Solve the following equation:

$$
4^{2 x+1}-\sqrt{29 \cdot 4^{2 x+1}+2^{2 x+1}+2 \cdot 8^{2 x}}=21 \cdot 2^{2 x}-4
$$
The equation can be rewritten as:

$$
4 \cdot 4^{2 x}-21 \cdot\left(2^{2}\right)^{x}+4=\sqrt{116 \cdot 4^{2 x}+2 \cdot\left(2^{2}\right)^{x}+2 \cdot 4^{3 x}}
$$

since $8=4^{3 / 2}$. Let $4^{x}=y$, then

$$
4 y^{2}-21 y+4=\sqrt{116 y^{2}+2 y+2 y^{3}}
$$

by squaring and reducing to zero:

$$
16 y^{4}-170 y^{3}+357 y^{2}-170 y+16=0
$$

Since $y=0$ is not a root of the equation, we can divide by $y^{2}$

$$
16\left(y^{2}+\frac{1}{y^{2}}\right)-170\left(y+\frac{1}{y}\right)+357=0
$$

and by introducing a new variable, we only need to solve a quadratic equation:

$$
\begin{gathered}
y+\frac{1}{y}=z, \quad y^{2}+\frac{1}{y^{2}}=z^{2}-2 \\
16 z^{2}-170 z+325=0, \quad z^{\prime}=65 / 8, \quad z^{\prime \prime}=5 / 2
\end{gathered}
$$

from each value, we get two $y$ values:

$$
y_{1}=8, \quad y_{2}=1 / 8, \quad \text { or } \quad \mathrm{y}_{3}=2, \quad \mathrm{y}_{4}=1 / 2
$$

The first two satisfy (1), but in the case of the latter two, the left side of (1) is negative, so they are not solutions to the original problem.

Finally, from $4^{x}=8$ we get $x_{1}=3 / 2$, and from $4^{x}=1 / 8$ we get $x_{2}=-3 / 2$.

Ferenc Halász (Budapest, Berzsenyi D. Gymnasium, III. grade)