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History of Hindu Mathematics: A Source Book v2.c17.s8: 10. Quadratic Equations

History of Hindu Mathematics: A Source Book v2.c17.s8

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10. Quadratic Equations

v2.c17.s8.1

**Early Treatment.** It has been shown before that the altar-construction of the Vedic Hindus involved the solution of the complete quadratic equation $ax² + bx = c,$ (1) as well as of the pure quadratic ax² = c. The equation that had to be solved was $7x² + ½x = 7½ + m,$ which gives $x = 1/28(√(841 + 112m) - 1),$ or $x² = 1/784 842 + 112m - 2√(841 + 112m).$ Simplifying and neglecting higher powers of m, we get $x² = 1 + 4m/29, approximately.$ Kâtyâyana gives the value² $x² = 1 + m/7.$

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¹ *L*, pp. 21f. See also *GT*, 123. If the rate of interest be given as r per p per t months, then p is the argument, t the time and r the fruit. *Cf.* Part I, p. 204; also pp. 225-226.

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² Datta, *Śulba*, p. 167.

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The geometrical solution of the simple quadratic equation $4b² - 4dh = - c²$ is found in the early canonical works of the Jainas (500-300 B. C.) and also in the *Tattvârthâdhigama-sûtra* of Umâsvâti (*c.* 150 B. C.),¹ as $h = ½(d - √(d² - c²)).$

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¹ Datta, "Geometry in the Jaina Cosmography," *Quellen und Studien Zur Ges. d. Math.*, Ab. B, Bd. 1 (1931), pp. 245-254.

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**Bakhshâlî Treatise.** The solution of the quadratic equation was certainly known to the author of the Bakhshâlî treatise (*c.* 200). In this work there are some problems of the following type: A certain person travels *s* *yojana* on the first day and *b* *yojana* more on each successive day. Another who travels at the uniform rate of *S* *yojana* per day, has a start of *t* days. When will the first man overtake the second ?

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If *x* be the number of days after which the first overtakes the second, then we shall have $S(t + x) = x s + ( (x - 1)/2 ) b ,$ or $bx² - 2(S - s) + bx = 2tS.$ Therefore $x = (√(2(S - s) + b² + 8btS) + 2(S - s) + b)/2b,$ which agrees exactly with the solution as stated in the Bakhshâlî treatise.

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"The daily travel [*S*] diminished by the march of the first day [*s*] is doubled; this is increased by the common increment [*b*]. That (sum) multiplied by itself is designated {as the *kṣepa* quantity}. The product of the daily travel and the start [t] being multiplied by eight times the common increment, the *kṣepa* quantity is added. The square-root of this {is increased by the *kṣepa* quantity; the sum divided by twice the common increment will give the required number of days}."

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Nearly the whole of the detailed working of the particular example in which S = 5, t = 6, s = 3 and b = 4, is preserved. It is substantially as follows: St = 30; S - s = 5 - 3 = 2; 2(S - s) + b = 8; 2(S - s) + b² = 64; 8St = 240; 8Stb = 960; 2(S - s) + b² + 8Stb = 1024; √1024 = 32; 32 + 8 = 40; 40 ÷ 8 = 5 = x.

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For another problem S = 7, t = 5, s = 5, b = 3; then $x = (7 + √889)/6.$

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The formula for determining the number of terms (n) of an A.P. whose first term (a), common difference (b) and sum (s) are known, is stated in the form $n = (√(8bs + (2a - b)²) - (2a - b))/2b.$

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The working of the particular example in which s = 60, a = 1, b = 1 is preserved substantially as follows: 8bs = 480; 2a = 2; 2a - b = 1; (2a - b)² = 1; 8bs + (2a - b)² = 481; n = ½(- 1 + √481), etc.

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² *BMs*, Folio 5, verso; Compare also Kaye's Introduction, pp. 37, 45.

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**Āryabhaṭa I.** To find the number of terms of an A.P., Āryabhaṭa I (499) gives the following rule:

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¹ *BMs*, Folio 5, recto.

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³ *BMs*, Folio 6, recto and verso.

