Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Of Cubic Equations. 95. A cubic equation is that in which the unknown quantity rises to 3 dimensions; and, like quadratics, or those of the higher orders, is either simple or compound. A simple or pure cubic

b equation is of the form

axb, or x=-; where x = A compound cubic equation is of the form

23 fax=b, 2 tax=b, or x taxi tbx= in each of which the known quantities a, b, c, may be either +

.

a

1

or

[ocr errors]

a.

Or either of the two latter of these equations may be reduced to the same form as the first, by taking away.its second term ; which is done as follows:

RULE. Take some new unknown quantity, and subjoin to it a third part of the coefficient of the second term of the equation with its sign changed; then, if this sum, or difference, as it may happen to be, be substituted for the original unknown quantity and its powers in the proposed equation, there will arise an equation wanting its second term.

Note. The second term of any of the higher orders of equations may also be exterminated in a similar manner, by substituting for the unknown quantity some other unknown quantity, and the 4th, 5th, &c. part of the coefficient of its second term, with the sign changed, according as the equation is of the 4th, 5th, &c.

power. 1. Required to exterminate the second term of the equation

3a 2+3ax=b, or 2+3ax?_=0. Here z=2

3 x=28_3az* +3a®z- Then 3ax +3aze-6a2z+32% ; whence 2-3a®z+20

5 -=0; or ze_3aʼz=6—20, in which equation the second powe er (24) of the unknown quantity is wanting.

2. Let the equation 23-127+3=16 be transformed into another that shall want the second term.

Here x2+4.
(2+4)=+12z* +48z+ 64
Then -1212+4)= - 12z2__962-192
(+ 2(2+4)

+ 3z + 12 Whence 28–452—1165-16; or z_452=100, which is an equation where z", or the second term, is wanting, as before.

3. Let the equation 20_6x=10 be transformed into another that shall want the second term.

Ans. 23-12z=26. 4. Let y:—154 +8ly=243 be transformed into an equation that shall want the second term.

Ans. 6.188.

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

3 64

9

[ocr errors]
[ocr errors]

[ocr errors]

20

[ocr errors]
[ocr errors]

= 0.

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

5. Let the equation z+*x +43-16=be transformed into another that shall want the second term. Ans. y+12=* Here y=x+5,

Here y-1, y=+15x+ 75x+125 x=yytity-65 -15= -15%

-150x-375 +4r= +4y— yta +8ly=

+ 81x+405 +2= Byt32 -243=

-243

23 +62—8850. Hence we have y +Hy-20. 6. Let x* +848–5x+10x—450 be transformed into another that shall want the second term. Ans. y-29yo +947-92—0.

Here =y-2,
X2 : 74 ?

8y + 24y - 32y + 16
8.23

8y-48y + 964 - 64 5x?

-57% + 203

- 20 +10%

107 4

4 We have x4 29y2 + 947 92 7. Let 2-3x8 +3.x2–5x2=0 be transformed into another that shall want the third term.

Ans. y*+-4y—6=0.
Here yte

x=,
: 74 + 4y + 6ey? + 4eRy + et
-3.23

Зу
34 - 9e y — 9ey - 3e3

+ 3y + 6ey + 3e
52

5 y - 52 2

2 Now, the value of e, by which the third term is taken away, is had by resolving the quadratic equation 6e-9e+3=0; the roots of which are found = 1, or s: hence, by substituting yt1 for 1, in the given equation, we find y + y - 4y—6=1, an equation

y

0 wanting the third term.

8. Let 3x8–2x+1=0 be transformed into another whose roots are the reciprocals of the former. Ans. 73-2y+3=1. 1

3 2 Here let x =- ; then 3x2–2x+1: +1=0, or 3-2y+ у

Y y=0; hence y: 24°+3=0; the roots of which are the reciprocals of the former.

9. Let *—128-$22-ti's be transformed into another, in which the coefficient of the highest term shall be unity, and the remaining terms integers. Ans. yt-3x+12y—1624+72=0. y ya y Зу

1 Here let x=

tis=y4—34 + 6 64

3.62 4.6 12y—162y+72-0, the equation required; the roots of which are 6 times those of the former.

+ 3x?

[ocr errors]

y

; then

[blocks in formation]
[ocr errors]

a

as also

[ocr errors]

Of the solution of Cubic Equations. 97. Rule. Take away the second term of the equation when necessary, as directed in the preceding rule. Then, if the numeral coefficients of the given equation, or of that arising from the reduction above mentioned, be substituted for a and b in either of the following formulæ, the result will give one of the roots, as required. *

2 tax.

b 82 a
4
27

4
72

ja 27

62 4

27 Where, it is to be observed that when the coefficient a, of the

a second term of the above equation, is negative,

in

27 3 the formula, will be negative, and if the absolute term 6 be negative, 16 in the formula will be negative ; but { will be positive. It may likewise be remarked, that when the equation is of the

ai

72 form 2 ax=+b, and

27
is greater than

4

or 4a greater than 277, the solution of it cannot be obtained by the above rule ; as the question, in this instance, falls under what is usually called the Irreducible Case of cubic equations. See index.

