Al-Jabr (continued)

In my previous post, I looked at the first two detailed examples provided by al-Khwarizmi in his compendium, the title of which gives us the word “algebra”. Al-Khwarizmi discussed three types of mathematical objects: Numbers (N, constants), roots (R, unknowns), and squares (S, squares of unknowns). Because he was limited to positive solutions, he was limited to six logical relationships (where $a$ and $b$ are quantities of numbers and roots): $aN = bR$, $aN = S$, $bR = S$, $S + bR = aN$, $S + aN = bR,$ and $bR + aN = S$. The first three he considered trivial enough to not warrant explicit diagrams; in my previous post, I gave illustrated examples for the fourth and fifth cases.

The most convoluted case, when solved geometrically, is the case where some quantity of roots and some quantity of numbers is equal to a root squared. In his example, “three roots and four of simple numbers are equal to a square”, that is, $3x + 4 = x^2$.

Al-Khwarizmi starts with a square representing the full value, shown here as $ADEF$. He divides this with $\overline{CG}$, so that $ACGF$ represents an area of $3x$ and $CDEG$ has an area of $4$. Since $AF = x$, it must be that $AC = 3$.

He locates $B$ at the midpoint of $\overline{AC}$, and creates the square $BCIH$ with a side length of $\frac{3}{2}$ and hence an area of $\frac{9}{4}$.

He locates $J$ such that $HJ = CD$, noting that $BD = BJ$ and $HJ = IK$. He further notes that $BDLJ$ is a square. Since $BC = CI$ and $CD = IK$, $AB = KG$. Since $GK = AB = BC$, the area of $HIKJ$ is equal to the area of $ELKG$.

Al-Khwarizmi returns now to the square comprised of the purple square, the green rectangle, and one of the blue rectangles. He notes that the green and blue, taken together, represents $4 = \frac{16}{4}$. The purple represents $\frac{9}{4}$, so the entire square $BDLJ$ has an area of $\frac{25}{4}$, and thus a side length of $BD = \frac{5}{2}$.

Meanwhile, since $AB = BC$, we know that $AB = \frac{3}{2}$, and so $AB + BD = AD = \frac{8}{2} = 4$, which is the positive solution of the equation.

Algebraically, this method says that, given $ax + b = x^2$, $x = \frac{a}{2} + \sqrt{\frac{a^2}{4} + b}$$= \frac{a + \sqrt{a^2 + 4b}}{2}$. Put another way, given $x^2 – bx – c = 0$, $x = \frac{-b + \sqrt{b^2 – 4c}}{2}$.  This is, of course, the larger root of the quadratic formula when $a = 1$, derived entirely through geometric reasoning.

This is where al-Khwarizmi concludes his introductory remarks on solving relationships between numbers, squares, and roots.