Page:The Algebra of Mohammed Ben Musa (1831).djvu/74

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If the instance be: “To find a number,^[1] one-third of which, when multiplied by one-fourth of it, restores the ^[1]number,”^[2] then the computation is: You multiply one-third of thing by one-fourth of thing, this gives one-twelfth of a square, equal to thing, and the square is equal to twelve things, which is the root of one hundred and forty-four.

If the instance be: “A number,^[1] one-third of which and one dirhem multiplied by one-fourth of it and two dirhems restore the number,^[1] with a surplus of thirteen dirhems;”^[3] then the computation is this: You multiply one-third of thing by one-fourth of thing, this gives half one-sixth of a square; and you multiply two dirhems by one-third of thing, this gives two-thirds of a root; and one dirhem by one-fourth of thing gives one-fourth of a root; and one dirhem by two dirhems gives two dirhems. This altogether is one-twelfth of a square and two dirhems and (42) eleven-twelfths of a thing, equal to thing and thirteen dir-

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1. “Square” in the original.
2. ${\begin{aligned}{\tfrac {x}{3}}\times {\tfrac {x}{4}}&=x\\x^{2}&=12x\\x&=12\end{aligned}}$
3. ${\begin{aligned}&({\tfrac {x}{3}}+1)-({\tfrac {x}{4}}+2)=x+13\\&{\tfrac {x^{2}}{12}}+{\tfrac {11}{12}}x+2=x+13\\&{\tfrac {x^{2}}{12}}={\tfrac {x}{12}}+11\\&x^{2}=x+132\\&x={\tfrac {1}{2}}+{\tfrac {23}{2}}=12\end{aligned}}$