Combining this equation with y+b, by addition and sub This method of solution applies only to those cubic equations which have one real root and two impossible ones. For let h+k, h − √ k, and 2h be the three roots of a cubic equation; then (125) that equation will be x3 (3 h2 + k) x + 2h (h1 — k) = 0. Now if we substitute — (3h + k) for a and h2 ke 2 + 1 h2 Now this expression is always impossible when k is positive, and when k is positive, the roots h+k and h√k are real; and on the contrary, when k is negative, the radical value above is real, and the roots h + √k and hk are impossible. Cardan's method of solution is hence useless in practice when the roots are all real, as they cannot be computed by means of it, on account of the occurrence of an imaginary or impossible quantity, and it is only when the cubic equation has one real root that it is applicable. BIQUADRATIC EQUATIONS.—SIMPSON'S METHOD OF SOLUTION. 132. Let the general form of a biquadratic equation x+p+qx3 +rx+8=0 be transformed into the form x + a x2 + bx + c = 0, where the third power of the unknown quantity is absent; then if we assume x2 + a x2 + bx + c = (x2 + y)2 — (hx + k)2 =x+(2y-h2) x2 - 2 h k x + y2 — k2, and equate the coefficients of the same powers of x on both sides, then 2y-ha, or h2 = 2y- -a • b, or 2hk-b y2 - k2 = c, or k2 = y2 - c (1) Now it is obvious that the square of (2) is equal to four times the product of (1) and (3); hence 1 2 1 4 (2 y − a) (y2 — c) = bo, or y' — ' ay' — c y = ¦ ¦ (b2 — 4 a c). 8 Remove the second term of this cubic, and find the value of y by the previous method; hence h = ± √ (2 y − a) and k = √(y2 — c) will be known, and they must have opposite signs, for 2 hk is a negative quantity by (2). Then the first side 1 + a x2 + bx + c of the first equation being equal to 0, we have (x2 + y)2 = (hx + k)2, or x2 + y = ± (hx + k) ; hence, since y, h, and k are all known, the solution of the quadratics x2 hx + y − k = 0, and x2 + hx + y + k = 0 will afford the four values of x. These methods may be applied to any of the examples which have already been given in (129) and (130). INDETERMINATE ANALYSIS. 133. We have seen that, when the conditions of a question furnish as many independent and consistent equations as there are unknown quantities to be determined, the values of these unknowns may be found; but if the number of unknown quantities exceeds the number of equations, the question will admit of various solutions, and it is hence said to be indeterminate. If integer values only are required, the number of solutions will be greatly restricted; and if negative values be excluded, the number of solutions in some cases may be very limited, while in others the number of positive integer solutions is unlimited. Indeterminate analysis of the first degree. Indeterminate equations of the first degree are of the form ax+by=c, or ax+by+cz=d, where a, b, c, d are given whole numbers, and x, y, z are limited to positive integer values. When the formula a x + by = c admits of integer values of x and y, the coefficients a and b cannot have a common divisor which is not also a divisor of c. For if a = m d, and b = me, then c, hence dx ±ey: If a and b are prime to. each other, the solution of an equation of the form a xby cis always possible, and the number of solutions is unlimited, but if the equation be of the form ax + by = c, the number of solutions is always limited, and in some cases no solution can be obtained. Let a x-by= c be the equation in which a is less than b; then we have x= by + c a where my and n are the nearest quotients, whether in excess or defect, arising from dividing by and c by a; hence d and e are each less than Let the last fraction, which must obviously be a whole number, be put=v, then we have dy ±e = a v, and as before a. where ƒ and g are each less than d. Put the last fractional expression, which must also be a whole number, v1, and repeat the process until we arrive at a fractional expression in which the coefficient of the subsidiary unknown quantity v, is unity. Let this fraction = p, then we have an expression of the form, v1 = hp ± k; and by reversing the steps, we get in succession the values of the subsidiary unknown quantities which have been employed, and finally those of x and y. EXAMPLES. 