Solving Quadratic Equations by Graphical Methods | 利用图像解一元二次方程

📚 Solving Quadratic Equations by Graphical Methods | 利用图像解一元二次方程

In your Edexcel IGCSE Mathematics course, you are expected to solve quadratic equations algebraically, but you must also know how to use graphs. A graph gives a visual picture of where the equation is true, and it is especially useful when exact answers are not easy to read or when a calculator is not allowed.

在 Edexcel IGCSE 数学课程中，你不仅要会用代数方法解一元二次方程，还要会用图像法。图像能够直观地展示方程何时成立，尤其在无法精确读数或不允许使用计算器时非常有用。

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1. The Shape of a Quadratic Graph | 二次函数图像的形状

A quadratic expression has the form ax² + bx + c, where a, b and c are constants and a ≠ 0. When y = ax² + bx + c is plotted, the graph is a curve called a parabola.

二次表达式具有 ax² + bx + c 的形式，其中 a、b、c 为常数且 a ≠ 0。将 y = ax² + bx + c 作图后，得到的曲线称为抛物线。

If a > 0, the parabola has a U shape and a lowest point called the minimum. If a < 0, the parabola has an n shape and a highest point called the maximum. The turning point lies on the axis of symmetry.

当 a > 0 时，抛物线呈 U 形，有最低点，称为最小值点；当 a < 0 时，抛物线呈 n 形，有最高点，称为最大值点。顶点位于对称轴上。

y = x² – 2x – 3

For this graph, the coefficient of x² is positive, so the curve opens upwards. Its axis of symmetry is x = 1, because the curve is symmetric about the vertical line x = 1.

对于这个图像，x² 的系数为正，所以曲线开口向上。它的对称轴是 x = 1，因为曲线关于直线 x = 1 对称。

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2. Roots as x-Intercepts | 根就是图像与 x 轴的交点

Solving a quadratic equation f(x) = 0 means finding all values of x for which f(x) = 0. On the graph of y = f(x), this condition is exactly the same as y = 0.

解一元二次方程 f(x) = 0，就是求出所有使 f(x) = 0 的 x 值。在 y = f(x) 的图像上，这个条件正好等价于 y = 0。

The x-intercepts are the points where the curve meets the x-axis, because at those points y = 0. Therefore, the real roots of a quadratic equation are the x-coordinates of the points where the curve crosses the x-axis.

图像与 x 轴的交点就是 x 轴截距，因为在这些点上 y = 0。因此，一元二次方程的实数根，就是抛物线穿过 x 轴时对应点的 x 坐标。

ax² + bx + c = 0

If the curve crosses the x-axis at x = p and x = q, then p and q are the solutions. If the curve does not touch the x-axis, the equation has no real solutions.

如果曲线在 x = p 和 x = q 处与 x 轴相交，那么 p 和 q 就是方程的解。如果曲线与 x 轴没有交点，则方程没有实数解。

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3. Making a Table of Values | 列表取值

To draw a quadratic graph, you should start with a table of values. Choose x-values that include the region around the turning point, so that the shape of the curve is clear.

要绘制二次函数图像，首先应列出数值表。选取的 x 值应包含顶点附近的区域，这样才能清楚地看出曲线的形状。

For the quadratic y = x² – 2x – 3, use integer x-values from x = -2 to x = 4.

例如，对于二次函数 y = x² – 2x – 3，可取 x = -2 到 x = 4 的整数值。

x	-2	-1	0	1	2	3	4
y	5	0	-3	-4	-3	0	5

For example, when x = 2, y = 2² – 2(2) – 3 = 4 – 4 – 3 = -3. You should always check at least one value by substitution.

例如，当 x = 2 时，y = 2² – 2(2) – 3 = 4 – 4 – 3 = -3。你应该至少用代入法检查一个数值。

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4. Drawing the Curve | 绘制曲线

Plot each point from the table on a coordinate grid. Use a clear scale on both axes, and label the axes with x and y.

在坐标网格上描出表格中的每一个点。两个坐标轴都要使用清晰的刻度，并标出 x 轴和 y 轴。

Join the points with a smooth curve. Do not join them with straight line segments. The curve should be continuous and symmetric about its axis of symmetry.

用平滑曲线连接各点。不要用直线段连接。曲线应是连续的，并且关于对称轴对称。

The graph of y = x² – 2x – 3 should pass through the points in the table. It has a minimum point at x = 1, and it crosses the x-axis at x = -1 and x = 3.

y = x² – 2x – 3 的图像应经过表格中的各点。它在 x = 1 处有最低点，并在 x = -1 和 x = 3 处穿过 x 轴。

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5. Reading the Solutions | 从图像读取解

Now look at the graph of y = x² – 2x – 3. The curve crosses the x-axis where y = 0.

现在观察 y = x² – 2x – 3 的图像。曲线在 y = 0 处与 x 轴相交。

x² – 2x – 3 = 0

From the graph, the x-intercepts are x = -1 and x = 3. Therefore the solutions are:

从图像可以看出，x 轴截距为 x = -1 和 x = 3。因此方程的解为：

x = -1 或 x = 3

Check: when x = -1, (-1)² – 2(-1) – 3 = 1 + 2 – 3 = 0. When x = 3, 3² – 2(3) – 3 = 9 – 6 – 3 = 0. Both values satisfy the equation.

