Page:The Algebra of Mohammed Ben Musa (1831).djvu/67

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of the other half thing; there remains one dirhem, equal to half a thing. Double it, then you have one thing, equal to two dirhems. Consequently, the other number^[1] is four.

If some one say: “I have divided ten into two parts; I have multiplied the one by ten and the other by itself, and the products were the same;” ^[2] then the computation is this: You multiply thing by ten; it is ten things. Then multiply ten less thing by itself; it is a hundred (37) and a square less twenty things, which is equal to ten things. Reduce this according to the rules, which I have above explained to you.

In like manner, if he say: “I have divided ten into two parts; I have multiplied one of the two by the other, and have then divided the product by the difference of the two parts before their multiplication, and the result of this division is five and one-fourth;"^[3] the computation will be this: You subtract thing from ten; there remain ten less thing. Multiply the one by the other, it is ten things less a square. This is the product of the multiplication of one of the two parts by the other. At

↑ “Square” in the original.
↑ ${\begin{aligned}10x=(10-x)^{2}=100-20x+x^{2}\\x=15-{\sqrt {225-100}}=15-{\sqrt {125}}\end{aligned}}$
↑ ${\begin{aligned}{\tfrac {x(10-x)}{(10-2x)}}=5{\tfrac {1}{4}}\\10x-x^{2}=52{\tfrac {1}{2}}-10{\tfrac {1}{2}}x\\20{\tfrac {1}{2}}x=x^{2}+52{\tfrac {1}{2}}\\x=10{\tfrac {1}{4}}-7{\tfrac {1}{4}}=3\\\end{aligned}}$

[1] “Square” in the original.

[2] ${\begin{aligned}10x=(10-x)^{2}=100-20x+x^{2}\\x=15-{\sqrt {225-100}}=15-{\sqrt {125}}\end{aligned}}$

[3] ${\begin{aligned}{\tfrac {x(10-x)}{(10-2x)}}=5{\tfrac {1}{4}}\\10x-x^{2}=52{\tfrac {1}{2}}-10{\tfrac {1}{2}}x\\20{\tfrac {1}{2}}x=x^{2}+52{\tfrac {1}{2}}\\x=10{\tfrac {1}{4}}-7{\tfrac {1}{4}}=3\\\end{aligned}}$

[1]

[2]

[3]