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⁴ *BMs*, Folio 65 verso. Working of this example has been continued on folios 56, verso and recto, and 64, recto.

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“The sum of the series multiplied by eight times the common difference is added by the square of the difference between twice the first term and the common difference; the square-root (of the result) is diminished by twice the first term and (then) divided by the common difference: half of this quotient plus unity is the number of terms.”\1

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\1 *Ā*, ii. 20.

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That is to say, $n = ½ (√(8bs + (2a - b)²) - 2a)/b + 1 .$

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The solution of a certain interest problem involves the solution of the quadratic $tx² + px - Ap = 0.$ Āryabhaṭa gives the value of x in the form\2 $x = (√(Apt + (p/2)²) - p/2)/t.$

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\2 *Ā*, ii. 25; vide Part I, pp. 219f.

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Though Āryabhaṭa I has nowhere indicated any method of solving the quadratic, it appears from the above two forms that he followed two different methods in order to make the unknown side of the equation ax² + bx = c, a perfect square. In one case he multiplied both the sides of the equation by 4a and in the other simply by a.

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**Brahmagupta’s Rules.** Brahmagupta (628) has given two specific rules for the solution of the quadratic. His first rule is as follows:

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“The quadratic: the absolute quantities multiplied by four times the coefficient of the square of the unknown are increased by the square of the coefficient of the middle (*i.e.*, unknown); the square-root of the result being diminished by the coefficient of the middle

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and divided by twice the coefficient of the square of the unknown, is (the value of) the middle."¹

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¹ *BrSpSi*, xviii. 44. It will be noted that in this rule Brahmagupta has employed the term *madhya* (middle) to imply the simple unknown as well as its coefficient. The original of the term is doubtless connected with the mode of writing the quadratic equation in the form $ax²+bx+0=0x²+0x+c,$ so that there are three terms on each side of the equation.

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$i.e., x = (√(4ac+b²)-b)/2a.$

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The second rule runs as:

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"The absolute term multiplied by the coefficient of the square of the unknown is increased by the square of half the coefficient of the unknown; the square-root of the result diminished by half the coefficient of the unknown and divided by the coefficient of the square of the unknown is the unknown."²

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$i.e., x = (√(ac+(b/2)²)-(b/2))/a.$

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The above two methods of Brahmagupta are identical with those employed before him by Āryabhaṭa I (499). The root of the quadratic equation for the number of terms of an A.P. is given by Brahmagupta in the first form:³

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³ *BrSpSi*, xii. 18.

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$n = (√(8bs+(2a-b)²)-(2a-b))/2b.$

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² *BrSpSi*, xviii. 45.

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For the solution of the quadratic Brahmagupta uses also a third formula which is similar to the one now commonly used. Though it has not been expressly described in any rule, we find its application in a few instances. One of them is an interest problem: A certain sum (p) is lent out for a period (t₁); the interest accrued (x) is lent out again at this rate of interest for another period (t₂) and the total amount is A. Find x.

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The equation for determining x is $t₂/pt₁x² + x = A.$

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Hence, we have $x = √((pt₁/2t₂)² + Apt₁/t₂) - pt₁/2t₂,$ which is exactly the form in which Brahmagupta states the result.¹

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¹ *BrSpSi*, xii. 15. Vide Part I, p. 220.

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There is a certain astronomical problem which involves the quadratic equation² $(72+a²)x² 24apx = 144((R²)/2 - p²),$ where a = agra (the sine of the amplitude of the sun), b = palabha (the equinoctial shadow of a gnomon 12 *aṅguli* long), R = radius, and x = koṇaśaṅku (the sine of the altitude of the sun when his altitude is 45°). Dividing out by (72+a²), we have $x² 2mx = n,$ where $m = 12ap/(72+a²), n = (144(R²/2 - p²))/(72+a²).$

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Therefore, we have $x = √(m²+n) ± m,$ as stated by Brahmagupta. This result is given also in the *Sûryasiddhânta*³ (*c.* 300) and by Śrîpati (1039).⁴

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² *BrSpSi*, iii. 54-55.

