we easily get a = 0.418919, y= 3.912226, z=4.044884, correct to six decimal places. Similarly, if we reject y2, and then 22, on account of their supposed minuteness, as before, we shall get two other sets of answers, expressed by = 4.003766, y=-0.007099, 2=4.245989, and x=4.173264, y=4.287022, z=-0.330331. Since the equations are not affected by changing the signs of x, y, and z, it follows that, if the signs of the values of x, y, and z in the preceding answers are changed, the results will satisfy the equations. Remarks.-1st. This example, under the form 2 + yz = a=16, y2+xz = b = 17, 22 + xy=c=18, was proposed by Dr. Pell, near the close of the 17th century, and solved in a general manner by Dr. Wallis, who reduced its solution to that of an equation of the fourth degree, by eliminating y and z from the equations. 2d. When 16, 17, and 18 are put for a, b, and c, it is clear that our solution must be equivalent to that of Wallis, obtained from the solution of his equation of the fourth degree. Hence, if an equation containing one unknown letter is reduced to equations of lower degrees, containing more than one unknown letter, it is manifest that their solution will be equivalent to that of the reduced equation. 17. To find the roots of a3 7x+7 = = 0. By putting -7=y or a2 = 7+ y, the equation becomes xy+70, which shows that x and y must have unlike signs; consequently, the solution of the proposed equation is reduced to that of the equations 7+ y and xy+7=0, such that x and y must have contrary signs. = 1. After a few trials, we find that, by putting y = 2.29, we shall get a2 = 9.29, whose negative square root gives x= -3.047, and thence xy + 7 becomes 6.97763+ 7, a positive result; and in like manner, y = 2.30 gives x 3.049, and thence xy +7 becomes 7.0127+7, a negative result; consequently, 2.29 and 2.30 are limits to y, and 3.047, 3.049 are the corresponding limits of x. Putting x = 3.048 +p and y= = 2.295+q, and proceeding as in the preceding examples, we get q=0.004696 — 6.096p and 2.295p-3.048q +0.0048400. Solving these equations, we have p=0.000917 and q = 0.000894, and thence x= If we put x = -- - 3.048917 + P and 3.048917 and y = 2.295894. y = = 2.295894 +9, and proceed as before, and so on, then x can be found to any required degree of exactness. 2. Because y=4.1 gives = 1.70, and xy + 7 = − 6.977, and that y=4.2 gives = 1.67 and xy +7= — 7.014 +7, it follows that the equation has another root, which lies between 1.70 and 1.67; and proceeding as before, this root will be found to be 1.692021 +. In like manner, since y=5.1 gives a = 1.378 and xy +7= 7.02787, and that y= and ay 5.2 gives x = 1.341 76.9732 +7, it follows that the equation has another root, which lies between 1.37 and 1.34; and proceeding as above, this root will be found to be 1.356895 +. 18. To find the real roots of 3 5+8−1=0. 52 By putting a 5x+8=y, we have xy-1=0, or xy = 1; consequently, we have to satisfy the equations ∞2 +8=y and xy=1, noticing that, by the rule of Descartes, the equation has no negative roots. After a few trials, it will be found that y= 7.4 in the first equation gives x = 0.123 nearly, and the second equation gives xy = 0.9102 instead of 1. In like manner, y = 7.3 gives x = 0.145, and xy becomes xy=1.0585; consequently, 0.123 and 0.145 are limits to the value of x. Hence, the root is easily found; and since r and y must (each) be positive, it is easy to perceive that the remaining roots of the equation are imaginary. ANOTHER GENERAL METHOD OF DEVELOPING THE ROOTS OF EQUATIONS. (1.) Let A+ A1æ2¬1 + Аçã2¬2 + ... + An-12 + An = 0, (1), stand for any equation; then, if r, 71, 72, T3 ・ ・ ・ ・ In−1 represent its roots, and is put for x, (1) is easily reduced to У 2 Ay" + A1y"−1z + А2y”-222 + .... + An-1y2”-1 + A„~” = A(y — zr). (y — zr1). (y — zr1) × (y — zr3) · (y − zr1) × (2.) Supposing any one of the factors in (2), as y - zr, is either 0, or very nearly equal to 0, then we shall have (either y x exactly or very nearly) yzr or == r. Hence, if y+t = - 2 and zvare (separately) put for y and z in (2), we shall easily get nAy1 + (n − 1)A1yn−2z + (n − 2)A ̧y13¬31⁄23 +.... +A-11 A(y zr1). (y-zr2) × (y-zr's) X.... X (y-1) nearly (by omitting the terms which involve the factor y zr, on account of its supposed minuteness), for the coefficient of the simple power of t; and in like manner we have Ay"-+2Ay"-z+3Ay"-8-2 + 4A1y”—123 + .... + nA2"-1 Ary-zr). (y— zr2). (y — zr3) × (y — zr1). (y-zr1) x.. × (y-zrn-1), for the coefficient of the simple power of v, where it is clear that, in these coefficients, we may put 2 = 1 and y = x. ― = .... (3.) Since the coefficient of v divided by that of t gives r for the quotient, it follows, if we know the value of x, which gives r (either exactly or very nearly), that the quotient resulting from the division of the coefficient of v by that of t, will (exactly or very nearly) equal the negative of the root r, which corresponds to xr, and that whether r is positive or negative, real or imaginary. Hence, if r is real