Let's write down the equation of the line that is tangent to the curve given by the equation $y=3 x^{4}-4 x^{3}$ at two different points.

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Write down the equation of the line that is tangent to the curve given by the equation $y=3 x^{4}-4 x^{3}$ at two different points.
I. solution. The graph of the function $f(x)=3 x^{4}-4 x^{3}$ is shown in Figure 1, where we have drawn the sought line with a dashed line.

Notice that if we perform the substitution $x \leftarrow x+\frac{1}{3}$ on the polynomials in question - graphically, this means translating the curve along the $x$-axis by $-\frac{1}{3}$ units - the resulting fourth-degree polynomial $f_{1}(x)=f\left(x+\frac{1}{3}\right)$ does not contain a third-degree term:1 1 This so-called Tschirnhaus substitution is also used as the first step in solving the general fourth-degree equation to eliminate the third-degree term.

$$
f_{1}(x)=3\left(x+\frac{1}{3}\right)^{4}-4\left(x+\frac{1}{3}\right)^{3}=3 x^{4}-2 x^{2}-\frac{8}{9} x-\frac{1}{9}
$$

By completing the square for the even-degree terms in $f_{1}(x)$, we get

$$
f_{1}(x)=3\left(x^{2}-\frac{1}{3}\right)^{2}-\frac{8}{9} x-\frac{4}{9}
$$

The graph of $y=3\left(x^{2}-\frac{1}{3}\right)^{2}$ (Figure 2) clearly touches the $x$-axis at two points, so the graph of $f_{1}(x)$ has a double tangent at $y=-\frac{8}{9} x-\frac{4}{9}$. We can now find the equation of the sought line by applying the inverse substitution $x \leftarrow x-\frac{1}{3}$; its equation is:

$$
y=-\frac{8}{9}\left(x-\frac{1}{3}\right)-\frac{4}{9}=-\frac{8}{9} x-\frac{4}{27}
$$

II. solution. The line with equation $y=a x+b$ touches the graph of the function $y=3 x^{4}-4 x^{3}$ at exactly two points if and only if the graph of the polynomial $p(x)=3 x^{4}-4 x^{3}-a x-b$ touches the $x$-axis at two points, i.e., it has two multiple roots. Since the curve is fourth-degree, each multiple root is double, and there are no other roots, so if the roots are $x_{1}$ and $x_{2}$, then the factorized form of $p(x)$ is

$$
p(x)=3\left(x-x_{1}\right)^{2}\left(x-x_{2}\right)^{2}
$$

Performing the operations on the right-hand side:

$$
\begin{aligned}
3 x^{4}-4 x^{3}-a x-b= & \\
& =3 x^{4}-6\left(x_{1}+x_{2}\right) x^{3}+3\left(x_{1}^{2}+4 x_{1} x_{2}+x_{2}^{2}\right) x^{2}-6 x_{1} x_{2}\left(x_{1}+x_{2}\right) x+3 x_{1}^{2} x_{2}^{2}
\end{aligned}
$$

From the equality of the coefficients,

$$
x_{1}+x_{2}=\frac{2}{3},(1) x_{1}^{2}+4 x_{1} x_{2}+x_{2}^{2}=0,(2) 6 x_{1} x_{2}\left(x_{1}+x_{2}\right)=a,(3)-3 x_{1}^{2} x_{2}^{2}=b .(4)
$$

From (2), $\left(x_{1}+x_{2}\right)^{2}=-2 x_{1} x_{2}$, and using (1), $x_{1} x_{2}=-\frac{2}{9}$. From (3), $a=-\frac{8}{9}$, and from (4), $b=-\frac{4}{27}$.

Thus, the equation of the sought line can only be $y=-\frac{8}{9} x-\frac{4}{27}$, and this line is indeed a solution, for the system of equations (1) and (2) has real solutions $\left(x_{1}=\frac{1}{3}-\frac{\sqrt{3}}{3}\right.$ and $\left.x_{2}=\frac{1}{3}+\frac{\sqrt{3}}{3}\right)$, and thus the above steps are reversible, and the factors of the product $(*)$ are polynomials with real coefficients.

Béla András Rácz (Fazekas M. Főv. Gyak. Gimn., 9th grade)

III. solution. If $a$ is any real number, then the equation of the tangent line to the curve at the point $(a ; f(a))$ is

$$
y=f^{\prime}(a)(x-a)+f(a)
$$

If $b \neq a$, then the tangent at the point $(b ; f(b))$ is exactly the same line if and only if

$$
f^{\prime}(a)=f^{\prime}(b)(1) f(a)-a \cdot f^{\prime}(a)=f(b)-b \cdot f^{\prime}(b) \cdot(2)
$$

Dividing (1) by $a-b \neq 0$, we get

$$
a^{2}+a b+b^{2}-a-b=0
$$

Similarly, from (2),

$$
9(a+b)\left(a^{2}+b^{2}\right)-8\left(a^{2}+a b+b^{2}\right)=0
$$

From (3), $a^{2}+a b+b^{2}=a+b$; using this in (4),

$$
a^{2}+b^{2}=\frac{8}{9},(5) \text { and substituting in (3) } a+b-a b=\frac{8}{9} .(6)
$$

The system of equations (5) and (6) directly leads to a fourth-degree equation; since both equations are symmetric, it is useful to introduce the new unknowns $u=a+b, v=a b$: thus, we get the system

$$
u^{2}-2 v=\frac{8}{9}, u-v=\frac{8}{9}
$$

Substituting $v=u-\frac{8}{9}$, we get the equation $u^{2}-2 u+\frac{8}{9}=0$. Its roots are $u_{1}=\frac{2}{3}$, $u_{2}=\frac{4}{3}$. Accordingly, $v_{1}=-\frac{2}{9}$ and $v_{2}=\frac{4}{9}$.

The sought values of $a$ and $b$ are now the solutions of the equations

$$
t^{2}-u_{1} t+v_{1}=0, \quad \text { and } \quad t^{2}-u_{2} t+v_{2}=0
$$

In the first case, $a=\frac{1}{3}-\frac{\sqrt{3}}{3}$ and $b=\frac{1}{3}+\frac{\sqrt{3}}