Page:The Algebra of Mohammed Ben Musa (1831).djvu/78

 sixty-nine six-hundred-and-twenty-fifths. Subtract from this the twenty-three dirhems and the one twenty-fifth connected with the square. The remainder is one-hundred-and-thirty-two and four-hundred-and-forty six-hundred-and-twenty-fifths. Take the root of this it is eleven dirhems and thirteen twenty-fifths. Add this to the moiety of the roots, which was twelve dirhems and twelve twenty-fifths. The sum is twenty-four. It is the number which you sought. When you subtract its third and its fourth and four dirhems, and multiply the remainder by itself, the number is restored, with a surplus of twelve dirhems.

If the question be: “To find a square-root, which, when multiplied by two-thirds of itself, amounts to five;” then the computation is this: You multiply one thing by two-thirds of thing; the product is two-thirds of square, equal to five. Complete it by adding its moiety to it, and add to five likewise its moiety. Thus you have a square, equal to seven and a half. Take its root; it is the thing which you required, and which, when multiplied by two-thirds of itself, is equal to five.

If the instance be: “Two numbers, the difference