Page:The Algebra of Mohammed Ben Musa (1831).djvu/81

( 65 )

it is the root of seven and a half multiplied by two-thirds of the root of seven and a half. Multiply then two-thirds by two-thirds, it is four-ninths; and four-ninths multiplied by seven and a half is three and a third. The root of three and a third is two-thirds of the root of seven and a half. Multiply three and a third by seven and a half; the product is twenty-five, and its root is five.

If the instance be: “A square multiplied by three of its roots is equal to five times the original square;”^[1] then this is the same as if it had been said, a square, which when multiplied by its root, is equal to the first square and two-thirds of it. Then the root of the square is one and two-thirds, and the square is two dirhems and seven-ninths.

If the instance be: “Remove one-third from a square, then multiply the remainder by three roots of the first square, and the first square will be restored.”^[2] Computation: If you multiply the first square, before (47) removing two-thirds from it, by three roots of the same, then it is one square and a half; for according to the statement two-thirds of it multiplied by three

---

1. ${\begin{aligned}x^{2}\times 3x&=5x^{2}\\x^{2}\times x&=1{\tfrac {2}{3}}x^{2}\\x&=1{\tfrac {2}{3}}\\x&=2{\tfrac {7}{9}}\end{aligned}}$
2. ${\begin{aligned}(x^{2}-{\tfrac {1}{3}}x^{2})\times 3x=x^{2}&\;\therefore \;{\tfrac {2}{3}}x^{2}\times 3x=x^{2}\\x^{2}\times 3x&=1{\tfrac {1}{2}}x^{2}\\x={\tfrac {1}{2}}&\;\therefore \;x^{2}={\tfrac {1}{4}}\end{aligned}}$

k