202. $x^{2}+10 x=39$ (solve geometrically).
202. Omar Khayyam's solution: "The square and ten roots are equal to thirty-nine. Multiply half the roots by itself. Add this product to the number (i.e., to 39), subtract half the number of roots from the square root. The remainder will be equal to the root of the square:

$$
\left(x=\sqrt{\left(\frac{10}{2}\right)^{2}+39}-5\right)
$$

Geometric solution: Let the square be $A B C D\left(x^{2}\right)$, increased by ten roots ($10 x$), it equals thirty-nine. The ten roots will be represented by the rectangle $C D E F$ (Fig. 51). The line $D E=10$. Divide it at point $K$ in half. The product of $E A$ and $A D$, equal to the rectangle $A B F E$, added to the square $D K$, will be equal to the square $A K[(10+x) x+25=$ $\left.=(x+5)^{2}\right]$. But the square $D K(25)$ is known, as well as the rectangle $A B F E$, representing the given number (39). Therefore, the square $A K$ and the line $A K$ will be known $(x+5)$, and when we subtract $D K$ from $A K$, the remainder $A D$ will be known.

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-125.jpg?height=296&width=858&top_left_y=803&top_left_x=608)

Fig. 51.