先把函数展开成 $y=4x-x^3$，这是负三次函数形状。截距由 $x(4-x^2)=0$ 得到。

The intersections with the coordinate axes occur when $y=0$:

So

The sketch should be a negative cubic curve crossing the axes at

交点满足两个 $y$ 相等。由于 $C_2$ 中有 $\frac{A}{x}$，乘以 $x$ 去分母即可。

At intersections,

Multiply by $x$:

Rearrange:

This is the required result.

令 $u=x^2$，方程变成 $u^2-4u+A=0$。要有 $4$ 个不同的 $x$ 值，需要这个关于 $u$ 的二次方程有两个不同的正根。

Let

Then

becomes

For $4$ distinct real $x$ values, this equation must have two distinct positive $u$ roots.

First, the product of the two $u$ roots is $A$, so

Also, the roots must be distinct:

Therefore

函数图像与分析法。 我们将方程 $x^4 - 4x^2 + A = 0$ 变形为 $4x^2 - x^4 = A$。令左侧函数为 $f(x) = 4x^2 - x^4$（其中 $x \neq 0$，因为 $C_2$ 在 $x=0$ 处无定义）。 要使方程有 4 个不同的实根，水平直线 $y = A$ 必须与曲线 $y = f(x)$ 有 4 个不同的交点。 通过对 $x^2$ 进行配方，可以得到 $f(x) = 4 - (x^2 - 2)^2$。 由于 $(x^2 - 2)^2 \ge 0$，该函数的最大值为 $4$（在 $x^2 = 2$ 即 $x = \pm\sqrt{2}$ 处取得）。 又因为 $x \neq 0$，排除原点处的局部极小值 $f(0) = 0$。 因此，为了保证有 4 个交点，直线 $y = A$ 必须介于极小值 $0$ 和最大值 $4$ 之间，即 $0 < A < 4$。

We can rewrite the equation $x^4 - 4x^2 + A = 0$ as:

Let $f(x) = 4x^2 - x^4$ for $x \neq 0$. For the equation to have 4 distinct real roots, the line $y = A$ must intersect the curve $y = f(x)$ at 4 distinct points.

We can complete the square for $f(x)$:

Since $(x^2 - 2)^2 \ge 0$, the maximum value of $f(x)$ is $4$, which occurs when $x^2 = 2 \implies x = \pm\sqrt{2}$.

For $x \neq 0$, we have $f(x) > 0$. Thus, the graph of $f(x)$ has a local minimum (excluding $x=0$) at $f(0) = 0$ and two maximum points at $(\sqrt{2}, 4)$ and $(-\sqrt{2}, 4)$.

For the horizontal line $y = A$ to intersect $y = f(x)$ at 4 distinct points, $A$ must lie strictly between the local minimum and the maximum value: