Ex. 9. Given x3-3x-4=0, to find the value of x. Ans. x=2.2; 1.1+/-.63; -1.1--.63, very nearly.

Ex. 10. Given x3+24x=250, to find the value of x. Ans. x=5.05. Ex. 11. Given z3-6z3+13z-12=0, to find the values of z. Ans. z=3, or 1-7. Ex. 12. Given 2x 3-12x2+36x=44, to find the value of x. Ans. 2.32748, &c.

§ III. RESOLUTION OF BIQUADRATIC EQUATIONS BY THE METHOD OF DES CARTES.

556. The same observation may be applied to biquadratic equations as was applied to cubic equations in (Art. 549), that, since the equation x+a'x3 + b2x2 +x+s=0 may be transformed, (Art. 540), into another which shall be deficient in its second term, and whose roots shall have a given relation to the roots of the given equation, the complete solution of a biquadratic equation will be effected, if we can arrive at the solution of it in the form

(1) ;

x2+ax + bx+c=0. where a, b, c, may be any numbers whatever, positive or negative.

557 In the solution of a biquadratic equation, after the manner of Des Cartes, the formula 4+ax2 +b+c is supposed to be the product of two quadratic factors, x2+px+q and x2+rx+s, in which p, q, r, s, are unknown quantities. Or, which is the same, the biquadratic equation x2+ax2+bx+c=0 is considered as produced by the multiplication of the two quadratics,

(2). · x2+px+q=0; x2+ræ+s=0

(3). 558. Hence, by the actual multiplication of the above two factors, we shall have

x +(p+r)x3+(s+q+pr)x2 +(ps+qr)x+qs=

+ax3

+be+c.

And, consequently, by equating the coefficients of

the like powers of x, (Art. 435), in this last equation, we shall have the four following equations,

p+r=0; s+g+pr=a; ps+qr=b; gs=c. Or, if p, which is the value of r in the first of these, be substituted for r in the second and third,

b

they will become, s+q=a+p2; s − q = - ; qs=c.

Ρ

Whence, subtracting the square of the second of these from that of the first, and then changing the sides of the equation, we shall have

p2

And, therefore, by multiplying by p2, and placing the terms according to the order of their powers, the result will give, po+2ap1 +(a2 —4c)p2=b2

6

(4). From which last equation, if there be put p2=z, we shall have, z3 +2az3 + (a2 —4c)z= b2

•

b

(5),

Hence, also, since s+q=a+p3, and s-q=-, there

will arise, by addition and subtraction,

2p

Ρ

b

2p

where p being known, s and q are likewise known. And, consequently, by extracting the roots of the two assumed quadratics, (2) and (3); or of their equals, x2+px+q=0, and x2-px+s=0; we shall have

(6)

(7);

which expressions, when taken in + and -, give the four roots of the proposed biquadratic, as was required.

559. It may be observed, that whichever of the values of the unknown quantity, in the cubic, or reduced equation (5), be used, the same values of x will be obtained.

560. To this we may farther add. that when the roots of the cubic, or reduced equation (5), are all real, then the roots of the proposed biquadratic

are all real also. But if only one root of the cubic equation (1) be real, and, therefore, (Art. 554), the other two imaginary; then the proposed biquadratic will have two real and two imaginary roots.

Ex. 1. Given the equation x3-3x2+6x+8=0, to find its roots, or the values of x.

Comparing this equation with x2+ax2+bx+c=0, we have a=-3, b=6, and c=8; therefore, z3+2az3+(a2 —4c) z—b2 — z3-622-23z-36=0. Let 2y+2, and substitute y+2 for z in the latter equation; the resulting equation is y3-35y-98= 0. Now, by comparing this last equation with x2+ ax=b, (Art. 549), we have a=—35, and b=98; therefore, (Art. 550),

y=3/[49++√(65856)]+[49—†√(65856)] =2/(49+28.514)+(49—28.514)=3(77.514)+

/(20.486)

=4.264+2.736=7.

