### Solving the simplified equations

Just as in modern algebra, medieval equations were simplified to standard forms. Where we have one type of simplified quadratic equation, \(ax^2+bx+c=0\), Arabic algebraists classified six types. This is because the rules for solving them call for operations on their "numbers" (coefficients), and only positive numbers were acknowledged in premodern mathematics. So there were three simple, two-term equations which we write as

(1) \(ax^2=bx\),

(2) \(ax^2=c\), and

(3) \(bx=c\),

and three composite, three-term equations corresponding to our

(4) \(ax^2+bx=c\),

(5) \(ax^2+c=bx\), and

(6) \(bx+c=ax^2\)

Abū Kāmil simplifies his equation translated above as

four ninths of a *māl* and four dirhams less two things and a third of a thing equal a thing and twenty-four dirhams [Abū Kāmil 2012, 379.14]

to

twenty dirhams and three things and a third of a thing equal four ninths of a *māl*,

which we would write as \[20+3{1\over 3}x={4\over 9}x^2.\] This is a type 6 equation.

Each of the six types was solved by its own rule. Solutions to the simple equations are trivial, requiring at most a single division to find the root or the māl. The rules for the composite equations, however, are more complex. Here are Ibn al-Bannāʾ's solutions to the three composite types from his *Condensed Book*. First he gives the solutions presuming that the number (coefficient) of the *māl* is 1, and after that he explains that this number must be set to one before following the procedure:

To work out the fourth type you halve the number of roots and you square the half, and you add it to the number and you take a root of the result and you subtract the half from it, leaving the root.

The sixth type is solved similarly, except that you add the half at the end to a root of the sum to get the root.

For the fifth type, you subtract the number from a square of half of the number of roots, and you take a root of the remainder. If you add it to the half, it gives a root of the greater [*māl*], and if you subtract it, it gives a root of the smaller *māl*. Whenever a square of the half is equal to the number, then the half is the root, and the *māl* is the number.

For each of the three composite types, whenever there is more than one *māl,* reduce it to one *māl,* and likewise reduce all the terms in the equation. And for each of them, whenever there is less than one *māl,* restore it to one *māl,* and likewise restore all the terms in the equation. [Ibn al-Bannāʾ 1969, 75.1].

If these rhetorical instructions are not clear enough, we can contort them into modern notation. There is always one positive solution to the type 4 equation \(x^2+bx=c\), which is \[x=\sqrt{c+\left({b\over 2}\right)^2}-{b\over 2}.\] The one positive solution to the type 6 equation \(x^2=bx+c\) is \[x=\sqrt{c+\left({b\over 2}\right)^2}+{b\over 2}.\] The type 5 equation \(x^2+c=bx\) might have two, one, or no positive solutions. If \(\left({b\over 2}\right)^2>c\) then the two solutions are \[x={b\over 2}\pm\sqrt{\left({b\over 2}\right)^2-c}.\] If \(\left({b\over 2}\right)^2=c\) then there is only one solution, \(x={b\over 2}\), and \(x^2=c\). Ibn al-Bannāʾ remarks in his algebra book that if the discriminant is negative then there is no solution, but he says nothing about that here. And last, if the number of *māl*s is not one, set it to one. For the type 4 equation \(ax^2+bx=c\), for example, one should work with the normalized version \[x^2+{b\over a}x={c\over a}.\]

Returning to Abū Kāmil's equation \[20+3{1\over 3}x={4\over 9}x^2,\] the \({4\over 9}x^2\) must first be restored to a full \(x^2\). He multiplies all terms by \(2{1\over 4}\) to get

forty-five dirhams and eight things and a fourth of a thing equal a *māl'* \(\left(45+8{1\over 4}x=x^2\right).\)

If he had bothered to spell out the subsequent steps, he would have shown us that we should take half of the \(8{1\over 4}\) to get \(4{1\over 8}\). Its square is \(17{1\over 64}\), and adding this to 45 gives \(62{1\over 64}\). Its square root is \(7{7\over 8}\), to which we add the \(4{1\over 8}\) to get 12, which is the value of the thing/root. Instead, he simply writes

You work it out the way I described to you to get twelve, which is the [unknown] quantity. [Abū Kāmil 2012, 379.20].

Because it is not apparent that these rules actually give the required values, many authors provided proofs that they work. The proofs in earlier books are based in geometry, while it is common for later proofs to be based in arithmetic.

**Figure 2**. Folio 51b from the Library of Congress manuscript of *Lifting the Veil*. The proofs begin on the 14th line, last word on the left. For an image of the title page of this work, see the *Convergence* article "Mathematical Treasure: Ibn al-Banna's Operations of Calculation".