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EXAMPLES.

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1. It is required to exterminate the second term of the equation 23 + 30x2 = b, or 203 + 3axo b=0.

За
Here x =%

a,
3
10 23 3aza + 3a2 a3
Then Зах?

+ +3a22. .

6aʻz + 3 b

-b

b=0,

Whence 23 3az + 223
Or 23

30+z=h - 2a, in which equation the second power (24), of the unknown quantity, is wanting. 2. Let the equation x3 - 123x2 + 3x =

1222 + 3x = -16, be transformed into another that shall want the second term.

Here x =% + 4.

(2 + 4) = 23 + 1222 + 482 + 64 Then

12(2 + 4)2 1222 962 192 +3(+4)=

+ 32 + 12

Whence 23 - 452 - 116 16

Or, 23 452 = 100 which is an equation where z’, or the second term, is wanting, as before.

3. Let the equation 2013 6x9 = 10, be transformed into another that shall want the second term.

Ans: 23-122 26. 4. Let y3 15y2 + 81y= 243, be transformed into án equation that shall want the second term.

Ans. 23 + 6 = 88.

which shall want the third term, by assuming x=y+e; and in the resulting equation, let 3e2 — 12e'+9, or c2-4e+3=0, in which the values of e are 1 and 3; then assume x=y +3, or y +1, and the resulting equation will be y3 +342_1=0, an equation wanting the third

term.

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1 4. The equation 6x2. 11x2+6x-1=0, by assuming x=

v! be transformed into y3 - 642 +lly-6=0; the roots of which are to be reciprocals of the former.

5. The equation 3.23 --- 13.22+14x + 16 = 0, by assuming x= be transformed into y3 1342+42y + 144 =0, the roots of which are three times those of the former. -Ed.

u

3, may

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3 7 9 1. Let the equation x3 +222 +

0, be transform

4 8 16 ed into another that shall want the second term.

11 3 Ans. y9 +16=

4° 6. Let the equation x4 + 8203 5xe + 10x - 4 4 = 0, be transformed into another, that shall want the second term.

Ans. y - 297 +94y - 92 = 0. 7. Let the equation x4 3x2 + 3.02 5x = 2 = 0, be transformed into another, that shall want the third term.

Ans. yo + y - 44 6 0. 8. Let the equation 3x3 20 + 1 = 0, be transformed into another, whose roots are the reciprocals of the former.

Ans. y3 2y2 + 3 = 0. 9. Let the equation 2014 - 3x3 + 30m2 - 3x += 0, be transformed into another, in which the coefficient of the highest term shall be unity, and the remaining terms integers.

Ans. y4 - 3y + 12ya — 162 y +72 = 0.

OF THE SOLUTION OF CUBIC EQUATIONS,

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.*

* If, instead of the regular method of reducing a cubic equation of the general form

x3 + ax2+bx+6=0, to another, wanting the second term, as pointed out in the preceding article, there be put <=i(y -- a), we shall have, by substitution and reduction, y3 +(96 - 3a2) y =9ab-270--2a3; where, since the value of y can be determined by either of the formule given in this rule, the value of x will also be known, being x =f(y-a). And if b=0, or the original equation be of the following form, 23 fax?+c = 0, the reduced equation will be y3-3a2y = 2a3----27c, where the value of y, being found as above, we shall have, as before, c=f(y-a), which formulæ, it may be observed, are more convenient, in some cases, than those resulting from the preceding article; as the coefficients, thus oke tained, are always integers; whereas, by the former method, they are frequently fractions.

202 + ax = b.

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1 1 0
tv
27

b2
3/ tv
2

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

a3 the second term of the above equation, is negative,

27' a also in the formula, will be negative; and if the absolute

as

2

2

6 term b be negative, in the formula will also be negative;

73 but will be positive.*

4 It may likewise be remarked, that when the equation is of the form

203

ax = Ib, * This method of solving cubic equations is usually ascribed to Cardan, a celebrated Italian analyst of the 16th century, but the authors of it were Scipio Ferreus and Nicholas Tartalea, who discovered it about the same time, independently of each other, as is proved by Montucla, in his Histoire des Mathematiques, Vol. I. p. 568, and more at large in Hutton's Mathematical Dictionary, Art. Algebra.

'The rule above given, which is similar to that of Čardan, may be demonstrated as follows: Let the equation, whose root is required, be 23 +-a.c =b,

And assume y+z=x, and 3Yx Then, by substituting these values in the given equation, we shall have 93 +34/22+3yz2+z3 +8 x(y+-x) = y3+23+ 3yz X (y+z) tax (y+z)=13+ z3 ma X (+z)tax(y + 2) =b, or y3 + z3 =b.