检验：当 x = -1 时，(-1)² – 2(-1) – 3 = 1 + 2 – 3 = 0。当 x = 3 时，3² – 2(3) – 3 = 9 – 6 – 3 = 0。两个值都满足原方程。

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6. Solving x² – 2x – 3 = 1 Using the Same Graph | 用同一图像解 x² – 2x – 3 = 1

Suppose you need to solve x² – 2x – 3 = 1. You already have the graph of y = x² – 2x – 3.

假设你需要解 x² – 2x – 3 = 1。你已经有了 y = x² – 2x – 3 的图像。

Since the left side is exactly y, the equation is satisfied when y = 1. Draw the horizontal line y = 1 on the same axes.

由于等号左边正好是 y，所以当 y = 1 时等式成立。在同一坐标轴上画水平直线 y = 1。

The x-coordinates of the points where the line y = 1 meets the curve are the solutions. From a carefully drawn graph, you would read approximately:

直线 y = 1 与曲线交点的 x 坐标，就是方程的解。从绘制精确的图像上，可以近似读出：

x ≈ -1.24 或 x ≈ 3.24

Algebraically, x² – 2x – 3 = 1 gives x² – 2x – 4 = 0, so x = 1 ± √5. These exact values are approximately -1.24 and 3.24.

用代数方法计算，x² – 2x – 3 = 1 可化为 x² – 2x – 4 = 0，所以 x = 1 ± √5。这些精确值约为 -1.24 和 3.24。

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7. Solving by Drawing a Straight Line | 通过画直线解方程

Sometimes the equation to solve contains x on the right side. For example, use the graph of y = x² – 2x – 3 to solve:

有时需要解的方程右边含有 x。例如，利用 y = x² – 2x – 3 的图像解：

x² – 2x – 3 = 2x + 1

Notice that the left side is y. So the equation becomes y = 2x + 1. Draw the straight line y = 2x + 1 on the same axes.

注意，等号左边就是 y。因此方程变为 y = 2x + 1。在同一坐标轴上画直线 y = 2x + 1。

To draw the line, find two points: when x = -1, y = -1; when x = 4, y = 9. Plot these points and join them with a straight line.

画直线时，先求两个点：当 x = -1 时，y = -1；当 x = 4 时，y = 9。描出这两点并用直线连接。

The solutions are the x-coordinates of the points where this line intersects the parabola. From the graph, the readings are approximately:

方程的解就是直线与抛物线交点的 x 坐标。从图像上读取，近似为：

x ≈ -0.83 或 x ≈ 4.83

This method works because both equations are true at the same time only at the intersection points of the curve and the line.

这个方法成立的原因是，只有在曲线和直线的交点处，两个方程才能同时成立。

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8. Rearranging When the Equation Is Not in Standard Form | 当方程不是标准形式时如何变形

Sometimes you are given y = x² – 2x – 3, but the equation you must solve is 2x² – 4x – 5 = 0. You need to rearrange it so that the expression x² – 2x – 3 appears.

有时题目给出 y = x² – 2x – 3，但要求解的方程却是 2x² – 4x – 5 = 0。你需要把它变形，使 x² – 2x – 3 出现。

First divide every term by 2:

首先将每一项都除以 2：

x² – 2x – 2.5 = 0

Now rewrite the constant term so that x² – 2x – 3 appears:

现在改写常数项，使 x² – 2x – 3 出现：

(x² – 2x – 3) + 0.5 = 0

This means x² – 2x – 3 = -0.5. So draw the horizontal line y = -0.5 on the graph of y = x² – 2x – 3.

也就是说 x² – 2x – 3 = -0.5。所以在 y = x² – 2x – 3 的图像上画水平直线 y = -0.5。

The x-coordinates of the intersections are approximately x ≈ -0.87 and x ≈ 2.87. These are the solutions of the original equation 2x² – 4x – 5 = 0.

交点的 x 坐标约为 x ≈ -0.87 和 x ≈ 2.87。这就是原方程 2x² – 4x – 5 = 0 的解。

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9. No Real Roots and Repeated Roots | 无实数根与重根

Not every quadratic equation has real roots. If a quadratic graph lies completely above the x-axis, then y ≠ 0 for all x, so the equation has no real solutions.

并非所有一元二次方程都有实数根。如果二次函数图像完全位于 x 轴上方，那么对所有 x 都有 y ≠ 0，因此方程没有实数解。

For example, y = x² + 1 has its minimum at y = 1, so the graph never touches the x-axis. The equation x² + 1 = 0 has no real roots.

例如，y = x² + 1 的最低点为 y = 1，所以图像永远不会接触 x 轴。方程 x² + 1 = 0 没有实数根。

If the curve touches the x-axis at exactly one point, then the equation has one repeated root. For example, y = (x – 2)² has only one x-intercept at x = 2.

如果曲线与 x 轴恰好相切于一点，则方程有一个重根。例如，y = (x – 2)² 只有一个 x 轴截距 x = 2。

The discriminant b² – 4ac tells you the number of real roots: positive means two roots, zero means one repeated root, and negative means no real roots.

判别式 b² – 4ac 可以判断实数根的个数：大于零时有两个根，等于零时有一个重根，小于零时没有实数根。

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10. Limitations and Accuracy | 准确度与局限

The graphical method is visual and quick, but it gives approximate answers. The accuracy of your answers depends on the scale of the graph and on how carefully you read the intersections.

图像法直观且快速，但得到的是近似答案。答案的准确度取决于坐标轴的刻度以及你读取交点时的仔细程度。

If a root is not

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