Do the following equations have a common root?

$$
\begin{aligned}
& 3 x^{4}+13 x^{3}+20 x^{2}+17 x+7=0 \\
& 3 x^{4}+x^{3}-8 x^{2}+11 x-7=0
\end{aligned}
$$

Solve both equations.
I. The question can be reduced to the question of the existence of a common root of two simpler equations. Indeed, if a number is a root of (1) and (2), then it also makes the left side of the equation

$$
12 x^{3}+28 x^{2}+6 x+14=0
$$

zero, which results from their subtraction, as well as the left side of the equation

$$
6 x^{4}+14 x^{3}+12 x^{2}+28 x=x\left(6 x^{3}+14 x^{2}+12 x+28\right)=0
$$

which results from their addition. Furthermore, since the latter has an obvious root $x=0$, which does not satisfy either (1) or (2), the common root, if it exists, is also a root of the equation

$$
6 x^{3}+14 x^{2}+12 x+28=0
$$

obtained by separating the 0 root from (4), and therefore a common root of (3) and (5).

Similarly, the common root of (1) and (2) must also satisfy the equation

$$
-18 x-42=0
$$

which we formed by subtracting (3) from twice (5). Its only root is $x=-7 / 3$. This satisfies both (1) and (2), so it is their common root, and they have no other common roots.

II. According to these, $x+7 / 3$ is a root factor of both original equations. To determine their other roots, let's divide the left sides of (1) and (2) by the common root factor, $3 x+7$. Using the known

$$
x^{3}+1=(x+1)\left(x^{2}-x+1\right) \quad \text { and } \quad x^{3}-1=(x-1)\left(x^{2}+x+1\right)
$$

identities, the quotient can be transformed as follows:

$$
\begin{array}{ll}
\text { in the case of (1): } & \left(x^{3}+1\right)+2 x(x+1)=(x+1)\left(x^{2}+x+1\right) \\
\text { in the case of (2): } & \left(x^{3}-1\right)-2 x(x-1)=(x-1)\left(x^{2}-x+1\right)
\end{array}
$$

From this, we see that (1) has an additional root -1, and (2) has an additional root +1, and they have no more real roots, because setting the quadratic factors to 0, the (common) discriminant of the equations is -3, which is negative.

László Takács (Sopron, Széchenyi I. g. II. o. t.)

Remarks. 1. We can also look for the common root by factoring the left side of (3) or (5):

$$
\begin{gathered}
6 x\left(2 x^{2}+1\right)+14\left(2 x^{2}+1\right)=2(3 x+7)\left(2 x^{2}+1\right), \text { or } \\
6 x\left(x^{2}+2\right)+14\left(x^{2}+2\right)=2(3 x+7)\left(x^{2}+2\right)
\end{gathered}
$$

The quadratic factors do not yield real roots, but even if we consider the complex roots derived from them, they are not common roots.

Ágnes Hernádi (Budapest, Berzsenyi D. g. II. o. t.) Imre Kloknicer (Budapest, Bláthy O. Erősáramú Ip. Techn. III. o. t.)

2. The above procedure is simpler if we first separate the easily recognizable non-common roots, -1, and +1, from (1) and (2).