Squares and Numbers Equal to Roots

“A square and 21 in numbers are equal to 10 roots of the same square.”

What must be the amount of a square, which, when 21 dirhems are added to it, becomes equal to the equivalent of ten roots of that square?

Halve the number of the roots; Multiply this by the moiety the product itself; Solution is is

five. twenty-five. Subtract from this the twenty-one which are connected with the square root ; roots, it is is is four. Extract its Subtract this from the moiety of the two. which the remainder ; five ; the remainder is three. This is the root of the square which you required, and the square is nine. roots ; Or you may add the the sum is seven this ; which you sought for, root to the moiety of the is the root of the square and the square itself is forty- nine. When this case, you meet with an instance which refers you to try its solution by addition, and if that do not serve, then subtraction certainly will. both addition and subtraction will For in this case may be employed, which not answer in any other of the three cases in which

• 2d case, ex 1 --a-bx Example. #‘4-21 S = N " 25 ~ 21the that, when And know, the roots must be halved. number of 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 the square, then the number of dirhems connected with instance (8) is impossible;* but the product be equal to if 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, f as I have explained under the first case. Roots and Numbers are equal to Squares;^, for instance, " three roots and four of simple numbers are equal to a square." is is one and a Solution = Halve the roots the moiety ; the product this half. by itself; Multiply two and a quarter. Add this to the four the sum ;
• If in an equation, of the form the case in supposed the 2 x’ +a=bx, 2 (|) = (|)2=a, then* 2 f cx +a=bx is to be reduced \ 3d case z. equation cannot happen. ex to 2 x 2 --^~x bx + a 2 Example x = +4 = V'6f = 2j -f ij is a, If13 ( and a quarter. six Add half. Extract this to the square, and the square four. is is These are the two and a to This is the root of the a multiple or sub-multiple one entire square. which six cases I mentioned in the They have now been introduction to this book. plained. it is sixteen. Whenever you meet with it root ; its moiety of the roots, which was one and a half; the sum of a square, reduce ) ex- have shown that three among them do not I require that the roots be halved, and I have taught how they must be resolved. As for the other three, in which halving the roots is necessary, I think it expe- dient, more accurately, to explain them by separate chapters, in which a figure be given for each will case, to point out the reasons for halving. Demonstration of the Case = " a Square and ten Roots are equal The which to thirty -nine Dirhems"* figure to explain this a quadrate, the sides of are unknown. It represents the square, the which, or the root of which, you wish to know. the figure as one of sides A B, its each side of which roots ; and if may be This is considered you multiply one of these by any number, then the amount of that number may be looked upon as the are added to the square. number of Each the roots which side of the quadrate represents the root of the square; and, as in the instance,
• Geometrical illustration of the case, x 2 + io# = 39 (9)14 ( ) the roots were connected with the square, one-fourth of ten, that combine Thus with two and a to say, take half, and of the four sides of the figure. with each it is we may A the original quadrate new paral- B, four each having a side of the qua- lelograms are combined, drate as its length, and the number of two and a half as and they are the parallelograms C, G, T, its breadth K. We have now a quadrate of equal, though sides ; ; unknown but in each of the four corners of which a square two and a half piece of two and a half multiplied by wanting. In order to compensate for this want and to complete the quadrate, we must add (to that have already) four times the square of two and a is, twenty-five. figure, which we half, that We know (by the statement) that the first namely, the quadrate representing the square, together with the four parallelograms around represent the ten roots, bers. If to this we is it, which equal to thirty-nine of num- add twenty-five, which is the equivalent A B, completed, then we of the four quadrates at the corners of the figure by which the great figure D H know makes that this together of this great quadrate is its is sixty-four. root, that subtract twice a fourth of ten, that as is is, One side If eight. is five, from we eight, from the two extremities of the side of the great quadrate D H, then the remainder of such a side will be three, and that is the root of the square, or the side of the original figure we have halved A B. It must be observed, that the number of the roots, product of the moiety multiplied by and added the itself to the numberthirty-nine, in order to complete the great figure in four corners ; because the fourth of any plied by itself, and then by is four, number of the moiety of that number its multi- equal to the product multiplied by itself.* Accordingly, we multiplied only the moiety of the roots by instead of multiplying itself, This then by four. its the figure is fourth by itself, and = o. The same may also be explained We proceed from the the square. It is on two it becomes by another figure. B, which represents our next business to add to roots of the same. so that quadrate A it the ten We halve for this purpose the ten, five, and construct two quadrangles A B, namely, G and D, sides of the quadrate the length of each of them being five, as the moiety of the ten roots, whilst the breadth of each A B. side of the quadrate Then number of the by five = roots which two sides of the first this five A B. This is equal being half of the we have added quadrate. equal to a a quadrate remains opposite the corner of the quadrate to five multiplied is to each of the Thus we know that (10)the quadrate, which first quadrangles on its sides, the square, and the two is which are the ten roots, make In order to complete the great together thirty-nine. quadrate, there wants only a square of five multiplied (11) by five, or twenty-five. This we add to thirty-nine, in We extract its root, sixty-four. The sum S H. order to complete the great square eight, which is is one of By subtracting from the sides of the great quadrangle. same quantity which we have before added, namely five, we obtain three as the remainder. This is this the the side of the quadrangle square; it is itself is nine. A B, which represents the and the square the root of this square, This is the figure = H Demonstration of the Case = " a Square and twenty-one Dirhems are equal We to ten Roots"* represent the square by a quadrate length of whose side we do not know. This paralellogram is A D, such H B. The D, the To this we join a parallelogram, the breadth of which the sides of the quadrate A is equal to one of as the side H N. length of the two
• Geometrical illustration of the case, x*

