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History of Hindu Mathematics: A Source Book v2.c17.s9: 11. Equations of Higher Degrees

History of Hindu Mathematics: A Source Book v2.c17.s9

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11. Equations of Higher Degrees

v2.c17.s9.1

**Cubic and Biquadratic.** The Hindus did not achieve much in the solution of the cubic and biquadratic equations. Bhâskara II (1150) attempted the application of the method of the *madhyamâharaṇa* (elimination of the middle) to those equations also so as to reduce them by means of advantageous transformations and introduction of auxiliary quantities to simple and quadratic equations respectively. He thus anticipated one of the modern methods of solving the biquadratic. "If, however," observes Bhâskara II, "due to the presence of the cube, biquadrate, etc., the work (of reduction) cannot proceed any further, after the performance of such operations, for want of a root of the unknown side (of an equation), then multiplied by 200 and added to the square of the number, and then multiplied by 2 and subtracted from the fourth power of the number will become one myriad less unity? Tell that number if thou be conversant with the operations of analysis.

v2.c17.s9.2

*Solution.* "Here the number is *x*; multiplied by 200 it becomes 200*x*; added to the square of the number, becomes *x*² + 200*x*; this being multiplied by two, 2*x*² + 400*x*; by this being diminished the fourth power of the number, namely, this *x*⁴, becomes *x*⁴ — 2*x*² — 400*x*. This is equal to a myriad less unity. Equi-clearance having been made, the two sides will be *x*⁴ — 2*x*² — 400*x* = 0*x*⁴ + 0*x*² + 0*x* + 9999. Here on adding four hundred *x* plus unity to the first side, the root can be extracted, but on adding the same to the other side, there will be no root of it. Thus the work (of reduction) does not proceed. Hence here ingenuity (is called for). Here adding to both the sides four times the square of *x*, four hundred *x* and unity and then extracting roots, we get *x*² + 0*x* + 1 = 0*x*² + 2*x* + 100. Again, forming equation with these and proceeding as before, the value of *x* is obtained as 11. In similar instances the value of the unknown must be determined by the ingenuity of the mathematician."¹

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¹ *BBi*, pp. 64f.

v2.c17.s9.4

**Higher Equations.** Mahâvîra considered certain simple equations of higher degrees in connection with the treatment of the geometric series. They are of the type (*i*) *axⁿ* = *q*, (*ii*) a(xn - 1)/(x - 1) = p;

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where a is the first term of a G. P., q its *guṇadhana*, i.e., (n+1)th term, p its sum and x the unknown common ratio.

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To solve equation (i) Mahâvîra says, "That which on multiplication by itself as many times as the number of terms becomes equal to the *guṇadhana* divided by the first term, is the common ratio."[1]

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$i.e., x = [n]q/p.$ In other words x is the nth root of q/p. But how to find such a root he does not attempt to indicate. His rule for solving an equation of the type (ii) is as follows :

v2.c17.s9.8

"That by which the sum divided by the first term is divisible again and again, subtracting unity every time, is the common ratio."[2]

v2.c17.s9.9

The method will be better understood from the solution of the following example :

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"(Of a certain series in G. P.) the first term is 3, number of terms 6 and sum 4095. What is the common ratio ?"[3]

v2.c17.s9.11

Thus $3 (x⁶ - 1)/(x - 1) = 4095,$ or $3(x⁵ + x⁴ + x³ + x² + x + 1) = 4095,$ a quintic equation. Here dividing 4095 by 3 we get 1365. Now let us try with the divisor 4; we have (1365 - 1)/4 = 341; (341 - 1)/4 = 85; (85 - 1)/4 = 21; (21 - 1)/4 = 5; (5 - 1)/4 = 1; (1 - 1)/4 = 0. So the number 1365 is exhausted on 6 successive divisions by 4, in the way indicated in the rule. Hence x = 4. What suggested the method is clearly this : $a (xn - 1)/(x - 1) ÷ a = (xn - 1)/(x - 1); (xn - 1)/(x - 1) - 1 = x((xn-1 - 1)/(x - 1));$ which is divisible by x. However, the solution is obtained in every case by trial only.

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¹ ^1 *GSS*, ii. 97.

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² ^2 *GSS*, ii. 101.

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³ ^3 *GSS*, ii. 102; compare also Rangacarya's note thereto.

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Mahâvîra has treated some equations of the following general type:

v2.c17.s9.16

a_1√(b_1 x) & + a_2√(b_2(x - a_1√(b_1 x))) / & + a_3√(b_3{(x - a_1√(b_1 x)) - a_2√(b_2(x - a_1√(b_1 x)))}) / & + \dots + R = x; or (x - a_1 & √(b_1 x)) - a_2√(b_2(x - a_1√(b_1 x))) / & - a_3√(b_3{(x - a_1√(b_1 x)) - a_2√(b_2(x - a_1√(b_1 x)))}) / & - \dots = R.

v2.c17.s9.17

If there be r terms on the left hand side, then on rationalisation, we shall have an equation of 2rth degree in x. By proper substitutions, the equation will be ultimately reduced to a quadratic equation of the form $X - A√BX = R,$ whose solution is given by Mahâvîra as $X = (A + √(A² + 4R/B))/2² × B.$

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This result has been termed by him, the "essence" (*sâra*) of the general equation.¹ Mahâvîra gives two problems involving equations of the above type.

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¹ *GSS*, iv. 51, 52.

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(1) "(Of a herd of elephants) nine times the square-root of the two-thirds plus six times the square-root of the three-fifths of the remainder (entered the deep forest); (the remaining) 24 elephants with their round temples wet with the stream of exuding ichor, were seen by me in a forest. How many were the elephants (in the herd) ?"¹

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¹ *GSS*, iv. 54-5.

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If x be the number of elephants in the herd, then by the statement of the problem

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$9√(2x/3) + 6√(⅗(x - 9√(2x/3))) + 24 = x.$

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Put y = x - 9√(2x/3); then the equation becomes

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$y - 6√(3y/5) = 24.$

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Therefore y = 60 or 48/5.

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Hence x - 9√(2x/3) = 60,

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whence x = 150, 24.

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Again x - 9√(2x/3) = 48/5,

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whence x = ⅗(61 ± 3√385).

v2.c17.s9.31

Of the four values of x obtained above, only the value x = 150 can satisfy all the conditions of the problem; others are inapplicable. That will explain why Mahāvîra has retained in his solution only the positive sign of the radical.

v2.c17.s9.32

(2) "Four times the square-root of the half of a collection of boars went into a forest where tigers were at play; twice the square-root of the tenth part of the remainder multiplied by 4 went to a mountain; 9 times the square-root of half the remainder went to the bank of a river; boars numbering seven times eight were seen in the forest. Tell their number."²

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² *GSS*, iv. 56.

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If x be the total number of boars in the collection,

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$4√(x/2) + 8√(1/10(x - 4√(x/2)))$ $+ 9√(½(x - 4√(x/2)) - 8√(1/10(x - 4√(x/2))))$ $+ 56 = x.$ Put y = x - 4√(x/2); then $y - 8√(y/10) - 9√((y - 8√(y/10))/2) = 56.$ Again put z = y - 8√(y/10); then $z - 9√(z/2) = 56.$ Therefore z = ((9 + √(81 + 4.2.56))/2)² × ½ = 128. Then y - 8√(y/10) = 128; whence y = ((8 + √(64 + 10.4.128))/2)² × 1/10 = 160. Again x - 4√(x/2) = 160; hence x = ((4 + √(16 + 4.2.160))/2)² × ½ = 200.

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Note that according to the problem the positive value of the radical has always to be taken.

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