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³ *SûSi*, iii. 30-1.

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⁴ *SiŚe*, iv. 74.

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**Śrīdhara’s Rule.** Śrīdhara (*c.* 750) expressly indicates his method of solving the quadratic equation. His treatise on algebra is now lost. But the relevant portion of it is preserved in quotations by Bhāskara II¹ and others.² Śrīdhara’s method is :

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¹ *BBi*, p. 61.

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“Multiply both the sides (of an equation) by a known quantity equal to four times the coefficient of the square of the unknown; add to both sides a known quantity equal to the square of the (original) coefficient of the unknown: then (extract) the root.”³

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³ “Caturâhatavargasamai rūpaiḥ pakṣadvayaṁ guṇayet, Avyaktavargarūpairyuktau pakṣau tato mūlam.” This is the reading of Śrīdhara’s rule as stated by Jñānarāja and Sūryadāsa and accepted also by Sudhakara Dvivedi. But according to the reading of Kṛṣṇa (*c.* 1580) and Râmakṛṣṇa (*c.* 1648), which has been accepted by Colebrooke, the second line of the verse should be “Pūrvâvyaktasya kṛteḥ samarūpāṇi kṣipet tayoreva” or “add to them known quantities equal to the square of the original coefficient of the unknown.”

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That is, to solve ax² + bx = c, we have 4a²x² + 4abx = 4ac, or (2ax + b)² = 4ac + b². Therefore 2ax + b = √(4ac + b²). Hence x = (√(4ac + b²) - b)/2a.

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An application of this rule is found in Śrīdhara’s *Triśatikā*, in connection with finding the number of terms of an A.P.⁴ *i.e.*, n = (√(8bs + (2a - b)²) - 2a + b)/2b, where a is the first term, b the common difference and s the sum of n terms.

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² Jñānarāja (1503) in his *Bījagaṇita* and Sūryadāsa (1541) in his commentary on Bhāskara’s *Bījagaṇita*.

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⁴ *Triś*, R. 41.

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**Mahâvîra.** The only work of Mahâvîra (850) which is available now is the *Gaṇita-sâra-saṃgraha*. As it is admittedly devoted to arithmetic we cannot expect to find in it a rule for solving the quadratic. But there are in it several problems whose solutions presuppose a knowledge of the roots of the quadratic. One problem is as follows:

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“One-fourth of a herd of camels was seen in the forest; twice the square-root of the herd had gone to the mountain-slopes; three times five camels were on the bank of the river. What was the number of those camels?”¹

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¹ GSS, iv. 34.

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If x be the number of camels in the herd, then $¼x + 2√x + 15 = x.$ Or, in general, the equation to be solved is $a/bx + c√x + d = x,$ $or (1 - a/b)x - c√x = d.$

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Mahâvîra gives the following rule for the solution of this equation:

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“Half the coefficient of the root (of the unknown) and the absolute term should be divided by unity minus the fraction (*i.e.*, the coefficient of the unknown). The square-root of the sum of the square of the coefficient of the root (of the unknown) and the absolute term (treated as before) is added to the coefficient of the root (of the unknown treated as before). The sum squared is the (unknown) quantity in this *mûla* type of problems.”²

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² GSS, iv. 33.

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$i.e., x = (c/2)/(1 - a/b) + √(((c/2)/(1 - a/b))² + d/(1 - a/b))²,$

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which shows that Mahâvîra employed the modern rule for finding the root of a quadratic. His solution for the interest problem treated by Brahmagupta is exactly the same as that of the latter.¹ We shall presently show that he knew that the quadratic has two roots.

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¹ *GSS*, vi. 44.

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**Âryabhaṭa II.** The formula for the number of terms (n) of an A.P. whose first term (a), common difference (b) and sum (s) are known is given by Âryabhaṭa II (*c.* 950) as follows :²

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$n = (√(2bs + (a - b/2)²) - a + b/2)/b,$

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² *MSi*, xv. 50.

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which shows that for solving the quadratic he followed the second method of Âryabhaṭa I and Brahmagupta.

