Page:The Algebra of Mohammed Ben Musa (1831).djvu/71

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restored; its root must consequently be three, and the square itself nine.

If the instance be: “To find a square, four roots of which, multiplied by three roots, restore the square with a surplus of forty-four dirhems,”[1] then the solution is: that you multiply four roots by three roots, which gives twelve squares, equal to a square and forty-four dirhems. Remove now one square of the twelve on account of the one square connected with the forty-four dirhems. There remain eleven squares, equal to forty-four dirhems. Make the division, the result will be four, and this is the square.

If the instance be: “A square, four of the roots of which multiplied by five of its roots produce twice the square, with a surplus of thirty-six dirhems;”[2] then the solution is that you multiply four roots by five roots, which gives twenty squares, equal to two squares and thirty-six dirhems. Remove two squares from the twenty on account of the other two. The remainder is eighteen squares, equal to thirty-six dirhems. Divide now thirty-six dirhems by eighteen; the quotient is two, and this is the square.