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thing; it is two squares and two things, equal to one dirhem and a half. Reduce them to one square: that is, take the moiety of all you have. You say, therefore, one square and one thing are equal to three-fourths of a dirhem. Reduce this, according to what I have taught you in the beginning of this work.
If the instance be: “A number,[1] you remove one-third of it, and one-fourth of it, and four dirhems: then you multiply the remainder by itself, and the number,[2] is restored, with a surplus of twelve dirhems:” then the computation is this: You take thing, and subtract from it one-third and one-fourth; there remain five- twelfths of thing. Subtract from this four dirhems: (43) the remainder is five-twelfths of thing less four dirhems. Multiply this by itself. Thus the five parts become five-and-twenty parts; and if you multiply twelve by itself, it is a hundred and forty-four. This makes, therefore, five and twenty hundred and forty-fourths of a square. Multiply then the four dirhems twice by the five-twelfths; this gives forty parts, every twelve of which make one root (forty-twelfths); finally, the four
![{\displaystyle {\begin{aligned}(x-{\tfrac {1}{3}}x-{\tfrac {1}{4}}x-4)^{2}=x+12\\({\tfrac {5}{12}}x-4)^{2}=x+12\\{\tfrac {25}{144}}x^{2}+16-3{\tfrac {1}{3}}x=x+12\\{\tfrac {25}{144}}x^{2}+4=4{\tfrac {1}{3}}x\\x^{2}+23{\tfrac {1}{25}}=24{\tfrac {24}{25}}x\\{\sqrt {\left[({\tfrac {24{\tfrac {24}{25}}}{2}})^{2}-23{\tfrac {1}{25}}\right]}}+{\tfrac {24{\tfrac {24}{25}}}{2}}-x\\11{\tfrac {13}{25}}+12{\tfrac {12}{25}}=24=x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4ccffe59a28a8b245dfa76bb7f06ac353fc2056)
