History of Hindu Mathematics: A Source Book v2.c17.s17.t1: (i) Solution of ax² + bx + c = y

History of Hindu Mathematics: A Source Book v2.c17.s17.t1

Language: English

(i) Solution of ax² + bx + c = y

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For the general solution of the quadratic indeterminate equation $ax² + bx + c = y², 1$

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Bhâskara II gives the following particular rule :

“On taking the square-root of one side, if there be on the second side only the square of the unknown together with an absolute number, in such cases, the greater and lesser roots should be determined by the method of the Square-nature. Of these two, the greater root is to be put equal to the square-root of the first side mentioned before, and thence the value of the first unknown should be determined. The lesser will be the value of the unknown associated with the *prakṛti*. In this way, the method of the Square-nature should be applied to this case by the intelligent.”

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As an illustration of this rule Bhâskara II works out in detail the following example:

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“What number being doubled and added to six times its square, becomes capable of yielding a squareroot? O ye algebraist, tell it quickly.”

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*Solution.* “Here let the number be x. Doubled and together with six times its square, it becomes 6x² + 2x. This is a square. On forming an equation with the square of y, the statement is $6x² + 2x + 0y² = 0x² + 0x + y².$ On making equi-clearance in this the two sides are 6x² + 2x and y².

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“Then multiplying these two sides by 6 and superadding 1, the root of the first side, as described before, is 6x + 1.

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“Now on the second side of the equation remains 6y² + 1. By the method of the Square-nature, its roots are : the lesser 2 and the greater 5, or the lesser 20 and the greater 49. Equating the greater root with the square-root of the first side, *viz.*, 6x + 1, the value of x is found to be 2/3 or 8. The lesser root, 2 or 20, is the value of y, the unknown associated with the *prakṛti*. In this way, by virtue of (the multiplicity of) the lesser and greater roots, many solutions can be obtained."¹

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¹ *BBi*, pp. 100-1.

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² *BBi*, p. 101.

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¹ *BBi*, p. 101.

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In other words the method described above is this:

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Completing the square on the left-hand side of the equation ax² + bx + c = y², we have $(ax + ½b)² = ay² + ¼(b² - 4ac).$ Putting z = ax + ½b, k = ¼(b² - 4ac), we get $ay² + k = z². 1.1$ If y = l, z = m be found empirically to be a solution of this equation, another solution of it will be $y = lq ± mp,$ $z = mq ± alp,$ where ap² + 1 = q². Hence a solution of (1) is $x = - b/2a + 1/a(mq ± alp),$ $y = lq ± mp.$

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Now suppose x = r, when z = m; that is, let m = ar + b/2. Substituting in the above expressions, we get the required solution of (1) as $. x &= 1/2a(bq - b) + qr ± lp, / y &= lq ± (apr + ½bp); 1.2$ where ap² + 1 = q² and ar² + br + c = l².

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Thus having known *one* solution of ax² + bx + c = y², an infinite number of other solutions can be easily obtained by the method of Bhâskara II. The method is, indeed, a very simple and elegant one. It has been adopted by later Hindu algebraists. As the relevant portion of the algebra of Nârâyaṇa (1350) is now lost, we cannot reproduce his description of the method. Jñânarâja (1503) says :

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"(Find) the square-root of the first side according to the method described before and, by the method of the Square-nature, the roots of the other side, where the coefficient of the square of the unknown is considered to be the *prakṛti* and the interpolator is an absolute term. Then the greater root will be equal to the previous square-root and the other (*i.e.*, the lesser root) to the unknown associated with the *prakṛti*."

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The above solution (1.2), but with the upper sign only, was rediscovered in 1733 by Eluer.¹ His method is indirect and cumbrous. Lagrange's (1767) method begins in the same way as that of Bhâskara II by completing the square on the left-hand side of the equation.²

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¹ Leonard Euler, *Opera Mathematica*, vol. II, 1915, pp. 6-17 ; Compare also pp. 576-611.