factors by ma+ax + b, on the supposition that m is a factor of 6, and b a factor of 10. = Putting m 2, and proceeding as in the last question, we get the results 20, 10, and 48, having 2+ a + b, b, and 2a + b for their divisors. After a few trials, we find 2 +a+b=10, b = 5, and 2a+b= 4, which give a = 3; consequently, ma + ax + b becomes 2x2 + 3x + 5, which divides the given equation, and gives the quotient 32-7x+2; consequently, the equation is reduced to (2x2 + 3x + 5) × (3x2 7x + 2) = 0. Putting the factors equal to 0, we easily get 3√31 4 1 -3+1-31 4 2, and for the roots of the given equation. 3 Remarks.-1st. The preceding methods for finding the quadratic factors are essentially the same as those given by Newton, at p. 40, etc., of his Universal Arithmetic. 2d. It is clear that the preceding question can be solved by first finding the simple factors; but we have found the quadratic factor, for the purpose of showing that the quadratic factors may be found without a knowledge of the simple factors. 11. Given, the first term of an increasing geometrical progression equal to 2, and the sum of its first four terms equal to 80, to find the progression. Let a stand for the ratio, then the conditions give the equation a3+x2 + x = 39, whose solution gives x = 3; consequently, the progression is 2, 6, 18, 54, etc. 12. Divide $1000 between A and B, so that the cube of A's share shall equal the square of B's share. Let a2 and a stand for the shares of A and B, and one of the conditions is satisfied; and to satisfy the other, we must solve the equation a3 + x2 = 1000. Developing the root of this equation, by (b) of Sec tion XV., we get x = 10 + + etc. 3 90 4050 3645000 100.333. + 0.0111. - 0.0002469+0.00000027 etc., which, by taking five terms, gives x 9.677531, very nearly. Hence, the shares of A and B are easily found to be $93.6546 and $906.3455, very nearly. 13. To find a number, such that the sum of its square, cube, and fourth roots shall equal 88, supposing the square and fourth roots (each) to be taken with the positive sign. Let 12 stand for the number, then the conditions give the equation a1388, or x+x+x3-88=0. + = Let a stand for the root, then the first member of the last equation is exactly divisible by x--a; consequently, putting 1, 0, and 1 successively for x in the equation and divisor, the results 85, — 88, - 87 have 1 a, for corresponding divisors. a, and = -1- a 2; consequently, since the first member of the equation is divisible by - 2, we have 2 for its root, and 212 212 = 4096 is the required number., 14. To find a from the equation V — 5 — √ 8 — x2 = 2, without freeing the equation from radicals. Let a stand for a near value of x, and put x= a + y; then we shall approximately have x2 = a2 + 2ay, x3 = a3 + 3a2y, √ (@3 — 5); ; consequently, the given equation is reduced to α √ a3-5-√8 — a2 + +(√(5)2+(8-a2))y = 2 nearly, which gives y = (2 + √ 8 — a2 — † (a3 — 5)) ÷ α √(8 — a2) a2 (a3 — 5)2 + After a few trials, we find that it will answer our purpose to put a 2.8; which gives 1/8 a 0.4, Va3 - 5 = 2.568859, and 2 + √ 8 — a2 — √ a3-50.168859; we = quently, xa+y=2.8 0.020 2.78 very nearly. By putting 2.78 for a, and repeating the process, we shall get more nearly, and so on, to any desired degree of exactness. Remark. It is easy to perceive, from what is said at p. 494, how we may apply (b) (see p. 494) to find y, by a converging series, and thence to get a = a + y to any degree of exactness. 15. To find x and y from the equations Va-2y= 5 and Vx + 3 Vy = 17, without removing the surds. Let a and b stand for near values of x and y, and put x = a + p, y = b + q for their exact values; then we shall, ap proximately, have √x = √a+ Ny = √b + 2 Va Hence, the proposed equations are reduced to a + 3q = 216 3√ a 216 Solving these equations (regarding p and q as the unknown letters), we get p = (9A√☎ + 4B √) ÷ 2 √ a 34a2 4√6), and After a few trials, we find that it will answer our purpose to put a 105 and b 18, which give 5 - Va+2 √b = A = 0.005 very nearly, and 17-va-3b=B=-0.445 = Hence we deduce p2.403 and q = -1.117; and thence we have x=a+p=102.597, and y = b+q=16.883, for nearer values of x and y. Substituting these values of x and y in the given equations, we have V-2 Vy 4.997, and V + 3 √y = 17.004 very nearly. = If we put 102.597 and 16.883 for a and b in x = a + p and y=b+ 2, and repeat the process, we shall get more correct values of and y, and so on, to any required degree of exactness. Remarks.