Page:The Algebra of Mohammed Ben Musa (1831).djvu/25

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The solution is the same when two squares or three, or more or less be specified;^[1] you reduce them to one single square, and in the same proportion you reduce also the roots and simple numbers which are connected therewith. (6)

For instance, “two squares and ten roots are equal to forty-eight dirhems;”^[2] that is to say, what must be the amount of two squares which, when summed up and added to ten times the root of one of them, make up sum of forty-eight dirhems? You must at first reduce the two squares to one; and you know that one square of the two is the moiety of both. Then reduce every thing mentioned in the statement to its half, and it will be the same as if the question had been, a square and five roots of the same are equal to twenty-four dirhems; or, what must be the amount of a square which, when added to five times its root, is equal to twenty-four dirhems? Now halve the number of the roots; the moiety is two and a half. Multiply that by itself; the product is six and a quarter. Add this to twenty-four; the sum is thirty dirhems and a quarter. Take the root of this; it is five and a half. Subtract from this the moiety of the number of the roots, that is two and a half; the

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1. $cx^{2}+bx=a$ is to be reduced to the form $x^{2}+{\tfrac {b}{c}}x={\tfrac {a}{c}}$
2. ${\begin{aligned}2x^{2}+10x=48\\x^{3}+5x=24\\x={\sqrt {\left({\tfrac {5}{2}}\right)^{2}+24}}-{\tfrac {5}{2}}\\={\sqrt {6{\tfrac {1}{4}}+24}}-2{\tfrac {1}{2}}\\=5{\tfrac {1}{2}}-2{\tfrac {1}{2}}=3\end{aligned}}$

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