1. Given 7 x + 12y = 50, to find the values of x and y in whole numbers. Here the smaller coefficient is 7, and therefore we have Consequently we see that if x is to be a whole positive number, the value of p is limited to zero; hence when p=0, we have x=2 and y = 3, which are the only integer values of the two unknown quantities. 2. A person purchased between 50 and 60 horses and oxen; he paid 31 dollars for each horse, and 20 dollars for each ox, and he found that. the oxen cost him seven dollars more than the horses; how many of each did he buy? Let x denote the number of horses, and y the number of oxen; then 20 y 31x= 31 x + 7 y = =2x 20 20 9 x 20 7 v; then will 9 x 20 v + 7; hence In these expressions for x and y, we see that p may have all possible values from 0 to any extent; hence making p = 0, p = 1, p = 2, etc., x = 3, 23, 43, 63, 83, 103, etc.; we get y = 5, 36, 67, 98, 129, 160, etc.; = 36, therefore by the limitation in the question, we have x = 23 and y the respective numbers of horses and oxen that were purchased. The total number of solutions is unlimited. 3. How many ounces of gold, of 17 and 22 carats fine, must be mixed with 5 ounces of 18 carats fine, so that the composition may be 20 carats fine? = Let the number of ounces of 17 carats fine, and y the number of 22 carats fine, then by the condition of the question we get 17x+22y + 5 × 18 = 20 (x + y + 5) ; or 2 y 3x + 10; 134. If there be only one equation to determine three unknown quantities, as ax+by+cz=d; then by transposition, ax + by =dez, and by giving to z all its different integer values, we shall obtain in the usual manner all the corresponding values of x and y. The values of x and y cannot be less than unity; therefore the highest value of z cannot exceed the value derived from the equation 2= d-a-b C EXAMPLES FOR PRACTICE. 1. Given 14 x = 5y + 7, to find the least values of x and y. Ans. x 3, y=7. 2. Given 27 x + 16 y 1600, to find the least values of x and y. Ans. x 48, y = 19. = 3. Given 11x+5y = 254, to find all the possible values of x and y. Ans. x 19, 14, 9, 4; y = 9, 20, 31, 42. 4. In how many ways can 201. be paid without using any other coin than half-guineas and half-crowns. Ans. In 7 different ways. 2y+z5 and 2x + y = 7, to find the values 5. Given x of x, y, z. Ans. x = 5, 6, 7, 8 ... 6. A person bought 100 animals for 100l., namely, oxen at 57. each, sheep at 17., and fowls at 1 shilling each; how many of each kind did he purchase? Ans. 19 oxen, 1 sheep, and 80 fowls. VOL. I. N 7. I owe a person 3 shillings, and have nothing about me but guineas, and he has nothing but crowns; how must I discharge the debt? Ans. I must give 3 guineas, and receive 12 crowns. 8. A jeweller requires to mix gold of 14, 11, and 9 carats fine, so as to make a composition of 20 ounces of 12 carats fine; find the quantities of each kind of gold to form the required mixture. Ans. 8, 10, 2 ounces, or 10, 5, 5 ounces. 9. Given 5x7 y + 11 z = 224, to find all the possible values of x, y, and z in whole numbers. Ans. The number of ways is 59. INDETERMINATE COEFFICIENTS. 135. If two series of quantities be equal to each other, as a + bx + c x2 + dx3 +...=· A+B x + C x2+ Dx23+ . whatever be the value of x, and where the coefficients are independent of x; then will Aa, Bb, Cc, D=d, etc. For by transposition we have a A (Bb) x + (C − c) x2 + (D − d) x2 + . . . .; and if a is not equal to A, then a A will be some constant quantity; but the second side of this equation varies as x varies, and may be made less than the fixed quantity a- A by taking a sufficiently small value for x, which is absurd, and therefore we must have a = Á. The proposed equation, then, separates itself into two equations, viz., A = a, and bx + c x2 + dx23+. . . = B x + С x2 + Dx23 + Dividing each side of the latter by x gives b + c x + dx2 + ex2 + . . . = B+C+D x2 + Ex3 + which by similar reasoning separates into the two equations, Bb, and c + dx + ex2 + . . = C + D x + Ex2 + and the latter gives, as before, C = c, and so on; hence we have A = a, B = b, C : = c, D=d, etc. Let EXAMPLES. 1. Required the development of √(a− x) in a series. = Hence A a, 2 AB 1, 2 AC+ B' 0, 2 AD + 2BC= 0, etc.; = |