1. Given 2x3—122+36x=44, to find the value of x.

Here 2-6x*+18x=22, by dividing by 2; and, in order to ex. terminate the second term, put x=z+f=2+-2:

(2+2) S28+62 +122+8
Then -612+2)= +62—242—24 = 22,

18(2+2) = 182+36 * If, instead of the regular method of reducing a cubic equation of the general form 2 + ax'+bx+c=0, to another, wanting the second term, as pointed out in the preceding article, there be put x =

fly a), we shall have, by substitution and reduction, Y+(96–3a)y=9ab27c-20°; where, since the value of y can be determined by either of the formulæ given in this rule, the value of x will also be known, being ==}(y-a). And if b=0, or the original equation be of the following form, x'taxt=0, the reduced equation will be 28–3a*y=-27c, where the value of y being found as above, we shall have, as before, z=f(ya), which formulæ, it may be observed, are more convenient, in some cases, than those resulting from the preceding article; as the coefficients thus obtained are always integers ; whereas, by the for. mer method, they are frequently fractions.

[ocr errors]

3

[ocr errors]

7 (6+5.2915)+7(6+52915) -2.2435+2.2435

3

247

Whence z'+62+20=22, or z+6z=2; and, consequently, by substituting 6 for a and 2 for b, in the first formula, we shall have

Wi+w(+40)}+{}-^(+49)}; ♡{1+ (1+8)}+{i-vi1+8)=W1+9+ (1-19)=(1+3)+(1-3)=42; therefore 2 +2=74-32+2=2+ 1.587401 -1.259921–2.32748, Ans. 2. Given 2:46.x=12, to find the value of x. Here a being equal to -6, and b equal to 12, we shall have

2 by the formula z={6+w(36–8)}-7 {6+- (36—8}} 2

2 Ņ(6+128)+

(6+228

) 2

2 =»(11.2915)+

=

= 2.2435+ (41.2915) .8957=1392; therefore x3.1392, the answer.

3. Given 28-235-4, to find the value of 2.

Here a being -2, and b-4, we shall have, by the formula, R={-2+(

49)}+{-2-714-)}, or by reduction, (-2+13)-(2+3)=(-2+1.9245) -(2+1.9245)=31.0755—33.9245=-1226_1.5773

-1.9999, or -2, Ans. Note. When one of the roots of a cubic equation has been found, by the common formula as above, or in any

other other two roots may be determined as follows:

Let the known root be denoted by r, and put all the terms of the equation, when brought to the left hand side, =0; then, if the equation so formed be divided by ztr, according as r is positive or negative, there will arise a quadratic equation, the roots of which will be the other two roots of the given cubic.

4. Given 28—15x=4, to find the three roots or values of x.

Here z is readily found, by a few trials, to be equal to 4, and therefore, by division, we get 2-4)23—150—4(2*+4x+1;

42 Whence, according to the note above given, ze + 4x+1=1, or *** +4x= -1; the two roots of which quadratic are —2+13, and —2-73; and consequently 4, -2+13, and -2-3, are the three roots of the proposed equation. 1. Given x+3x?_6.x = 8, to find the root of the equation, or 2

-8, the value of x. See index.

Ans. 2. 2. Given 28—6x=6, or 2*+9x=6, or 20 +224_23x 70, or T-172+541350, or 23x=5, or 2+2=500, to find the value of ac in each expression. Answers, 34+/2, or 393, or 5.1346, or 14.954, or 3. 426, and 7.61728, as required.

way, the

#

.

[ocr errors]
[ocr errors]

a

or

3. Given 23–6x=4, to find the three roots of the equation, or the three values of x.

Ans. —2, 1+13, and 1-3. 4. Given 23_522 + 2x 12, to find the three roots of the equation, or values of x. Ans. -3,1 +5, and 1-5. 6. Find the root of the equation 2: +x=500, or 2346x=5.6.

Ans. x==7.89500828, and x=2.82526384. Solution of Cubic Equations by the Tables of Sines and Tangents.

98. Cubic Equations may often be more readily resolved by means of the following trigonometrical formulæ, than by the method before given; particularly when they fall under what is usually called the irreducible case, in which the common method is known to fail. 1. Given 20 tax=2b, to find the value of x.

2ar Put

tan. 7, and (ol tan. }z= tan. U,

2 Then x=ENAX cot. 2u; where the upper or under sign is

3 to be taken according as b is positive or negative. 2. Given -AX=#b, to find the value of x.

2ar This form has two cases, according as Name is less

36 3 greater than radius r, or 4a® less or greater than 2762. In the first of these cases, or when 4a 276,

2ar Put

sin. Z, and V (que tan. z)= tan. U ; . 36

37

2 Then z= N X cosec. 2u, where the upper or under sign is

3 to be taken, according as b is positive or negative. In the second case, or when 4a?<2763,

3br 3 Put ✓= cos. z, then x will have the three following values; 2a 2

2
X Fins

3
2
and X=F-NO

Š x cos. (60° +) where the upper sign is to be used when 6 is positive, and the under sign when it is negative, as in the other cases. 1. Given 2+18x=6, to find the only real value of x.

2ar Here a=18, and b=6; whence

=27 76= tan.z ;

36 3 and, consequently, by formula 1, we shall have

a

a

a

a

a

[ocr errors]

EN

vý x cos.ži X cos. (60°–

a

[ocr errors]

a

« ΠροηγούμενηΣυνέχεια »