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**Śrîpati's Rules.** Śrîpati (1039) indicates two methods of solving the quadratic. There is a lacuna in our manuscript in the rule describing the first method, but it can be easily recognised to be the same as that of Śrîdhara.

v2.c17.s8.68

"Multiply by four times the coefficient of the square of the unknown and add the square of the coefficient of the unknown; (then extract) the square-root......... divided by twice the coefficient of the square of the unknown, is said to be (the value of) the unknown."

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"Or multiplying by the coefficient of the square of the unknown and adding the square of half the coefficient of the unknown, (extract) the square-root. Then (proceeding) as before, it is diminished by half the coefficient of the unknown and divided by the coefficient of the square of the unknown. This (quotient) is said to be (the value of) the unknown."¹

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¹ *SiŚe*, xiv. 17-8, 19.

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i.e., \ &ax²+bx=c,\\ or \ &a²x²+abx+(b/2)²=ac+(b/2)².\\ Therefore &ax+b/2=√(ac+(b/2)²).\\ Hence \ &x=(√(ac+(b/2)²)-b/2)/a. \end{aligned}

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**Bhâskara II's Rules.** Bhâskara II (1150) says : "When the square of the unknown, etc., remain, then, multiplying the two sides (of the equation) by some suitable quantities, other suitable quantities should be added to them so that the side containing the unknown becomes capable of yielding a root (*pada-prada*). The equation should then be formed again with the root of this side and the root of the known side. Thus the value of the unknown is obtained from that equation."²

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² *BBi*, p. 59.

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This rule has been further elucidated by the author in his gloss as follows :

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"When after perfect clearance of the two sides, there remain on one side the square, etc., of the unknown and on the other side the absolute term only, then, both the sides should be multiplied or divided by some suitable optional quantity; some equal quantities should further be added to or subtracted from both the sides so that the unknown side will become capable of yielding a root. The root of that side must be equal to the root of the absolute terms on the other side. For, by simultaneous equal additions, etc., to the two equal sides the equality remains. So forming an equation again with these roots the value of the unknown is found."³

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³ *BBi*, p. 61.

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It may be noted that in his treatise on arithmetic Bhâskara II has always followed the modern method of dividing by the coefficient of the square of the unknown.¹

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¹ *L*, pp. 15ff.

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Jñânarâja (1503) and Gaṇeśa (1545) describe the same general methods for solving the quadratic as Bhâskara II.

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**Elimination of the Middle Term.** The method of solving the quadratic was known amongst the Hindu algebraists by the technical designation *madhyamâharaṇa* or "The Elimination of the Middle" (from *madhyama* = middle and *âharaṇa* = removal, or destroying, that is, elimination). The origin of the name will be easily found in the principle underlying the method. By it a quadratic equation which, in its general form, contains three terms and so has a middle term, is reduced to a pure quadratic equation or a simple equation involving only two terms and so having no middle term. Thus the middle term of the original quadratic is eliminated by the method generally adopted for its solution. And hence the name. Bhâskara II has observed, "It is also specially designated by the learned teachers as the *madhyamâharaṇa*. For by it, the removal of one of the two² terms of the quadratic, the middle one, (takes place)."³ The name is, however, employed also in an extended sense so as to embrace the methods for solving the cubic and the biquadratic, where also certain terms are eliminated. It occurs as early as the works of Brahmagupta (628).¹

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² Referring to the two terms on the unknown side of the complete quadratic. Or the text *varga-râśâvekasya* may be rendered as "of one out of the unknown quantity and its square." According to the commentators Sûryadâsa (1541) and Kṛṣṇa (1580), it implies "of one between the two square terms," *viz*., the square of the unknown and the square of the absolute number.

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³ *BBi*, p. 59.

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¹ *BrSpSi*, xviii. 2.

