Page images
PDF
EPUB

the roots of which will be the other two roots of the given cubic equation.

4. Given 203 15x = 4, to find the three roots or values

of x.

Here x is readily found, by a few trials, to be equal to 4, and therefore

* -- 4) * - 15x 4(202 + 4x + 1

203 4x?

[merged small][merged small][ocr errors][merged small][merged small][ocr errors]

-1;

Whence, according to the note already given,

m2 + 4x + 1 = 0, or 29 + 4x = the two roots of which quadratic are 2+13 and -2V3; and consequently

4, 2 + V3, and - 2-V3, are the three roots of the proposed equation.

EXAMPLES FOR PRACTICE.

1. Given 3 + 3x2 . 6x == 8, to find the root of the equation, or the value of x.

Ans. x = 2. 2. Given 23 + 2 = 500, to find the root of the equation, or the value of x.

Ans. X == 7.616789. 3. Given 23 3x2 = 5, to find the root of the equation, or the value of x.

Ans. x =

3.425101. 4. Given 23 6x = 6, to find the root of the equation, or the value of ą.

Ans. V 47 V 2. 5. Given 243 + 9x = 6, to find the root of the equation, or the value of x.

Ans. » 96. Given 23 + 2ac2 23 70, to find the root of thee quation, or the value of x.

Ans. X =

5.134599. 7. Given 2.3. - 17202 + 54x 350, to find the root of the equation, or the value of x.

Ans. X = 14.954068. 8. Given 23 6x -= 4, to find the three roots of the equation, or the three values of x.

Ans. 2,1 +V3, and 1 - V 3. 9. Given 23 5x2 + 2x = 12, to find the three roots of the equation, or the three values of x.

Ans. - 3,1 + V 5, and 1 - V.5.

V3.

OF THE SOLUTION OF CUBIC EQUATIONS,

BY CONVERGING SERIES.

This method, which, in some cases, will be found more convenient in practice than the former, consists in substituting the numeral parts of the given equation, in the place of the literal, in one of the following general formulæ, according to which it may be found to belong, and then collecting as many terms of the series as are sufficient for determining the value of the unknown quantity, to the degree of exactness required.

1. 2 + ax = 3.1
25
2.5 2752

2.5.8.11
1+
V [2(2762 + 4ao)]

6.9 2762

6.9.12.15

[ocr errors]

3

[ocr errors]

b

2s

* The method laid down in this article, of solving cubic equations by means of series, was first given by Nicole, in the Memoirs of the Academy of Sciences, an. 1738, p. 99; and afterwards, at greater length, by CLAIRAUT in his Elemens d’Algebre.

† With respect to the determination of the roots of cubic equations by means of series, let there be given, as above, the equation 23+ax =b, where the root by transposing the terms of each of the two branches of the common formula, is

v{v(+62 +2443) ++8}-{ v(*b2+ 37 (3) — 26}; or by putting, for the sake of greater simplicity, v (+62+273) =s, and reducing the expression, x = so { V(1+) - V(1- :)}

Hence, extracting the roots of the righthand inember of this equation by the binomial theorem, there will arise V(1+)=1+1(4)-36 +()

3.. 6.9.12 v(1+%)=1-1(3)-3.0)-26.06) (2)

And consequently, if the latter of these two series be taken from the former, the result, by making the first term of the remainder a multiplier, will give

{1+0+, &c.} where, since s= v (+62+2'7 as), (3)

2762 +-4as'

2057 () = (276212.cz)

V [2(2762 +423)]' Whence, also, by substitution, we have the above formula

(0)3.6.

2.5.8

3)-3.6:9.12

&c.

[ocr errors]

6s

2

2762

we shall have

2

[ocr errors]

26

&c. And

6s

653

( will

123 (27 (2744)

1+

2772

2.5.8.11.14.17 "2772 +

. Or 4a? 6.9.12.15.18.21 \2762 +- 403 26 2.5 2773

8.11
♡ [2(2762 + 40°)] 6.9\2752

12.15
2782
14.17 2762

20.23
B

C +
+4Q 12.21 6278 + 4a 24.27
2772

+, &c. 2762 +40 In which case, as well as in all the following ones, A, B, C, &c., denote the terms immediately preceding those in which they are first found.

2. 23 tb, where 1482 is supposed to be greater than tas, or 2762 > 4a3. B

2 2732_423 2.5.8 2782 423 = £23 I

.)

) 2 3.6 2782

3.6.9.12 2762 2.5.8.11.14 2762 - 4a?

-)3 &c.

