— с

C

a

2aA

· 12 B=+2A + q? - b= +12, and C=+

+

-22A + a2 6 22 = 73/2. Consequently the two quadratics become

22 X V2 + 1 + 3/2 = = 0, and m2 + x^2 + 1 - 3V2 = 0. The roots of these are x = 1/2 + V-7–312, and

=

x = - 1/2 +V-{+372, respectively. Ex. 2. Let 24 6203 58.02 1143 11 = 0 be given to find &.

Here the reducing cubic is A3 + 29A? + 182A 1256 = 0, and from this A=4. Whence B= +- 5/3, and C= + 3/3. Inserting these and reducing the quadratics, we get

x = i +173 +17+ * 13. Ex. 3. Given z + 2az3 – 37a222 38a3z + a4 = 0, to find z.

Dividing all by af, and putting = 2, we have 24 + 2x3 3702 382 + 1 =0.

The reducing cubic is 2A3 + 37A2 40A 399 = 0, and the roots (obtained by a method not yet explained, but they may be verified by substitution) are 3.5, -3, and —19. Either of these may be used in the quadratic equations, and the final result will in all cases be the same; and x can be obtained as before, and thence 2 = ax.

Ex. 4. Given 24 40x3 + 5a x2 - 4a3x + a4 = 0, to find a.

Ex. 5. Solve the three following equations by the methods explained in the notes, as well as by the general method. (1). 24 – 25x2 + 60x = 36.

Ans. 1, 2, 3, – 6. (2). 204 + 2q2+ 37%22 + 2qor - p = 0.

Ans. — 19+V -97q4 + 74. 9202 + 1522 2700 + 9 = 0.

Ans. 9+375 + V 78 + 54 5

4 Ex. 6. The following equations are proposed for solution

8203 1202 + 84x = 63. Ans. 2 + 17+ 711+ 17. (2). 204 + 3623 — 400x2 — 3168x + 7744 = 0.

Ans. – 9+ 137 +372 17 F 137. (3). 24 + 24m3 – 114x2 . 240 +1=0. Ans. + 197 — 14, and 2 + 75. 12x = 5.

Ans. 1 + 12, and -1+2 N - 1. (5). 24 = 12x3 + 4722 - 72x + 36=0.

Ans. 1, 2, 3, 6.

(3). 204

(1). 24.

(4). 204

The method of double position has been explained in the arithmetic, under the form best adapted to its use there ; which is, as the student will perceive by

k-kz. (4)

solving the same questions algebraically, in those cases where the equation of the question is of the first degree. In all other cases it furnishes but gradual approximations to the true answer; and in these, according to the circumstances of the given equation, there will be very different degrees of rapidity in the approaches to the three values of the unknown. In algebraical equations, the rate of approximation is generally considered to be such as to give at each operation about as many figures more as we had already obtained. In other classes of equations *, however, the approximation is much more slow; and were it not the only method which we have the means to employ in such cases, its great slowness, and its being superseded in all respects in its application to algebraical equations, would call for its total omission from this work : but as the modus operandi is so easily perceived in the application of it to algebraic equations, and some practice in the use of it so essential in future inquiries, it will be retained in nearly the same form as heretofore t.

Let ma + ax -1 + bx"-2 + .... hx = k ..... (1). Suppose two numbers x, and x, are found by trial, or otherwise nearly equal to x, and substitute them in (1); giving

20," + ax"-1 + bx,”—2 + . hx, = k (2)

X," + ax,"-1 + bx, "-+ . hx, = kg (3) Subtract (3) from (1) and from (2), and divide the results. Then (come — x,") + a(x) — 2,*-1) + 6(x*-? – ,"—?) + h(x - x2) (X;" — &") + a(x, "-1 – 2,*-1) + 6(x,”—2 — X.,-2) +... h(x, — x) ki - k, But each compound term of the numerator is divisible by x — x,

X

and each one of the denominator by X,

- Kg. Whence (k, – ka) (x x) 2," + 2,"? (x2 + a) + 3," (x2 + ax, + b) +

(5) (k — k.) (22 – X,) 2-1 + 21–2 (x, + a) + 2014–3 (+ ax, + b) +

Now if the second side of this equation were unity, the solution of (5) for x would be accurate ; but as it is only an approximation, a and x, being only nearly equal, the value of a determined on the hypothesis of x = x, will give a result differing from the truth by a small quantity dependent on this inequality.