and positive, the coefficients of t and v must have unlike signs; and if is real and negative, the coefficients of t and v must have like signs. x in the coefficients of t and (4.) If we put z = 1 and y v, and then multiply the coefficient of t by x, and add that of v to the product, and divide the sum by n, it is clear that the result will agree with the first member of (1). x r Hence, if a differs but little from r, and we divide the first member of (1) by the coefficient of t, the quotient will evidently equal very nearly; consequently, if we asssume x, and find that the quotient resulting from the division of the coefficient of v by that of t is numerically less. than the assumed value of x, it follows that too great a numerical value has been assumed for x, and vice versa. (5.) Regarding r as being the least root of (1), r1 as the next greater root, and so on (considering the numerically greatest negative root as being the least, and so on); then, 33 supposing to increase from being less than r, and to pass through the roots, it is clear, from an inspection of the coefficients of t and v, that their signs will not be changed by the passage of a through r, and that their signs will be changed by the passage of x through r1, since r and r1 will be interchanged, and so on. Hence, if the signs of the coefficients of t and v do not change, it follows, if (1) is of an even degree, that it has no real root, and if (1) is of an odd degree, that it has only one real root. Hence, too, if x (during its increase) introduces one change of the signs of the coefficients of t and v, it is clear that (1) will (generally) have two real roots, one of which will be less than the greater value of a; and if a (during its increase) produces another change of the signs of the coefficients of t and v, it is also clear that (1) has (generally) three real roots, two of which will be less than the greater value of x, and so on; noticing that, in what has been said, the roots of (1) are supposed to be unequal. (6.) If (1) has equal roots, it is clear that if we put the greatest common divisor of the coefficients of t and v equal to 0, and solve the resulting equation, and then increase the number of equal roots of each kind thus found by unity, that we shall get the number of equal roots of each kind contained in (1). EXAMPLES. 1. To find the roots of a2 Here 2x 5 and 5x+12 are the coefficients of t and v, which are suitable for finding the positive roots; and if we put x for x, they will become - 2x - 5 and 5x + 12, which are adapted to the negative roots; noticing that e must be positive. Since 25 and 52 + 12 have unlike signs when x is positive, it follows that the equation has no negative roots, which is in accordance with the rule of Descartes, since the given equation is complete and has no permanence of signs. To find the positive roots from 2x5 and -5x+12, we suppose a to increase from x = 0, and to pass through them; then, as they become -5 and 12 when a 0, it follows that one root must be found from 2x 5 and 5x + 12, when the value of the first has the sign -, and the second the sign +. Consequently, putting 1 and 2 successively for x, we find that by putting 2 for a they become 1 and 2, which give 21-2, and of course 2 is a positive root of the given equation. Because 2-5 and -5x+12 must change their signs before the remaining root can be found, we put 3 for x, and thence get 2x 51 and -5x+12 - 3, which give - 3÷1 = −3, and of course 3 is the remaining positive root. Hence, the given equation is reduced to a 5x + 6 = (x -2). (x-3)= 0, which is clearly as it ought to be. 2. To find the real roots of x2 2x+3=0. Here the coefficients of t and v are 2x 2 and 2x + 6, which clearly show that the equation has no negative roots. Also, since the coefficients show that the positive roots (if there are any) must be greater than 1, it follows that there can be no positive roots; for if there can be any positive roots greater than 1, then such a positive value can be assigned to x as will change the signs of both coefficients; which is clearly impossible when a is greater than 1. 3. "To find the real roots of Here the coefficients of t and v become 3x2 10x 18 and 5x36x + 216, which will enable us to find the positive roots; and changing x into, they become 3x2+10x -18 and 5x2 + 36x + 216, which, by assigning positive values to x, will enable us to find the negative roots. After a few trials, we find that by putting 4 for a in 3x2 + 10x18 and 5x2 + 36x + 216, they become 70 and 280, which give 280704; consequently, 4 is a negative root of the given equation, and it is clear that it is the only negative root. = To find the positive roots from 3-10-18 and - 5a2 36x + 216, we observe that when x = 0 they become -18 and 216, and that when a 7, they become 59 and 281; consequently, since the coefficients change their signs between x = 0 and x=7, it is clear that the positive roots must lie between x = 0 and x = 7. After a few trials, we find that by putting = 3, the coef |