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**Two Roots of the Quadratic.** The Hindus recognised early that the quadratic has generally two roots. In this connection Bhâskara II has quoted the following rule from an ancient writer of the name of Padmanâbha whose treatise on algebra is not available now:

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“If (after extracting roots) the square-root of the absolute side (of the quadratic) be less than the negative absolute term on the other side, then taking it negative as well as positive, two values (of the unknown) are found.”²

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² “Vyaktapakṣasya cenmûlamanyapakṣarṇarûpataḥ Alpaṁ dhanarṇagaṁ kṛtvâ dvividhotpadyate mitiḥ”—*BBi*, p. 67.

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Bhâskara points out with the help of a few specific illustrations that though these double roots of the quadratic are theoretically correct, they sometimes lead to incongruity and hence should not always be accepted. So he modifies the rule as follows:

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“If the square-root of the known side (of the quadratic) be less than the negative absolute term occurring in the square-root of the unknown side, then making it negative as well as positive, two values of the unknown should be determined. This is (to be done) occasionally.”³

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³ *BBi*, p. 59; also compare the author’s gloss on the same (p. 61).

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*Example 1.* “The eighth part of a troop of monkeys, squared, was skipping inside the forest, being delightfully attached to it. Twelve were seen on the hill delighting in screaming and rescreaming. How many were they ?”⁴

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⁴ *BBi*, p. 65.

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*Solution.* "Here the troop of monkeys is *x*. The square of the eighth part of this together with 12, is equal to the troop. So the two sides are¹ $¹₆₄x² + 0x + 12 = 0x² + x + 0.$ Reducing these to a common denominator and then deleting the denominator, and also making clearance, the two sides become $x² - 64x + 0 = 0x² + 0x - 768.$ Adding the square of 32 to both sides and (extracting) square-roots, we get $x - 32 = ± (0x + 16).$ In this instance the absolute term on the known side is smaller than the negative absolute term on the side of the unknown; hence it is taken positive as well as negative; the two values of *x* are found to be 48, 16."

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*Example 2.* "The fifth part of a troop of monkeys, leaving out three, squared, has entered a cave; one is seen to have climbed on the branch of a tree. Tell how many are they?"²

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*Solution.* "In this the value of the troop is *x*; its fifth part less three is ¹₅ x - 3; squared, ¹₂₅ x² - ⁶₅ x + 9; this added with the visible (number of monkeys), ¹₂₅ x² - ⁶₅ x + 10, is equal to the troop. Reducing to a common denominator, then deleting the denominator and making clearance, the two sides become $x² - 55x + 0 = 0x² + 0x - 250.$ Multiplying these by 4, adding the square of 55, and extracting roots, we get $2x - 55 = ± (0x + 45).$ In this case also, as in the previous, two values are obtained: 50, 5. But, in this case, the second (value) should not be accepted as it is inapplicable. *People have no faith in the known becoming negative.*"

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¹ We have here followed the modern practice of writing the two sides of an equation in a line with the sign of equality interposed, at the same time, preserving the other salient feature of the Hindu method of indicating the absent terms, if any, by putting zeros as their coefficients.

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² *BBi*, pp. 65ff.

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¹ *BBi*, pp. 66f.

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The implied significance of this last observation is this: If the troop consists of only 5 monkeys, its fifth part will be 1 monkey. How can we then leave out 3 monkeys? Again, how can the remainder be said to have entered the cave? It seems to have also a wider significance.

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*Example 3*. "The shadow of a gnomon of twelve fingers being diminished by a third part of the hypotenuse, becomes equal to fourteen fingers. O mathematician, tell it quickly."

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*Solution*. "Here the shadow is (taken to be) x. This being diminished by a third part of the hypotenuse becomes equal to fourteen fingers. Hence conversely, fourteen being subtracted from it, the remainder, a third of the hypotenuse, is x - 14. Thrice this, which is the hypotenuse, is 3x - 42. The square of it, 9x² - 252x + 1764, is equal to the square of the hypotenuse, x² + 144. On making equi-clearance, the two sides become $8x² - 252x + 0 = 0x² + 0x - 1620.$ Multiplying both these sides by 2 and adding the square of 63, the roots are $4x - 63 = ± (0x + 27).$ On forming an equation with these sides again, and (proceeding) as before, the values of x are 45/2, 9.

