History of Hindu Mathematics: A Source Book v2.c17.s17.t2: (ii) Solution of ax² + bx + c = a'y² + b'y + c'

History of Hindu Mathematics: A Source Book v2.c17.s17.t2

Language: English

(ii) Solution of ax² + bx + c = a'y² + b'y + c'

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Bhâskara II has treated the more general type of quadratic indeterminate equations : $ax² + bx + c = a'y² + b'y + c'. 2$ His rule in this connection runs as follows :

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² Additions to *Elements of Algebra* by Leonard Euler, translated into English by John Hewlett, 5th edition, London, 1840, pp. 537ff.

"If there be the square of the unknown together with the (simple) unknown and an absolute number, putting it equal to the square of another unknown its root (should be investigated). Then on the other side (find)

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the roots by the method of the Square-nature, as has been stated before. Put the lesser root¹ equal to the root of the first side and the greater root equal to that of the second."²

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¹ The meaning of the terms ‘lesser root’, ‘greater root’, etc., as used here, will be clear from the illustration and the general solution given below.

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He further elucidates the rule thus:

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“In this case, on taking the square-root of the first side, there remain on the other side the square of the unknown and the (simple) unknown with or without an absolute number. In that case forming an equation of the second side with the square of another unknown, the roots (should be found). Of these (roots just determined), making the lesser equal to the root of the first side (of the given equation) and the greater to the root of the second side, the values of the unknowns should be determined.”

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*Example.* “Say what is the number of terms of a series (in A. P.) whose first term is 3, the common difference is 2; but whose sum multiplied by 3 is equal to the sum of a different number of terms.”³

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³ *BBi*, p. 104.

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*Solution.* “Here the statements of the series are: first term =3, common difference =2, number of terms =x; first term =3, common difference =2, number of terms =y. The two sums are (respectively) x²+2x, y²+2y. Making three times the first equal to the second, the statement for clearance is $3x²+6x=y²+2y.$ After the clearance, multiplying the two sides (of the equation) by 3 and superadding 9, the square-root of the first side is 3x+3. On the second side of the equation stands 3y²+6y+9. Forming an equation of this with z², and similarly multiplying the sides by 3 and superadding -18, the root of it is 3y+3. Then the roots of the other side, 3z²-18, by the method of the Square-nature are the lesser =9 and greater =15, or the lesser =13 and greater =57. Equating the lesser root with the square-root of the first side, namely, 3x+3, and the greater root with the square-root of the second side, namely, 3y+3, the values of x, y are found to be (2, 4) or (10, 18). So in every case."

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² *BBi*, p. 104.

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In general, on completing the square on the left-hand side, equation (2) becomes $(ax+½b)² = aa'y² + ab'y + ac' + (¼b² - ac).$

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Put $ax+½b=z,$ @(2.1) and then complete the square on the right-hand side. Thus the given equation is finally reduced to $aa'z² - β = w²,$ @(2.2) where $w = aa'y + ½ab',$ @(2.3) and $β = a²a'c' + (¼b² - ac)aa' - (½ab')².$

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Now, if z = l, w = m be a solution of the equation (2.2), another solution will be z = lq ± mp, w = mq ± aa'lp; where a'ap² + 1 = q². Substituting in (2.1) and (2.3), we get . x &= -b/2a + 1/a(lq ± mp), / y &= -(b')/(2a') + 1/(aa')(mq ± aa'lp). } @(2.4)

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Now, let l = ar + ½b and m = aa's + ½ab'. Substituting in the above expressions, we get the required solution of (2) in the form : $. x &= 1/2a(qb ± pab' - b) + qr ± pa's, / y &= 1/(2a')(qb' ± pa'b - b') + qs ± par ; (2.5)$ where aa'p² + 1 = q², and ar² + br + c = a's² + b's + c'.

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The form (2.5) shows that having found empirically one solution of ax² + bx + c = a'y² + b'y + c' Bhâskara could find an infinite number of other solutions of it.

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Jñânarâja (1503) says :

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"If on the other side be present the square as well as the linear power of the unknown together with an absolute term, put it equal to the square of another unknown and then determine the lesser and greater roots. The lesser root will be equal to the first square-root and the greater to the second square-root."