* Or, 3.6.9.12.15.18 2779 bs 2 27.62 403 5.8 ,2772 400

4003

-) A
2
3.6 2762

9.12 2762
11.14 2772 403 17.20,2772-423
-)c

-)D

&c. 15.18 27.72

21.24 2762 In which case the upper sign must be taken when b is

[ocr errors]

&c. }:

{1

276

276

,

:)*

[ocr errors]

The root, as found by the common formula, when properly reduced, is
b
2

2

V (162 6

b

2 Or, putting, as in the last case, V (162-- az az), or its equal

b 2762-4a3

b , =

1 2762

2 Whence, extracting the roots of the righthand member of this equation,

2.5.8 there will arise 3/(1+s)=1+is-- $2+

+ 3.6

3.6.9 3.6.9.12 2.5

2.5.8 &c. (1-5)=1--ts

$4 ---, &c. 3.6 3.6.9 3.6.9.12 And, consequently, by adding the two series together, and taking the 1st

25 2 term of the result as a multiplier, we shall have x =

士 + 23

3.6 2.5.8 2.5 8.11.14

2762_-4a3

&c. 3.6.9.12 3,6.9..2.15.18

2762 for its equal s, we get the above expression.

[ocr errors]

$3

[ocr errors]
[ocr errors]

54

positive, and the under sign when it is negative; and the same for the first root in the two following cases :

3. 2003

ax = £b, where 782 is supposed to be less than a', or 2762 < 4a'.

b

2 4a - 2782 2.5.8 403 2772 x=23/1+

( -)

(

-) 2

3.6 -2762 3.6.9.12 2762 2.5.8.11.14 40° -2762

( -3 &c. Or, 3.6.9.12.15.18

2772

{1

* This expression is obtained from the last series, by barely changing the signs of the numerator and denominator in each of its terms; which does not alter their value.

Hence, in order to determine the other two roots of the equation, let

that above found, or its equivalent expression, v {46+v(+62— 1 23)}

7*

H
H

H

) Then, according to the formula that has been before given for these roots in the former part of the present article, we shall have 2 =

2 V-3

3/1627 a3] ib2 2

2 putting V (162- ztry a3) =s, and reducing the expression, x = 7

“土

[ocr errors]
[ocr errors]

{V(1+s) – V(1—-s)}. Whence, extracting the cube

[ocr errors]

2
roots of the righthand member of this equation, there will arise

2
2.5

2.5.8
V (1+8)=1+is-82+

S4 +, &c.
3.6 3.6.9 3.6.9.12
2
2.5

2.5.8
V (1-3)=1-is-

$4, &c. 3.6 3.6.9

3.6.9.12 And, consequently, by taking the latter of these series from the former, and making the first term of the remainder a multiplier, we shall have

[ocr errors]

$3

[ocr errors]
[ocr errors]

6.9.12.15547

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

(sbytv-3 2.5

2.5.8.11 2.5.8.11.14.17 { it=-52+

+, c. 2 3 6.9

6.9.12.15.18.21 2

2762423

2762_403 But since sa

I
b
29
2762

2762 423_2762 4a3_276212

V ;-3V. 2762 2752

3 3 2762-4a3

423_-2762
3
2762

962 V(4a3-2762)

if these values be substituted for their equals, in the last 93 262 series, the result will give the above expressions for the two remaining roots of the equation.

[ocr errors]
[ocr errors]

士2

[ocr errors]

1+

(la 1270)

6
2 403

2762 5.8 40 2762

A
2
3.6 2752

9.12 2762
11.14/403

2772

17.20 1403 2762 B +

С 15.18

Dt, &c. 2772

21.24 2762 which series answers to the irreducible case, and must be used when 2ais less than 27b2.

And if the root thus found be put = r, the other two roots may be expressed as follows:

V(4a - 27) 2.5 403 2769 2.5.8.11 X=I-+

1

+
2 9V 2/2

6.9
2762

6.9.12.15 4Q? 2782 2.5.8.11.14.17 14a3 2752

&c. 2762

3.6.9.12.15.18.21 2762

[ocr errors]
[ocr errors]

(1',

(2

[ocr errors]
[ocr errors]

Or,

于2

于士

)

B

V (4a3 -- 2762)
2.5 14a 2752

8.11 1

A+

A
93/262
6.9 2762

12.15 1403 - 2772

14.17/4Q3 - 2762 20.23 / 4a - 2762 2782

ct 18.21 2772

24.27 2762 D-, &c.

Where - žr, or + ür, must be taken according as b is positive or negative ; and the double signs I must be considered as +- in one case, and -- in the other, as usual.

4. 2003 where 162 is still supposed to be less than 2,a', or 2763 < 4a 26

2.5 2782 {i

+ 3 [2(4a -- 2764)]? 6.9 \4a_2762) 2.5.8.11 2782

2.5.8.11.14.17 2762 6.9 12.15 \4ď - 2752 6.9.12.15.18.21 \4a2762

[ocr errors]

AX

[merged small][merged small][merged small][ocr errors][merged small]

* By transposing the terms of the common formula, as in the first case,

we shall have x = v{V(162—4, 23)+86} - 3/ {v (+b2— 34, 23) —-}6}. Or, by putting, for the sake of simplicity as before, V. (+82 - *, 48)=s, and reducing the equation «="${v(1+) »

}

(1

28 Whence, extracting the roots of the righthand number, as in the former instances, h 2 b 12 2.5 b\3

2.5.8

4 + 2s.

3.6\2s/ 3.6.9 28 3.6.9.12 &c.

« PreviousContinue »