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(Thus) the value of the shadow is 45/2 or 9. The second value of the shadow is less than 14, so, on account of impracticability, it should not be accepted. Hence it has been said 'twofold values occasionally.' This will be an exception to what has been stated in the algebra of Padmanâbha, *viz*...."

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**Known to Mahâvîra.** It has been stated before that Mahâvîra (850) knew that the quadratic has two roots. We shall now substantiate it by the following rules and illustrations from his work.

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"One-sixteenth of a collection of peacocks multiplied by itself, was on the mango tree ; 1/9 of the remainder multiplied by itself together with 14 were on the *tamâla* tree. How many were they ?"¹

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¹ *GSS*, iv. 59.

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If x be the number of peacocks in the collection, the problem leads to the quadratic equation $x/16 × x/16 + 15x/(16 × 9) × 15x/(16 × 9) + 14 = x.$

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This is a particular case of the type of equations contemplated by the author $a/b x² - x + c = 0.$

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The following rule has been given for its solution.

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"The quotient of its denominator divided by its numerator, less four times the remainder, is multiplied by that denominator (as divided by the numerator). The square-root of this should be *added to* and *subtracted from* that denominator (as divided by the numerator) ; half that is the total quantity."²

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² *GSS*, iv. 57.

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Thus $x = (b/a ± √((b/a - 4c)b/a))/2.$

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Certain other problems¹ lead to equations of the type $(a/bx d)² + c = x.$ The solution is given as

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¹ *GSS*, iv. 62-4.

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"Half the denominator divided by its numerator is increased or diminished by the quantity to be subtracted or added. The square of this is diminished by the square of the quantity to be subtracted or added and by the remainder. The square-root of the result *added to or subtracted from* the square-root (of the square obtained before) and divided by the fractional part, will be the value (of the unknown)."²

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*i.e.*, x = (b/2a ± d) ± √((b/2a ± d)² - d² - c) ÷ a/b.

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We need not add further instances to prove that Mahâvîra recognised both the roots of a quadratic equation.³ There are, however, a few problems in which he has taken into consideration only one of the roots.⁴ For instance, take the equation on page 66, $¼x + 2√x + 15 = x,$ or ¾x - 2√x = 15. Therefore √x = 4/3 ± √((4/3)² + 4.15/3), or √x = (4/3 ± 14/3) = 6 or -10/3. ∴ x = 36. The root can not be negative, hence the negative value of the radical is neglected in the rule.

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² *GSS*, iv. 61.

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³ More instances will be found in *GSS*, vi. 29ff.

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⁴ See *GSS*, iv. 33-52.

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**Brahmagupta.** The existence of two roots of a quadratic equation appears to have been known also to Brahmagupta (628). In illustration of his rules for the solution of the quadratic, he has stated two problems involving practically the same equation.

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(1) “The square-root of the residue of the revolution of the sun less 2 is diminished by 1, multiplied by 10 and added by 2: when will this be equal to the residue of the revolution of the sun less 1, on Wednesday ?”¹

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¹ *BrSpSi*, xviii. 49.

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(2) “When will the square of one-fourth the residue of the exceeding months less three, be equal to the residue of the exceeding months ?”²

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Following Pṛthûdakasvâmî let us take in *Example* 1 the residue of the revolution of the sun to be x² + 2; then by the question $10(x - 1) + 2 = x² + 1,$ $or x² - 10x = - 9.$ In *Example* 2, put 4x for the residue of the exceeding months; then $(x - 3)² = 4x,$ $or x² - 10x = - 9.$ Now, by the second rule of Brahmagupta, retaining both the signs of the radical, we get $x = 5 ± √(25 - 9) = 9 or 1.$ As shown by Pṛthûdakasvâmî, the first value is taken for *Example* 1 and second value for *Example* 2. Thus it appears that Brahmagupta uses sometimes the positive and at other times the negative sign with the radical. Hence it seems that Brahmagupta knew that a quadratic equation has two roots, though from considerations of utility in his problems, he retains only one of them.

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² *BrSpSi*, xviii. 50.

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