And if, from the square of this last equation, there be taken 4 times the cubé of the equation yz - ša, we shall have y6 -- 24323 +- 26 =b2-+a3, or y3 -- 23=V (62 +27 a3).

But the sum of this equation and y3+23=5, is 2y3 =b+v (b2+ 7a3), and their difference is 223 = 6-0 (63+ ax); whence y=v [b+1 (163+2723)], and z=V[3b-V(162+ 27 a3)].

From which it appears, that ytz, or its equal x, is V [13+1 (+62 +27 au)]+ V (--v (+62+ay as)], which is the theorem; Or, since z is

it will be yt-z=y3y

Зу

ta V[364 v (+62 +34 93)] — V146+ V (+62+ 24 a3)]

the same as the rule.

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a

a

or

a

62 and

is greater than 27

or 4a greater than 2762 the solu

4 tion 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.*

EXAMPLES.

1. Given 2 1222 + 36x == 44, to find the value of x.

Here 23 - 6x2 + 18x = 22, by dividing by 2.
And, in order to exterminate the second term,

6
Pui x =%+ =2+2,

3

(2+2) = 23 + 62a + 122 +8. Then -612 + 2)2

) =

622 242 --- 24 = 22, 18 (+ 2) =

18z + 36

Whence 23 + 6% + 20 = 22, or 23 + 62 = 2, And consequently, by substituting 6 for a and 2 for 1, in the first formula, we shall have 52 216

216 + ✓ + +

✓ + 217

27

* It may here be farther observed as a remarkable circumstance in the history of this science, that the solution of the Irreducible Case above-mentioned, except by means of a table of sines, or by infinite series, has hitherto baffled the united efforts of the most celebrated Mathematicians in Europe; although it is well known that all the three roots of the equation are, in this case, real; whereas in those that are resolvable by the above formula, only one of the roots is real; so that, in fact, the rule is only applicable to such cubics as have two equal, or two impossible roots.

The reason why the assumptions, made in the note to the former part of this article with respect to the solution of the equation x3 — ax= b, are found to fail in the case in question (and it does not appear that any other can be adopted) is, that the two auxiliary equations, 3yz and yu+z3=b, which in this case become 3yz =d, and y3 +23=b, or y3z3 = and ys+- 23=1, cannot take place together; being inconsistent with each other.

For the greatest product that can be formed of the two quantities 33 +23 is when they are all equal to each other; or since y3-+- 23=b, when each of these b; in which case their product is = 162. But, as above shown, y323 by the question; therefore, when> 27

27 62

thetwo conditions are incompatible with each other; and conequently 4 the solution of the problem, upon that supposition, can only be obtained by imaginary quantities.

12

a3

217?

a3

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3

3

V[1 + 1 (1+ 8)] + V[1 – V(1 + 8)] = V(1 + V 9) +

V(1 - V9) = V(1 + 3) + V (1-3) = 4-2, Therefore c=%-+ 2 4- 32 +2=2+1.587401

1.25992! = 2.32748, the answer. 2. Given X3 -- 6x 12, to find the value of x.

Here a being equal to - 6, and b equal to 12, we shall have, by the formula,

-- 2 * =* [6+v (36 - 8)]

V{6+v (36 - 8)}

2 v (6 +128) +

=V (6 + 5.2915) + v (6 + V 28) 2

2 =V (11.2915) +

2.2435 + (6 +- 5.2915)

V (11.2915) 2

2.2435 +.8957 = 3.1392, 2.2435

Therefore x = 3.1392, the answer. 3. Given 23

· 4, to find the value of x. Here a being

2, and b

4, we shall have, by ihe formula, v=v{-2+v(4-9)} + v{ – 2 – v(4 –},

10

10 by reduction, V(-2+5 v 3) - V(2+5 v 3) = v(- 2 + 1.9245) – V (2 + 1.9245)= V(-:0755 — 3V 3.9245 .41226 -1.5773 : 1.9999, or 2;

Therefore 2 = 2, the answer.* Note.When one of the roots of a cubic equation has been found, by the common formula as above, or in any other way, the other two roots may be determined as follows:

Let the known root be denoted by, and put all the terms of the equation, when brought to the lefthand side, = 0; then, if the equation, so formed, be divided by x Er according as r is positive or negative, there will arise a quadratic equation,

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Or

* When the root of the given equation is a whole number, this method only determines it by an approximation of Is, in the decimal part, which sufficiently indicates the entire integer; but in most instances of this kind, its value may be readily found, by a few trials, from the equation itself. Or if, as in the above example, the roots, or numeral values of 3/

10 (-2+ -27-V3), and -V (2+5v3), be determined according to

VV3 the rule laid down in Surds, Case 12, the result will be found equal to - 2 as it ought.

9

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