## Task 10/62

It is

$$
\begin{array}{r}
-20=-20 \\
25-45=-20 \\
25-45=16-36 \\
5^{2}-2 \cdot 5 \cdot \frac{9}{2}=4^{2}-2 \cdot 4 \cdot \frac{9}{2}
\end{array}
$$

If we add $\left(\frac{9}{2}\right)^{2}$ to both sides of the equation, we get

$$
5^{2}-2 \cdot 5 \cdot \frac{9}{2}+\left(\frac{9}{2}\right)^{2}=4^{2}-2 \cdot 4 \cdot \frac{9}{2}+\left(\frac{9}{2}\right)^{2} \rightarrow\left(5-\frac{9}{2}\right)^{2}=\left(4-\frac{9}{2}\right)^{2}
$$

Taking the square root on both sides, we get

$$
5-\frac{9}{2}=4-\frac{9}{2} \quad \text { and by adding } \frac{9}{2} \text { to both sides, we get } \quad 5=4
$$

Where is the mistake?
The error does not lie in the typos smuggled in as an April Fool's joke, but in the square root extraction. The equation

$$
\left(5-\frac{9}{2}\right)^{2}=\left(4-\frac{9}{2}\right)^{2}
$$

is still correct, as $\left(5-\frac{9}{2}\right)^{2}=\frac{1}{4}$ and $\left(4-\frac{9}{2}\right)^{2}=\frac{1}{4}$. However, by extracting the square root, the positive value (on the left side of the equation) and the negative root value (on the right side) are set equal to each other.