注意要正确配对 x 和 y 的值。在这样对称的问题中，每个 x 可以与每个 y 配对，但一般情况下，你必须用原方程检验每一对组合。
7. Avoiding Common Algebraic Errors | 避免常见代数错误
One frequent mistake is forgetting to multiply all terms when substituting. For example, substituting y = 2x + 1 into y = x² + 3x − 2 is simple, but if the quadratic has a coefficient such as 2y = x² + 3, you must substitute carefully into 2(2x + 1) = x² + 3.
一个常见错误是代入时忘记将所有项相乘。例如，将 y = 2x + 1 代入 y = x² + 3x − 2 很简单，但如果二次方程有系数，比如 2y = x² + 3，你必须小心地代入 2(2x + 1) = x² + 3。
Another common error is ignoring negative signs when squaring. If x = y − 3, then x² = (y − 3)² = y² − 6y + 9, not y² − 9. Always expand brackets correctly.
另一个常见错误是在平方时忽略负号。如果 x = y − 3，则 x² = (y − 3)² = y² − 6y + 9，而不是 y² − 9。务必正确展开括号。
Finally, after solving for x, some students forget to substitute back to find y. The solution coordinate pair is incomplete without both x and y values.
最后，解出 x 后，有些学生忘记代回求 y。没有 x 和 y 的完整坐标对，解就是不完整的。
8. Efficient Checking of Solutions | 快速检验解的方法
After obtaining your pairs, check them quickly by substitution into both original equations. This does not take long and catches sign or arithmetic errors. For example, if your solution is x = 2, y = 1, verify that both the linear and quadratic equations are satisfied.
得到坐标对后，应迅速将两者代入原方程组进行检验。这不需要很长时间，却能发现符号或算术错误。例如，如果你的解是 x = 2，y = 1，就验证线性方程和二次方程是否都满足。
A more time-efficient approach is to check only the linear equation if that was the one not used for substitution. Since the quadratic equation determined the x-values, and the linear equation gives y, the linear equation is the one most likely to reveal an error.
更省时的方法是，如果线性方程不是用于代入的那个方程，就只需要检验线性方程。因为二次方程决定了 x 值，而线性方程给出 y 值，所以线性方程最容易暴露错误。
In exams, never skip this final check unless you are absolutely confident. One missing negative sign can change all your answers.
在考试中，除非你非常有把握，否则不要跳过这最后一步检验。一个漏写的负号就可能改变所有答案。
9. Working with Non-Monic Quadratics | 处理非首一二次式
A non-monic quadratic has a leading coefficient not equal to 1, such as 2x² − x − 3 = 0. After substitution, you may need to multiply through by a common denominator or divide by a common factor before solving.
非首一二次式的首项系数不等于 1，例如 2x² − x − 3 = 0。代入后，你可能需要乘以公分母或先约去公因子，然后再求解。
For instance, if substituting y = (x − 1)/2 into y = x² − 1 leads to (x − 1)/2 = x² − 1. Multiply both sides by 2 to obtain x − 1 = 2x² − 2, then rearrange to 2x² − x − 1 = 0.
例如，将 y = (x − 1)/2 代入 y = x² − 1 得到 (x − 1)/2 = x² − 1。两边乘以 2，得 x − 1 = 2x² − 2，整理得 2x² − x − 1 = 0。
When the coefficient of x² is not 1, factorising can still work if the factors are found by trial. Alternatively, complete the square or use the quadratic formula directly.
当 x² 的系数不是 1 时，通过尝试找到因子仍然可以因式分解。或者，使用配方法或直接使用求根公式。
10. Handling Parameters | 含参数问题的处理
Some questions include a parameter, such as k, and ask for the range of k for which the line and curve intersect. Substitute the linear equation into the quadratic, then form a new quadratic in x. Its discriminant must be ≥ 0 for real intersections.
有些问题包含参数，例如 k，并询问直线与曲线相交时 k 的取值范围。将线性方程代入二次方程，然后构造一个关于 x 的新的二次方程。要使交点存在，其判别式必须满足 Δ ≥ 0。
For example, if y = 2x + k intersects y = x² − 3x, then substituting yields 2x + k = x² − 3x, so x² − 5x − k = 0. The discriminant is Δ = (−5)² − 4(1)(−k) = 25 + 4k. For two points of intersection we require 25 + 4k > 0, so k > −25/4.
例如，如果 y = 2x + k 与 y = x² − 3x 相交，代入得 2x + k = x² − 3x，即 x² − 5x − k = 0。其判别式为 Δ = (−5)² − 4(1)(−k) = 25 + 4k。若要有两个交点，需要 25 + 4k > 0，即 k > −25/4。
If the question asks for tangency, set Δ = 0 and solve for the parameter. This type of question rewards methodical working and clear presentation.
如果问题要求相切，则令 Δ = 0 并解出参数。这类题目考查有条理的解题步骤和清晰的表达。
11. Practice Questions with Worked Solutions | 练习与解答
Example 1: Solve the simultaneous equations:
示例 1：解下列联立方程：
y = x² + 2x − 1, y − 2x = 3
From y − 2x = 3, we get y = 2x + 3. Substituting gives 2x + 3 = x² + 2x − 1. Simplify to x² − 4 = 0, so x = ±2. Then for x = 2, y = 7; for x = −2, y = −1.
由 y − 2x = 3 得 y = 2x + 3。代入得 2x + 3 = x² + 2x − 1。化简为 x² − 4 = 0，所以 x = ±2。当 x = 2 时，y = 7；当 x = −2 时，y = −1。
Example 2: Find the set of values of k for which the curves y = x² and y = kx − 1 touch each other.
示例 2：求使得曲线 y = x² 与 y = kx − 1 相切的 k 的取值范围。
Substituting gives x² = kx − 1, so x² − kx + 1 = 0. For tangency, Δ = 0: (−k)² − 4 × 1 × 1 = k² − 4 = 0, hence k = ±2.
代入得 x² = kx − 1，即 x² − kx + 1 = 0。相切要求 Δ = 0：(−k)² − 4 × 1 × 1 = k² − 4 = 0，因此 k = ±2。
12. Summary | 总结
The most efficient method for quadratic simultaneous equations is substitution when one equation is linear, and elimination when both equations are quadratic with matching terms. Always rearrange carefully, solve the resulting quadratic accurately, and pair the x and y solutions correctly.
解二次联立方程时，若其中一个方程是线性的，最有效的方法是代入法；若两个都是二次方程且项的结构匹配，则消元法更高效。务必仔细整理方程，准确求解一元二次方程，并正确配对 x 和 y 的解。
Use the discriminant to predict the number of solutions, and always check your final answers. With deliberate practice, solving these problems will become quick and reliable.
使用判别式预判解的个数，并始终检验最终答案。通过有针对性的练习，求解这类问题会变得既快速又可靠。
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