-1st. We have given this and the preceding question for the purpose of showing how, from the proposed equations, we may proceed to find the unknown letters, without any previous reduction of the equations. 2d. It is clear that a similar process can be applied to any number of equations which contain as many unknown letters as there are equations, provided the unknown letters are real. For if, by trial, we find values of the unknown letters which nearly satisfy the equations, then we can, as in the present question, find the corrections of the assumed values of the unknown letters to any degree of exactness that may be required. 3d. Thus, it is evident (after we have assumed suitable values for the unknown letters) that we may regard the solution of equations as being reduced to the solution of equations of the first degree. 16. To find the approximate values of x, y, and z, from the equations + yz = 16, y2 + xz = 17, 22 + xy = 18. 1. Since the right members of the equations do not differ much from equality, it is clear that we may solve them on the supposition that x, y, and z are nearly equal to each other; consequently, by adding them, and putting 6 for the sum of the terms in the first member of the resulting equation, we shall get 6 51 or ť2=8.5, whose positive square root gives t2.9 for the first approximation to the values of x, y, and z. Putting 2.9+p, y = 2.9+q, 22.9+r, and rejectx = ing p, q, 2, PL, pr, and qr (on account of their supposed minuteness), we get a 8.41 + 5.8p, y = 8.41 + 5.8q, z2--8.415.8, ay = 8.41 + 2.9(p + q), xz = 8.41 + 2.9(p + r), and yz8.41 + 2.9g+r); consequently (by substituting from these equations), the given equations are easily reduced to 5.8p2.9q+r)=0.82, 5.8q + 2.9(p+r) = 0.18, and 5.8r+2.9(p + q) = 1.18, or 2p + q r = −0.282, 2q +p +r=0.062, and 2r+p+q= 0.406. p = have x By adding these equations, we get p+q+r=0.046, which being subtracted from the successive equations gives 0.328, q=0.016, and r = 0.360; consequently, we 2.9+ p = 2.572, y = 2.9+ q2.916, and 2 = 2.9 + r = 3.260, which are correct to one decimal place in each. If we now put x = 2.5 +p, y 2.9+q, and z = 3.2 +r, and proceed as before, we shall get more correct values of x, y, and z, and so on, to any degree of exactness required. It may be added that the values of x, y, and z, to six decimal places, are 2.525513, y= 2.969152, and z = x = 3.240580. 2. The given equations also admit of three more answers, as is manifest from the following considerations. Thus, if we at first reject x2, on the supposition that x is very small (so that may at first be omitted), the equations will become yz=16, y2+xz = 17, and 2 + xy = 18. Because these equations are nearly satisfied by putting y = 4 and 2 = 4, we shall assume y = 4+ q and z = 4+r; then, rejecting q2, r2, qx, rx, and qr (on account of their supposed minuteness), we shall have y = 16 +8q, z2 = 16 + 8r, xz = 4x, xy=4x, and yz 16 + 4(q + r). Hence, by substitution, the equations will be reduced to 16 + 4(q + r) = 16, 16 + 8q + 4x= 17, and 16 + 8r + 4x = 18, or q=r, 8q + 4x = 1, 8r+4x=2; consequently, by adding the last two of these (since q + r = 0), we shall get 8x= = 3, or x = = 0.375 nearly, for which we may clearly put x = 0.4. 3 8 Substituting 0.4 for x in the equations 8q + 4x = 1 and 8r 42, they become Sq + 1.61 and 8r+ 1.6 = 2, thence y=4+9 and 2=4+r become y=3.925 and z=4.05. Hence, we have x 0.4, y = 3.925, z= 4.05, which are (each) correct to one decimal place. = If we put = 0.4 +p, y 3.9+q, and z = 4.0 + r, in the given equations, we can find x, y, and z, to any required degree of exactness; and by continuing the approximation, |