If X, X, and X , agree to p places of figures, the numerator and denominator will generally agree to np - 2 places, and hence the quotient will not differ from unity till about np 2 places. This, with equations of a low degree, and at the outset of the work, gives nearly the rate of approximation already spoken of: but it is evidently much more rapid in higher equations at the outset, and in all after a few steps in the approximation have been made. When, however, the given equations involve radicals, it is difficult to investigate without considerable detail the extent of the approximation.

Adopting then as an approximation that the right side of (5) is unity, we have

k - kg
X – X2 =
(2, - x)

(6)
ki ka

1-2

N-2

1-3

Such as, for instance, 10* = 50, in which x = : 1.69897; or out = 100, in which x = 3.597285. This application of the method requires, however, a knowledge of the principles of logarithms and the use of tables, the foundation and structure of which have not yet been explained. On this account, the solution of such equations will be reserved for a Supplement to the use of the tables referred to.

+ This method is due to John Bernoulli, see Butler's Mathematics, vol. ii. p. 155.

a

From this we have a closer approximation to x-than either & or X, were, viz. :

k — kg

(2,
22) + X2

(7)
ki kg
Hence, putting this formula into words, we have the following

=

RULE.

excess, and

1. Find, by trial, two numbers, as near to the true root as you can, and substitute them separately in the equation instead of the given quantity; and find how much the terms collected together, according to their signs + or -,

differ from the absolute known term of the equation, marking each error + if in

if in defect. 2. Multiply the difference of the two numbers, found or taken by trial, by either of the errors, and divide the product by the difference of the errors, having regard to the algebraical laws of the signs.

3. Add the quotient last found to the number belonging to that error, when its supposed number is too littl but subtract it when too great, and the result will give the true root nearly.

4. Take this root and the nearest of the two former, or any other that may be found nearer ; and, by proceeding in like manner as above, a root will be obtained still nearer than before. And so on, to any degree of exactness required t.

Notes. 1. It is best to employ always two assumed numbers that shall differ from each other only by unity in the last figure on the right hand; because then the difference, or multiplier, is only 1. It is also best to use always the least error in the above operation.

2. It will be convenient also to begin with a single figure at first, trying several single figures till there be found the two nearest the truth, the one too little, and the other too great; and in working with them, find only one more figure. Then substitute this corrected result in the equation, for the unknown letter, and if the result prove too little, substitute also the number next greater for the second supposition ; but contrariwise, if the former prove too great, then take the next less number for the second supposition ; and in working with the second pair of errors, continue the quotient only so far as to have the corrected number to four places of figures. Then repeat the same process again with this last corrected number, and the next greater or less, as the case may require, carrying the third corrected number to eight figures; because each new operation commonly doubles the number of true figures. And thus proceed to any extent that may be wanted.

3. The actual labour of finding the errors which result from the suppositions will be greatly abridged by the application of Problem I. of the Chapter on the Solution of Equations, in those cases where all, or nearly all, the terms are present, and the equation ordered according to descending powers of the unknown quantity. To that rule, for the sake of avoiding repetition, the student is at once referred. See p. 208.

EXAMPLES.

Ex. 1. Find the value of x in the equation 23 + 202 + x – 100 = 0. Here k 100, and it is easily discovered that the root lies between 4 and 5. Substitute these numbers for x, and X1, and adopt the algorithm referred to in note 3. Then, 1+1+1 100 (4 = X9

1+1+1 100 (5 = X1 4' 20 + 84 = k;

5 30 155 = ki

=

71

5 21 - 16 = kg - k

6 31 55 = k, – k Hence x, – Xe = 1, ki - kg = 71, k – k, = 16, and we have x = x2 + k - k,

16.1 (201 - 2.) = 4+ = 4.2 nearly. ki - k

Again, suppose X, = 4.2 and x = 4.3. Then, proceeding as before, we have 1+1+1 – 100 (4:2 = x,

1+1+1 - 100 (4:3 = X 4.2 208 91:36

4.3 21.2 95:16 1.04 4.568

5:3 1:59 7.137

5•2 22.84 — 4:072 = k, – k

23.79 + 2 297 = k, – k Hence X, – 0,='l, k, — k, = 6 369, and k — k, = 4:072. Hence, proceeding as before, we have x = 4.264 nearly.