21 iox17 ( its is length H equal to the line is figures together that ) has equal sides and angles, and one of plied by a unit by two is the root of the quadrate, or multiplied therefore, that a square H As C is it is stated, and twenty-one of numbers are we may conclude equal to ten roots, the line sides multi- its twice the root of the same. it is We know C. ten of numbers; for every quadrate that the length of equal to ten of numbers, since the line C D represents the root of the square. We now divide the line C H into two equal parts at the point G the line G C is then equal to H G. It is also evident that (12) the line G T is equal to the line C D. At present we

add to the line G T, in the same direction, a piece equal to the difference between to complete the square. K equal to C Then G and G the line T, in order TK becomes M, and we have a new quadrate of equal sides and know that the line angles, namely, the T K is five quadrate this is ; length also of the other sides M T. We consequently the the quadrate itself

is twenty-five, this being the product of the multiplication of half the times five number of is the roots twenty-five. by themselves, We have perceived for five that the H B represents the twenty-one of numbers which were added to the quadrate. We have then cut off a piece from the quadrangle H B by the line K T quadrangle (which is one of the sides of the quadrate only the part TA remains. M T), so that At present we take from K M the piece K L, which is equal to G K it appears that the line T G equal to M L rnore- the line then ; is D ;18 ( K L, which has been cut off from G; consequently, the quadrangle over, the line is equal to equal to K ) T A. Thus K M, MR is evident that the quadrangle it is M H T, R, is equal to augmented by the quadrangle H B, which represents the twenty-one. T was found to be equal to The whole quadrate the quadrangle M MT, we now If twenty-five. subtract from this quadrate, HT and M R, which are equal the quadrangles to twenty-one, there remains a small quadrate K which represents the difference between twenty-five This twenty-one. the line R four ; is G, which and G equal to is number two from (13) subtract this its ; that to say, three, is ginal square. line C G, which roots, then the C if But A, is the line R, which is is sum which is C is is the figure the root to a larger square. 3\r is L K is the line to the by the the line However, to this square, then the

you G, which number of seven, represented likewise be equal to ten roots of the by the root of the ori- the moiety of the you add twenty-one If two. you add the number two if and root, represented the moiety of the roots, then the remainder A C R, sum same square. will Here19 ( Demonstration of the Case

) " three Roots Simple Numbers are equal to and four of a Square"* Let the square be represented by a quadrangle, the sides of which are unknown to us, though they are equal among drate This themselves, as also the angles. is the qua- A D, which comprises the three roots and the four of numbers mentioned in drate one of its sides, In every qua- this instance. multiplied by a unit, is its root. We now cut off the quadrangle H D from the quadrate A D, and take one of its sides H C for three, which is the number of the roots. It follows, then, that the The same is quadrangle equal to H B represents Now the four of numbers which are added to the roots. we halve the side G the point H T, which ; C H, which is R D. equal to three roots, at from this division we construct the square is the product of half the roots (or one and (14) a half) multiplied by themselves, that We add is to say, two and G T a piece equal to the line A H, namely, the piece T L accordingly the line G L becomes equal to A G, and the line K N equal to T L. Thus a new quadrangle, with equal sides and angles, arises, namely, the quadrangle G M and we find that the line A G equal to M L, and the same line A G equal to G L. By these means the line C G remains equal to N R, and the line M N a quarter. then to the line ; ; is is equal to T equal to the H B L, and from the quadrangle quadrangle K L is cut a piece off,

• Geometrical illustration of the 3d case, 2 x' 30= -f- 4But we know that the quadrangle A R represents the four of numbers which are added to the three roots. The quadrangle A N and K L the quadrangle A gether equal to the quadrangle are to- R, which represents the four of numbers. We have seen, prises the and a also, that the quadrangle G M com- product of the moiety of the roots, or of one half, multiplied by itself; that is to say two and a quarter, together with the four of numbers, which are AN represented by the quadrangles and K L. There remains now from the side of the great original quadrate A D, which represents the whole square, only the moiety of the roots, that line G If C. is to say, we add G M, which is is the line A This it B which is C G, or the moiety of half, makes four, C, or the root to a square, which A D. Here follows was which we were desirous to explain. (15) A G, and a represented by the quadrate the figure. namely, the being equal to two and together with this, the three roots, namely, one half, the line this to the root of the quadrate a half; then one and a M JLWe have observed that every question which requires equation or reduction for one of the book. I now have its solution, will refer which cases six you have proposed in I to this also explained their arguments. Bear them, therefore, in mind.