Page:The Algebra of Mohammed Ben Musa (1831).djvu/59

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thing, it is one square. Subtract this from a hundred and a square minus twenty things, and you have a hundred, minus twenty things, equal to forty dirhems. Separate now the twenty things from a hundred, and add them to the forty; then you have a hundred, equal to twenty things and forty dirhems. Subtract now forty from a hundred; there remains sixty dirhems, equal to twenty things: therefore one thing is equal to three, which is one of the two parts.

If the question be: “I have divided ten into two parts, and having multiplied each part by itself, I have put them together, and have added to them the difference of the two parts previously to their multiplication, and the amount of all this is fifty-four;”^[1] then the computation is this: You multiply ten minus thing by itself; it is a hundred and a square minus twenty things. Then multiply also the other thing of the ten by itself; it is one square. Add this together, it will be a hundred plus two squares minus twenty things. It was stated that the difference of the two parts before multiplication should be added to them. You say, therefore, the difference between them is ten minus two things.

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1. ${\begin{aligned}(10-x)^{2}+x^{2}+(10-x)-x&=54\\100-20x+2x^{2}+10-2x&=54\\100-22x+2x^{2}&=54\\55-11x+x^{2}&=27\\x^{2}+28&=11x\\x={\tfrac {11}{2}}\pm {\sqrt {{\tfrac {121}{4}}-28}}={\tfrac {11\pm 3}{2}}&=7\;{\text{or}}\;4\end{aligned}}$