On trial, this is found to be a very small matter too little. Take therefore again, x = 3.59727, and next = 3.59728, and repeat the operation as follows: First, Supp. x = 3.59727.1 Second, Supp. x = 3.59728 Log. of 3.59727 is 0.555973 Log. of 3.59728 is 0-555974 3.59727 X log. 3-59728 X log. of 3.59727 = 1.9999854 the true number 2.0000000 of 3-59728 = 1.9999953 the true number 2.0000000 error, too little, -0.0000146 error, too little, -0.0000047 -0.0000047 0.0000099 diff. of the errors. Then, As 0000099: 00001 :: 0000047: 00000474747* the cor. 3.59728000000 added to gives nearly the value of x = 3.59728474747 Ex. 5. To find the value of x in the equation x3 + 10x3 + 52 = 260. Ans. x 4.1179857. Ex. 6. To find the value of x in the equation x3 2x 50. Ans. 3.8648854. The Author has here followed the general rule in finding as many additional figures as were known before: viz. The 6 figures 3.59728, but as the logarithms here used are to 6 places only, we cannot depend on more than 6 figures in the answer; we have no reason, therefore, to suppose any of the figures in 00000474747, to be correct. The log. of 3.59728 to 15 places is 0'55597 42431 34677 the log. of 359727 to 15 places is 0-55597 30358 47267 which logarithms multiplied by their respective numbers give the following products: 1.99999 50253 435122 both true to the last figure. 1.99998 51226 62298 Therefore, the errors are 49746 56488 148773 37702 99026 81214. Now since only 6 additional figures are to be obtained, we may omit the last three figures in these errors; and state thus: as diff. of errors 9902681: diff. of sup. 1 :: error 4974656: the correction 502354, which united to 3.59728 gives us the true value of x= 3·59728502354. Ex. 7. To find the value of x in the equation x2 + 2x2 23 x = 70. Ans. x 5·13457. Ans. 17 22 14.95407. Ex. 8. To find the value of x in the equation x3 +54x350. Ex. 9. To find the value of x in the equation 4 -3x2 10000. Ans. x 10.2609. 75x Ex. 10. To find the value of x in the equation 2x4 - 16x +40 x2 30x =- 1. Ans. x 1·284724. -- Ex. 11. To find the value of x in the equation x + 2x4 +3x3 + 4x2 + 5x = 54321. Ans. x 8.414455. = Ex. 12. To find the value of x in the equation x Ans. x 8.6400268. 7x3+11x3 3x11, to find x. Given 2x - (3x2 — 2 √✓ x + 1)3 — (x2 − 4x √ x + 3√/x)# = 56. Ans. x 18-360877. To resolve Cubic Equations by Carden's Rule. THOUGH the foregoing general method, by the application of Double Position, be the readiest way, in real practice, of finding the roots in numbers of cubic equations, as well as of all the higher equations universally, we may here add the particular method commonly called Carden's Rule, for resolving cubic equations, in case any person should choose occasionally to employ that method. The form that a cubic equation must necessarily have to be resolved by this rule, is this, viz. z3+ az = b, that is, wanting the second term, or the term of the 2d power z3. Therefore, after any eubic equation has been reduced down to its final usual form, x3 + px2 + qx = r, freed from the co-efficient of its first term, it will then be necessary to take away the 2d term pa2; which is to be done in this manner : Takep, or of the coefficient of the second term, and annex it, with the contrary sign, to another unknown letter z, thus z p; then substitute this for x, the unknown letter in the original equation x3 + px2 + qx = r, and there will result this reduced equation 23+ az=b, of the form proper for applying the following, or Carden's rule. r take c = a, and d = 1b, by which the reduced equation takes this form z3 + 3c z = 2d. Then Then substitute the values of c and d in this form z=d+ √ (d2 + c3) + d −√ (d2 + c3), C v/d + √ (d3 √ c3) — Vd+√(d3+c3), and the value of the root z, of the reduced equation 23 + az = b, will be obtained. Lastly, take x = 2 — P, which will give the value of x, the required root of the original equation x3px2 + qx =r, first proposed. = One root of this equation being thus obtained, then depressing the original equation one degree lower, after the manner described p. 260 and 261, the other two roots of that equation will be obtained by means of the resulting quadratic equation. Note. When the co-efficient a, or c, is negative, and c3 is greater than da, this is called the irreducible case, because then the solution cannot be generally obtained by this rule. Ex. To find the roots of the equation x3-6x2 + 10x = 8. First, to take away the 2d term, its co-efficient being - 6, its 3d part is 2; put therefore x z + 2; then 6x2 = +10x = 2z+4=8 → 2z =4 = 4, c=- &, d = 2. ་ Theref. /d+(d2 +c3)=3/2+√(4−{})=√/2+√200 3/2+5o√3 = 1.57735 and ad-(d+c3)= 3/2—√(4−q)=√/2+√= 3/2—4° √3 = 0.42265 10 then the sum of these two is the value of z = 2. Hence x z + 24, one root of x in the eq. x3 10x = 8. 6x2+ To find the two other roots, perform the division, &c. as in p. 261, thus: Hence x2-2x=-2, or x2 2x+1== ± √ −1; x = i + √ −1 or=1-√ other sought. Ex. 2. To find the roots of x3 Ans. x 3, or = 3 + ✔ Ex. 3. To find the roots of x3 9x2+28x= 30. 1, or = 3√ −1. 7x3 + 14x = 20. Ans. x = 5, or 1+√3, or = 1 − √ — 3. OF SIMPLE INTEREST. As the interest of any sum, for any time, is directly proportional to the principal sum, and to the time; therefore the interest of 1 pound, for 1 year being multiplied by any given principal sum, and by the time of its forbearance, in years and parts, will give its interest for that time. That is, if there be put r = the rate of interest of 1 pound per annum, t = the time it is lent for, and a = the amount or sum of principal and interest; then is prt the interest of the sum p, for the time t, and conseq. p+prt or p× (1 + rt) = a, the amount for that time. From this expression, other theorems can easily be deduced, for finding any of the quantities above mentioned; which theorems, collected together, will be as below: 1st, a=p+prt, the amount. For Example. Let it be required to find, in what time any principal sum will double itself, at any rate of simple interest. = In this case, we must use the first theorem, a = p + prt, in which the amount a must be made 2p, or double the principal, that is, p + prt 1 and hence t = -. = 2p, or prt = P, or rt = 1; Here, Here, r being the interest of 11. for 1 year, it follows, that the doubling at simple interest, is equal to the quotient of any sum divided by its interest for 1 year. So, if the rate of interest be 5 per cent. than 100 ÷ 5 = 20, is the time of doubling at that rate. Or the 4th theorem gives at once a-p 2p - P pr pr 1040 COMPOUND INTEREST. BESIDES the quantities concerned in Simple Interest namely, p = the principal sum, r the rate or interest of 11. for 1 year, a = the whole amount of the principal and interest, t = the time, there is another quantity employed in Compound Interest, viz. the ratio of the rate of interest, which is the amount of 11. for 1 time of payment, and which here let be denoted by R, viz. R=1+r, the amount of 11. for 1 time. Then the particular amounts for the several times may be thus computed, viz. As 1. is to its amount for any time, so is any proposed principal sum, to its amount for the same time; that is, as 11. : R 11. 11. R :: P : PR, the 1st year's amount, and so on. Therefore, in general, pRta is the amount for the t year, or t time of payment. Whence the following general theorems are deduced: |