Page:The Algebra of Mohammed Ben Musa (1831).djvu/54

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In the present question the quotient is four and the divisor is thing. Multiply, therefore, four by thing; the result is four things, which are equal to the sum to be divided, which was ten minus thing. You now reduce it by thing, which you add to the four things. Then we have five things equal to ten; therefore one thing is equal to two, and this is one of the two portions. This question refers you to one of the six cases, namely, “roots equal to numbers.”

Fourth Problem.

I have multiplied one-third of thing and one dirhem by one-fourth of thing and one dirhem, and the product was twenty.^[1]

Computation: You multiply one-third of thing by one-fourth of thing; it is one-half of a sixth of a square. Farther, you multiply one dirhem by one-third of thing, it is one-third of thing; and one dirhem by one-fourth of thing, it is one-fourth of thing; and one dirhem by one dirhem, it is one dirhem. The result of this is: the moiety of one-sixth of a square, and one-third of thing, and one-fourth of thing, and one dirhem, is equal to twenty dirhems. Subtract now the one dirhem from

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1. ${\begin{aligned}({\tfrac {1}{3}}x+1)({\tfrac {1}{4}}x+1)&=20\\{\tfrac {x^{2}}{12}}+{\tfrac {1}{3}}x+{\tfrac {1}{4}}x+1&=20\\{\tfrac {x^{2}}{12}}+{\tfrac {7}{12}}x&=19\\x^{2}+7x&=228\\x={\sqrt {{\tfrac {49}{4}}+228}}-{\tfrac {7}{2}}&=12\end{aligned}}$