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3. Let the equation x3 -6x2=10, be transformad lato another, that shall want the second term.

Ans. y3 - 12y=26 4. Let y3 – 15y2 +81y=243, be transformed into an equation that shati want the second term.

Ang. 23 +6x==88 3 7 9 x2

-0, be trans8

16 formed into another, that shall want the second term.

11 3 Ans. yö+164

4 6. Let the equation 2x3 - 3x2 + 4x --50, be trans

. formed into another, that shall want its second term.

5. Let the equation x3+ ***+

4

OF THE SOLUTION OF CUBIC EQUATIONS.

RULE

Take away the second term of the equation when necessary, as directed in the preceding role. 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.

x3 + ax=b 62

( 6 62 -+

27

27

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62

ja
-+
4

b ba a3

+

2 4 27 Where it is to be observed, that when the coefficient a, of the second term of the above equation, is negative, a3 27 3

M ?

a

, as also , in the formula, will be negative ; and if

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b 2

b2

the absolute b be negative, ğ, in the formula, will, also, be negative; but

will be positive. (e) It may, likewise, be remarked, that when the equation is of the form

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(e) 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 Nicolas Tartalea, who discovered it about the same time, independently of each other, as is proved by Montucla, in his Historire des Mathematiques, Vol. 1. p. 568, and more at large in Hutton's Mathematical Dictionary, Art. Algebra.

The rule above given, which is similar to that of Cardan, may be demonstrated as follows: Let the equation, whose root is required, be x3 tax=b.

And assume y +z=x, and 3yz Then, by substituting these values in the given equation, we shall have y3 +3y2z+3y23 +23+uX (y+z)=y3 +23+3y2x(+8)+ ax (y+2)=y3+23-ax(4+2)+ax(y+z)=b, or

y3 +235o. And if, from the square of this last equation, there be taken 4 times the cube of the equation yz=-ga, we shall bave y6-2y3z3 +26 -b2 +3723, or

y3–23=v(62 +2qa3) But the sum of this equation and y3+z3=b, is 2y3 =b+v(62+ a3) and their difference is 2x3=6-v(62 +423); whence y= 36+ (463+27a3), and z=fb-v (462 +24a3).

From which it appears, that ytz, or its equal x, is = 36+ (+62 +1793)+36-(+62 +27a3), which is the theo.

rem.

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Or, since z is =

Зу * 16+ (162 + 4a3

it will be y+z=y-,

3y ja

-, the same as the 6+(769 +343

rule.

a 3

62 and is greater than or 4a3 greater than 2762, the 27

4' 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. (f)

EXAMPLES.

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1. Given 2x3 – 12x2 +36x=44, to find the value of se.

Here x3 – 6x2 + 18x=22, by dividing by 2.
And, in order to exterminate the second term,

6
Put x=zt =2+2,

3
(2+2)=23 +62 +12z+8
Then -6(2+2)= 622 - 242-24 =22
18(2+2) =

182 +36

3

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X=

Whence 23 +62+20=22, or 23 +6=2, And, consequently, by substituting 6 for a, and 2 for b, in the first formula, we shall have, 2 4 216,

,4 216 V{+ve+

+ ; 4

4 27 žitư (1+8)+71-v(1+8)=3/1+9+

1-v=V1+3+31-3=4-2, Therefore x=2

=2+2=74-82+2=2+1.587401 1.259921=2.32748, the answer.

4+379)}+v{-v+

)}=

27

(f) It may here be farther observed, as a remarkable circum. stance 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 impossible roots.

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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 x=36+(36–8)

v{6+7(3€ -8)}

2
V6+28+

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

2
=>/11.2915+

=2.2435+ V(6+5.2915)

(11.2915) 2

=2.2435+.8957==3.1392 2.2435

Therefore x=3.1392, the answer. 3. Given x3 – 2x=-4, to find the value of x.

Here a being =-2, and b=-4, we shall have,.by the formula, 8

8

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27

x=V{-2+v(4- )}+V4-2-v(4- )}, by reduction, ==(-2+"0/3)-0(2+ /3)=

10

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3-2+1.9245 - 32+1.9245=-.0755-
33.9245=-.4226_1 5773=- 1.9999, or - 2

Therefore x=-2, the answer. (g) 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 r, and put all the

(8) When the root of the given equation is a whole number, this method only determines it by an approximation of 9s. in the decima part, which sufficiently indicates the entire integer; but in most instances of this kind, its value may be more readily found, by a few trials, from the equation itself.

terms of the equation, when brought to the left hand side, =0; then if the equation, so formed, be divided by x Fr, 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 equation.

4. Given x3 – 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

x-4)33 -- 15x - 4(x2 + 4x +.1

x3 4x2

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Whence, according to the note above given,

22 +4x+1=0, or x2 + 4x=-1; the two roots of which quadratic are -2+3 and 2 -V3; and consequently

4, -2+73, and -2-73, are the three roots of the proposed equation.

5. Given x3 +3x2 - 6x=8, to find the root of the equation, of the value of x.

Ans. x=2 6. Given x3+x=500, to find the root of the equation, or the value of x.

Ans. x=7.617 7. Given 23-4802 200, to find the root of the equation, or the value of x.

Ans =47.9128 8. Given x3 - 6x=6, to find the root of the equation, or the value of x.

Ans. x=4+2 9. Given x3 +-9x=6, to find the root of the equation, or the value of x.

Ans. x=3/9-13

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