Page:The Algebra of Mohammed Ben Musa (1831).djvu/83

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( 67 )

square, which, when added to the root of the same square, less one root, is equal to two dirhems. Subtract from this one root of the square, and subtract also from the two dirhems one root of the square. Then two dirhems less one root multiplied by itself is four dirhems and one square less four roots, and this is equal to a square less one root. Reduce it, and you find a square and four dirhems, equal to a square and three roots. Remove square by square; there remain three roots, equal to four dirhems; consequently, one root is equal to one dirhem and one-third. This is the root of the square, and the square is one dirhem and seven-ninths of a dirhem. (48)

Instance: “Subtract three roots from a square, then multiply the remainder by itself, and the square is restored.”[1] You know by this statement that the remainder must be a root likewise; and that the square consists of four such roots; consequently, it must be sixteen.