The Compendious Book on Calculation by Completion and Balancing/Of the six problems

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The Algebra of Mohammed Ben Musa (1831)
by Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī, translated by Friedrich Rosen

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Abū ʿAbdallāh Muḥammad ibn Mūsā al-Khwārizmī4187940The Algebra of Mohammed Ben Musa1831Friedrich Rosen

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OF THE SIX PROBLEMS.

Before the chapters on computation and the several (25) species thereof, I shall now introduce six problems, as instances of the six cases treated of in the beginning of this work. I have shown that three among these cases, in order to be solved, do not require that the roots be halved, and I have also mentioned that the calculating by completion and reduction must always necessarily lead you to one of these cases. I now subjoin these problems, which will serve to bring the subject nearer to the understanding, to render its comprehension easier, and to make the arguments more perspicuous.

First Problem.

I have divided ten into two portions; I have multiplied the one of the two portions by the other; after this. I have multiplied the one of the two by itself, and the product of the multiplication by itself is four times as much as that of one of the portions by the other. ^[1]

Computation: Suppose one of the portions to be thing, and the other ten minus thing: you multiply thing by ten minus thing; it is ten things minus a square. Then multiply it by four, because the instance states “four times as much.” The result will be four times the product of one of the parts multiplied by the other. This is forty things minus four squares. After this you multiply thing by thing, that is to say, one of the portions by itself. This is a square, which is equal to forty things minus four squares. Reduce it now by the four squares, and add them to the one square. Then the equation is: forty things are equal to five squares; and one square will be equal to eight roots, that is, sixty-four; the root of this is eight, and this is one of the two portions, namely, that which is to (26) be multiplied by itself. The remainder from the ten is two, and that is the other portion. Thus the question leads you to one of the six cases, namely, that of “squares equal to roots.” Remark this.

Second Problem.

I have divided ten into two portions: I have multiplied each of the parts by itself, and afterwards ten by itself: the product often by itself is equal to one of the two parts multiplied by itself; and afterwards by two and seven-ninths; or equal to the other-multiplied by itself, and afterwards by six and one-fourth.^[2]

Computation: Suppose one of the parts to be thing, and the other ten minus thing. You multiply thing by itself, it is a square; then by two and seven-ninths, this makes it two squares and seven-ninths of a square. You afterwards multiply ten by ten; it is a hundred, which much be equal to two squares and seven-ninths of a square. Reduce it to one square, through division by nine twenty-fifths;^[3] this being its fifth and four-fifths of its fifth, take now also the fifth and four-fifths of the fifth of a hundred; this is thirty-six, which is equal to one square. Take its root, it is six. This is one of the two portions; and accordingly the other is four. This question leads you, therefore, to one of the six cases, namely, “squares equal to numbers.”

Third Problem.

I have divided ten into two parts. I have afterwards divided the one by the other, and the quotient was four.^[4]

Computation: Suppose one of the two parts to be (27) thing, the other ten minus thing. Then you divide ten minus thing by thing, in order that four may be obtained. You know that if you multiply the quotient by the divisor, the sum which was divided is restored. In the present question the quotient is four and the divisor is thing. Multiply, therefore, four by thing; the result is four things, which are equal to the sum to be divided, which was ten minus thing. You now reduce it by thing, which you add to the four things. Then we have five things equal to ten; therefore one thing is equal to two, and this is one of the two portions. This question refers you to one of the six cases, namely, “roots equal to numbers.”

Fourth Problem.

I have multiplied one-third of thing and one dirhem by one-fourth of thing and one dirhem, and the product was twenty.^[5]

Computation: You multiply one-third of thing by one-fourth of thing; it is one-half of a sixth of a square. Farther, you multiply one dirhem by one-third of thing, it is one-third of thing; and one dirhem by one-fourth of thing, it is one-fourth of thing; and one dirhem by one dirhem, it is one dirhem. The result of this is: the moiety of one-sixth of a square, and one-third of thing, and one-fourth of thing, and one dirhem, is equal to twenty dirhems. Subtract now the one dirhem from these twenty dirhems, there remain nineteen dirhems, equal to the moiety of one-sixth of a square, and one-third of thing, and one-fourth of thing. Now make your square a whole one: you perform this by multiplying all that you have by twelve. Thus you have one square and seven roots, equal to two hundred and twenty-eight dirhems. Halve the number of the roots, and multiply it by itself; it is twelve and one-fourth. Add this to the numbers, that is, to two hundred and twenty-eight; (28) the sum is two hundred and forty and one quarter. Extract the root of this; it is fifteen and a half. Subtract from this the moiety of the roots, that is, three and a half, there remains twelve, which is the square required. This question leads you to one of the cases, namely, “squares and roots equal to numbers.”

Fifth Problem.

I have divided ten into two parts; I have then multiplied each of them by itself, and when I had added the products together, the sum was fifty-eight dirhems.^[6]

Computation: Suppose one of the two parts to be thing, and the other ten minus thing. Multiply ten minus thing by itself; it is a hundred and a square minus twenty things. Then multiply thing by thing; it is square. Add both together. The sum is a hundred, plus two squares minus twenty things, which are. equal to fifty-eight dirhems. Take now the twenty negative things from the hundred and the two squares, and add them to fifty-eight; then a hundred, plus two squares, are equal to fifty-eight dirhems and twenty things. Reduce this to one square, by taking the moiety of all you have. It is then: fifty dirhems and a square, which are equal to twenty-nine dirhems and ten things. Then reduce this, by taking twenty-nine from fifty; there remains twenty-one and a square, equal to ten things. Halve the number of the roots, it is five; multiply this by itself, it is twenty-five; take from this the twenty-one which are connected with the square, the remainder (29) is four. Extract the root, it is two. Subtract this from the moiety of the roots, namely, from five, there remains three. This is one of the portions; the other is seven. This question refers you to one of the six cases, namely “squares and numbers equal to roots.”

