Page:The Algebra of Mohammed Ben Musa (1831).djvu/28

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the number of the roots must be halved. And know, that, when in a question belonging to this case you have halved the number of the roots and multiplied the moiety by itself, if the product be less than the number of dirhems connected with the square, then the (8) instance is impossible;^[1] but if the product be equal to the dirhems by themselves, then the root of the square is equal to the moiety of the roots alone, without either addition or subtraction.

In every instance where you have two, squares, or more or less, reduce them to one entire square, ^[2] as I have explained under the first case.

Roots and Numbers are equal to Squares;^[3] for instance, “three roots and four of simple numbers are equal to a square.” Solution: Halve the roots; the moiety is one and a half. Multiply this by itself; the product is two and a quarter. Add this to the four; the sum is

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1. If in an equation, of the form $x^{2}+a=bx,({\tfrac {b}{2}})^{2}<a$, the case supposed in the equation cannot happen. If $({\tfrac {b}{2}})^{2}=a$, then $x={\tfrac {b}{2}}$
2. $cx^{2}+a=bx$ is to be reduced to $x^{2}+{\tfrac {a}{c}}={\tfrac {b}{c}}x$
3. ${\begin{aligned}{\text{3d case}}\;cx^{2}&=bx+a\\{\text{Example}}\;x^{2}&=3x+4\\x^{2}&={\sqrt {({\tfrac {3}{2}})^{2}+4}}+{\tfrac {3}{2}}\\&={\sqrt {1{\tfrac {1}{4}}^{2}+4}}+1{\tfrac {1}{2}}\\&={\sqrt {2{\tfrac {1}{4}}+4}}+1{\tfrac {1}{2}}\\&={\sqrt {6{\tfrac {1}{4}}}}+1{\tfrac {1}{2}}\\&={\sqrt {2{\tfrac {1}{2}}}}+1{\tfrac {1}{2}}=4\\\end{aligned}}$