Ex. 8. Given x3-6x2+10x-8=0, to find the values of x. Ans. x=4, or 1±√-1. Ex. 9. Given x3—3x —4—0, to find the values 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. 2. Ans. x=5.05. Ex. 11. Given z3-622+132-12=0, to find the values of Ans. z=3, or + √7. Ex. 12. Given 2x3—12x2+36x=44, to find the value of c 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 x2+a'x3+b2x2+r'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 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 x2+ax2+bx+c is supposed to be the product of two quadratic factors, x2+px+q and x+x+s, in which p, q, r, s, are unknown quantities. Or, which is the same, the biquadratic equation x1+ax2+bx+c= O is considered as produced by the multiplication of the two quadratics, (2) · (3). x2+px+9=0; x2+rx+s=0 . 558. Hence, by the actual multiplication of the above two factors, we shall have x2+(p+r)x3+(s+1+pr)x2+(ps+gr)x+qs= x4 + bx+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, 1 ptr=0; stq+pr=a; ps+qr=b; qs=c. Or, if p, which is the value of r in the first of these, be substituted for r in the second and third, they will become, b s+q=a+p2; s-q=-; gs=c. P 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 p And, therefore, by inultiplying by p2, and placing the terms according to the order of their powers, the result will give, p+2ap1+(a2-4c)p2=b2. . . (4). From which last equation, if there be put p2=z, we shall have, z3+2az1+(a3—4c)z=b2 b (5). Hence, also, since s+q=a+p, and s-q=-, there will arise, by addition and subtraction, 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+8=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 x-3x2+6x+8=0, to find its roots, or the values of x. Comparing this equation with x1+ax2+bx+c=0, we have a=-3, b=6, and c=8; therefore, z3+2az2+(a2-46)z—b2—z3—622+23%−36=0. Let zy+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)]+3/[49—}✓✓(65856)] 466) (49+28.514)+(49-28.514)=3/(77.514)+/20. 4.264+2.736=7. Hence, z=y+2=9, and p2=z=9, or p=+3; ..(Art. 559), taking p=3, s=-3+2+1=3+1=4, and q= -3+21=2. Consequently, by substituting these values for p, q, and s, 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, 1-1/-7. Ex. 2. Given x2-6x2—17x+21=0, to find the values of x. Ans. x=3, or 1; or −2±√ ~3. Ex. 3. Given the equation 1-4x3-8x+32=0, to find its roots, or the values of x. Ans. 4, or 2; or 1-3. Ex. 4. Given the equation x2-6x3+5x2+2x-10=0, to find its roots, or the values of x. roots, or values of x. Ans. 1, or +5; or 1±√- 1. Ex. 5. Given x4-9x3+30x2-46x+24=0, to find the Ans. x=1, or 4; 2±√−2. Ex. 6. Given x2+16x3+99x2+228x+144=0, to find the roots, or values of x. Ans. x=-1,-3; or -6-12. −1, −3; 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. § IV. RESOLUTION OF NUMERAL EQUATIONS BY THE METHOD OF DIVISORS. 561. Since the last term (v) of the equation (A)=x+ Axm-1+Bxm-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 2, 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 3—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 x2 -x2-2x+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 gives x±√7 ; the three roots of the given equation, therefore, are -2, + {√7}}-{√—7. 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 (A)=0, be—a, 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 a+2. a+1, a, α-- - 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 r3+x1 − 14x3—6x2+20x+48=0, to find the roots of the equation, or values of x. Num. Results.| Divisors. -1 1, 2, 3, 4, 6, 8, 12, 24, 48, Progress. 3 4 3 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 divided by it, and the quotient is x2+ 2x+2=0, whose roots are-1+-1; the five roots of the proposed equation are, therefore, 2, 3, -4, -1+√−1, -1-√-1. 564. If the highest power of the unknown quantity has any coefficient prefixed to it, let the equation be assumed of the form (nx+a)p=0, and substitute 2, 1, 0,-1,-2, successively for x, as, in the former instance. Then, as before, the divisors of the several results, arising from this substitution, will be the terms of the arithmetical series. 2n+a, n+a, a, -n+a, and -2n+a; where the common difference n must be a divisor of the first term of the equation, or otherwise the operation would not succeed. Hence, in this instance, the progressions must be so taken out of the divisors, that their terms shall differ from each other by some aliquot part of the coefficient of the first term. Therefore, if the terms of these series, standing opposite to 0, be divided by the common difference, the quotient thus arising, taken in and, according as the progression is increasing or decreasing, will generally be the roots of the equation. It is necessary to continue the series 2, 1, 0, −1, −2, far enough to show whether the corresponding progression may not break off, after a certain number of terms; which it never can do when it contains a real rational root. Ex. 3. Given 2x3-3x2+16x—24-0, to find the roots of the equation or values of x. Substituting 2, 1, 0, -1, -2, successively, for x, as in the. former case, we shall have Where the progression is ascending, the number to be tried is, therefore, 3, which is found to be a root of the equation. Let the given equation be divided by x-3, and the quotient is 2x2-16=0, whose roots are +2/2; the three roots of the proposed equation are, therefore, -3, +2/2,—2√2. Ex. 4. Given x2+x3—29x2 — 9x+180=0, to find the roots of the equation. Ans. 3, 4, —3, and —5. Ex. 5. Given xa—4x3- 8x+32=0, to find the roots of the equation, or values of x. Ans. x2, or 4; or 1-3. |