Sixth Problem.

I have multiplied one-third of a root by one-fourth of a root, and the product is equal to the root and twenty-four dirhems.^[7]

Computation: Call the root thing; then one-third of thing is multiplied by one-fourth of thing; this is the moiety of one-sixth of the square, and is equal to thing and twenty-four dirhems. Multiply this moiety of one-sixth of the square by twelve, in order to make your square a whole one, and multiply also the thing by twelve, which yields twelve things; and also four-and-twenty by twelve: the product of the whole will be two hundred and eighty-eight dirhems and twelve roots, which are equal to one square. The moiety of the roots is six. Multiply this by itself, and add it to two hundred and eighty-eight, it will be three hundred and twenty-four. Extract the root from this, it is eighteen; add this to the moiety of the roots, which was six; the sum is twenty-four, and this is the square sought for. This question refers you to one of the six cases, namely, “roots and numbers equal to squares.”

↑ ${\begin{aligned}x^{2}&=4x(10-x)=40x-4x^{2}\\\;5x^{2}&=40x\\\;x^{2}&=8x\\\;x&=8;(10-x)=2\end{aligned}}$
↑ ${\begin{aligned}10^{2}&=2\times 2{\tfrac {7}{9}}\\100&=x^{2}\times {\tfrac {25}{9}}\\{\tfrac {2}{25}}\times 100&=x^{2}\\36&=x^{2}\\6&=x\end{aligned}}$
↑ ${\tfrac {9}{25}}={\tfrac {1}{5}}\times {\tfrac {4}{5}}+{\tfrac {1}{5}}$
↑ ${\begin{aligned}{\tfrac {10-x}{x}}&=4\\10-x&=4x\\10&=5x\\2&=x\end{aligned}}$
↑ ${\begin{aligned}({\tfrac {1}{3}}x+1)({\tfrac {1}{4}}x+1)&=20\\{\tfrac {x^{2}}{12}}+{\tfrac {1}{3}}x+{\tfrac {1}{4}}x+1&=20\\{\tfrac {x^{2}}{12}}+{\tfrac {7}{12}}x&=19\\x^{2}+7x&=228\\x={\sqrt {{\tfrac {49}{4}}+228}}-{\tfrac {7}{2}}&=12\end{aligned}}$
↑ ${\begin{aligned}x^{2}+(10-x)^{2}&=58\\2x^{2}+20x+100&=58\\x^{2}-10x+50&=29\\x^{2}+21&=10x\\x=5\pm {\sqrt {25-21}}&=5\pm 2=7\;{\text{or}}\;3\end{aligned}}$
↑ ${\begin{aligned}{\tfrac {x}{3}}\times {\tfrac {x}{4}}&=x+24\\{\tfrac {x^{2}}{12}}&=x+24\\x^{2}&=12x+288\\x=6+{\sqrt {36+288}}&=6+18=24\end{aligned}}$

[1] ${\begin{aligned}x^{2}&=4x(10-x)=40x-4x^{2}\\\;5x^{2}&=40x\\\;x^{2}&=8x\\\;x&=8;(10-x)=2\end{aligned}}$

[2] ${\begin{aligned}10^{2}&=2\times 2{\tfrac {7}{9}}\\100&=x^{2}\times {\tfrac {25}{9}}\\{\tfrac {2}{25}}\times 100&=x^{2}\\36&=x^{2}\\6&=x\end{aligned}}$

[3] ${\tfrac {9}{25}}={\tfrac {1}{5}}\times {\tfrac {4}{5}}+{\tfrac {1}{5}}$

[4] ${\begin{aligned}{\tfrac {10-x}{x}}&=4\\10-x&=4x\\10&=5x\\2&=x\end{aligned}}$

[5] ${\begin{aligned}({\tfrac {1}{3}}x+1)({\tfrac {1}{4}}x+1)&=20\\{\tfrac {x^{2}}{12}}+{\tfrac {1}{3}}x+{\tfrac {1}{4}}x+1&=20\\{\tfrac {x^{2}}{12}}+{\tfrac {7}{12}}x&=19\\x^{2}+7x&=228\\x={\sqrt {{\tfrac {49}{4}}+228}}-{\tfrac {7}{2}}&=12\end{aligned}}$

[6] ${\begin{aligned}x^{2}+(10-x)^{2}&=58\\2x^{2}+20x+100&=58\\x^{2}-10x+50&=29\\x^{2}+21&=10x\\x=5\pm {\sqrt {25-21}}&=5\pm 2=7\;{\text{or}}\;3\end{aligned}}$

[7] ${\begin{aligned}{\tfrac {x}{3}}\times {\tfrac {x}{4}}&=x+24\\{\tfrac {x^{2}}{12}}&=x+24\\x^{2}&=12x+288\\x=6+{\sqrt {36+288}}&=6+18=24\end{aligned}}$

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The Compendious Book on Calculation by Completion and Balancing/Of the six problems

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