Hence, z=y+2=9, and p2=z=9, or p=±3; .. (Art. 559), taking p=3, s=−}+&+i=3+1=4. and 9=-3+-1=2. Consequently, by substituting these values for p, q, and r, in the equations (2), (3), we shall have

x2+3x+2=0, and x2-3x+4=0; ..x, and x=-7;

so that the four roots of the given equation are -1, -2, +7, -7.

3

2

Ex. 2. Given x46x2-16x+21=0, to find the values of x. Ans. x=3, or 1; or --2±√-3. Ex. 3. Given the equation - 4x3-8x+32=0, xa to find its roots, or the values of x.

Ans. 4, or 2; or -1±√-3.

Ex. 4. Given the equation 1—6x3+5x2+2x10=0, to find its roots, or the values of x.

Ans. 1, or +5; or 1-1. Ex. 5. Given x4-9x3 +30x2-46x+24=0, to find

the roots, or values of x.

Ex. 6. Given x4-16x3+99x2 +228x+144=0, to find the roots, or values of x.

Ans. x=-1,

3; or -6±√-12. Ex. 7. What two numbers are those, whose product, multiplied by the greater, is equal to 1; and if from the square of the greater, added to six times the lesser, the cube of the lesser be subtracted, the remainder shall be 8.

Ans. √2±√(1+√2), +√2±√(1−√2).

§ IV. RESOLUTION OF NUMERAL EQUATIONS BY THE

METHOD OF DIVISORS.

561. Since the last term (v) of the equation (A)= xTM+AxTM··1+вxm--2 Tx+v=o, is equal to the product of all its roots, (Art. 534), it is evident, that if any of those roots be whole numbers, they will be found among the divisors of that term. To discover, therefore, whether any of the roots of a given equation be whole numbers, we have only to find all the divisors of its last term, and substitute each of them, first with the sign and then with the sign, for x, in the given equation, such of them as reduce the equation to 0=0, will be roots of the equation.

562. Or, if the divisors of the last term should be too numerous, the equation may be transformed into another, that shall have its last term less than that of the former; which is done by increasing or diminishing the roots by 1, or some other quantity, as in (Art. 539).

Ex. 1. Given a3—x2 —2x+8=0, to find the roots of the equation, or values of x.

Here the divisors of its last term, are 1, 2, 4, 8; substitute 1, 2, 4, 8, and -1, -2, -4, -8, for x in the given equation, and -2, will be found to be the only one of these numbers which gives the result 0; -2 therefore is the only integral root of the equation. Hence, (Art. 529), x+2 will divide x-x-2x+8 without a remainder; let this division be made, and

the quotient being put equal to 0, we shall have x2. 3x+4=0, a quadratic equation which contains the other two roots. The solution of this quadratic x gives±√7; the three roots of the given equation, therefore, are -2, +-7,3-1√7. 3

563. The integral roots of any numeral equation of the kind above mentioned, may also be found, by NEWTON'S Method of Divisors, which is founded upon the following principles.

Let one of the roots of the equation (▲)=0, be —ɑ, or, which is the same, let the proposed equation be represented under the form (x+a)p=0, where the binomial x+a denotes one of the divisors, or factors, of which the equation is composed, and p the product of the rest. Then, if three or more terms of the arithmetical series, 2, 1, 0, -1, -2, be successively substituted for x, the divisors of the results, thus obtained, will be

P

a+2, a+1, a, a-1, and a-2.

And as these are also in arithmetical progression, it is plain that the roots of the given equation, when integral, will be some of the numbers in such a series.

Whence, if a progression of this kind, whose common difference is 1, can be found among the divisors of the results above mentioned, by taking one number out of each of the lines, that term of it which answers to the substitution of 0 for x, taken in + or

according as the series is increasing or decreasing, will generally be a root of the equation.

Ex. 2. Given x+x4-14x36x2+20x+48=0, to find the roots of the equation, or values of x. Num. Results.

Divisors.

Progress.

-1

36

3

4 3

1, 2, 3, 4, 6, 9, 12, 18, 36, Here the numbers to be tried are 2, 3, 4, all of which are found to succeed; so that the equation has three integral roots; namely, 2, 3, 4. The equation whose roots are 2, 3, 4, is (x-2).(x-3).(x+ 4)=x3-x2-14x+24=0, let the given equation be