Working, thirdly, with 4.264 and4.265 in the same manner, we have x = 4.2644299 more nearly.

So long as the first three figures only appear in the approximations, and the equation is only cubic, the table of squares and cubes at the end of this volume will facilitate the work of finding the values of the substitutions. Ex. 2. Find the value of ac in 23 15x2 + 63x = 50.

Ans. 1.0280392317 ... Ex. 3. Let it be required to find the value of x in the equation

V 144x2 – (ac2 + 20)2 + ✓196x2 — (x2 + 24)2 = 114.
By a few trials it is soon found that the value of x is but little above 7.
pose, therefore, first that x is = - 7, and then x = 8.
First, when x2 = 7.

Second, when X, = 8.
47.906

V 144m2

(2a + 20) 46.476 65.384 V196x2 (x2 + 24) 69.283

Sup

.

.

.

ky – k, = 2:469. Hence by the formula we find x = :72, nearly

Suppose again x = 7.2, and then, since this turns out too great, take 7•1 and 7.2 for the trial numbers, as follows: First, when X2 = 7.2,

Second, when 21 = 7.1 47.990 = V14422 (ac2 + 20)2 = 47.973 66.402 = V196m2 (oc2 + 24)2 = 65.904

kg = 114.392
k= 114.000

113.877 = ki
114.000 =k

kg - k= '392

- 123 = ki – k

'1.'123 Hence x = x2 + , — 2) =

= 7.124 nearly. - k2

•515 Ex. 4. Resolve the equation 23 +10x2 +5x=260. Ans. x=4:11798574108. Ex. 5. To find the value of x in the equation 23—2x=50. Ans. 3.864854. Ex. 6. To find x in the equation x3 + 2x2 — 23x=70. Ans. x=5:13457. Ex. 7. To find x in the equation 23—17x* + 54x=350. Ans. x=14.95407. Ex. 8. To find x in the equation 24—3x2 ——75x=10000. Ans. x=10.2609. Ex. 9. Find a in 2x4 — 16x3 +40x2 — 30x=-1.

Ans, x=1.284724. Ex. 10. Resolve 205 +2x4+3x3+4x2 + 5x=54321.

Ans. x=8414455. Ex. 11. Given 2x4—723 +11** --3x=ll, to find x. Ans. x = 1.8375506.

Ex. 12. To find the value of x in the equation (3x? — 2 7x + 1)! – (22 — 4« Vx+3 123% = 56. Ans, x = 18-360877.

XC

X

=

THE NUMERICAL SOLUTION OF ALGEBRAIC

EQUATIONS.

The solution of algebraic equations having general or literal coefficients has never been effected beyond the fourth degree; and methods to this extent have already been explained in this work, at pages 170, 185, 198, and 200. A considerable number of the properties of the roots of algebraic equation without limitation as to degree, when taken in connexion with the coefficients, have been investigated; but as these have been viewed more with a prospect of literal than mere numerical solution, they become in reference to our present object rather matters of curiosity than of utility. It is not proposed, then, to enter upon the general theory of equations in this work further than it conduces to the solution of those whose coefficients are numerically given.

As, whatever may be the number of equations simultaneously given and containing the same number of unknowns, these equations can be reduced to a single one containing only one unknown, the inquiry will in this place be restricted to this single equation, whether originally so given or obtained from the system of coexisting equations by elimination.

DEFINITIONS AND NOTATION.

1. An algebraic equation is any one which contains positive integer powers of the unknown quantity. The following is the general type, n being a positive