formula (2), x=23/2.8(1+; .16 70.56 14936.1408 Now, by comparing this value of x, with 3/(a+b' ✓-1)+(a−b'-1), we have a' 2.8, and b'.4; ... substituting these values for a' and b' in the above .2560 &c.) =2.82(1+.00227-.00002, &c.)=2.826345 nearly, Here three terms of the series are sufficient, on account of its converging so rapidly, to give an approximate value of x, which is exact enough for all practical purposes. And, in fact, the value may be still found more accurate by continuing the series to five or six terms. Ex. 4. Given x-324-2z-8=0, to find the values of z. Let z2x+1, and the equation will be transformed into x3-5x=12; .. since a=-5, and b=12, x=3/[6+√(36—125)]+3⁄4//[6−√(36—125 )] =(6+5.6009)+(6—5.6009) = 2.26376+.736 24=3. 27 27 And, consequently, 22x+1=4, or z=±2. 555. Two roots of the proposed equation, therefore, are 2 and 2; divide z-3z4-222-8 by 224, and the quotient is z+z2+2;.. (Art. 529), z1+22+2=0, whose roots are z=√(√7); Hence four roots of the proposed equation are imaginary. 4 It may be observed that, in general, all equations, as 23m+az2m+b+c=0, may be reduced to one of the third degree, by putting zTM=x—ļa. - Ex. 5. Given x3+30x=117, to find the values of x. Ans. x 3, or 1-3. Ex. 6. Given x3 +9x=270, to find the values of x. Ans. x=6, or -3±6-1. 3 Ex. 7. Given x3-36x=91, to find the values of x. Ex. 8. Given x3-6x2+10x-8=0, to find the values of x, Ans. 4, or 1-1 Ex. 9. Given x3-3x-4=0, to find the value of x. Ans. *=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 z3-6z2+13z-12=0, to find the values of z. Ans. z=3, or -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++ax3·+b2x2 +r+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 x2+ax +bx+c=0 . . (1); 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 x1+ax2 +bx+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, • (3). (2) . . . . x2 +px+q=0; x2+rx+s=0 558. 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= + 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, p+r=0; s+q+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, b they will become, s+q=a+p2; s−q=-; qs=c. 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 6 b2 p2 4qs, or 4c. 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 (4). From which last equation, if there be put p2=z, we shall have, z3+2a23+(a2-4c)z= b2.. (5), b Hence, also, since s+q=a+p3, 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+s=0; we shall have (); (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 r1-3x2+6x+8=0, to find its roots, or the values of x. Comparing this equation with x+ax2+bx+c=0. we have a=-3, b=6, and c=8; therefore, 23+2az2+(a2 —4c) z—b2=z3-6za-23z-36=0. Let z=y+2, aud 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+ ar=b, (Art. 549), we have a=-35, and 6=98; therefore, (Art. 550), y=/[49+√(65856)]+[49—↓√(65856)] =(49+28.514)+3(49—28.514)=2/(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=−3+&+1=3+1=4. and 9-1=2. Consequently, by substituting these values for p, q, and r, in the equations (2), (3), we shall have x2+3x+2=0, and x-3x+4=0; so that the four roots of the given equation are -1, ·2, } + { √ −7, -7. - Ex. 2. Given x-6x-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, to find its roots, or the values of x. Ans. 4, or 2; or 1-3. 3 Ex. 4. Given the equation x2-6x3+5x2+2x10=0, to find its roots, or the values of x. Ans. 1, or +5; or 1±√-1. Ex. 5. Given x1-9x3 +30x2-46x+24=0, to find the roots, or values of x. Ans. x=1, or 4; 2±√2. Ex. 6. Given x-16x3+99x2+228x+144=0, to find the roots, or values of x. Ans. 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.、 m Ans. —√2±√(1+√2), +√2±√(1−√2). IV. RESOLUTION OF NUMERAL EQUATIONS BY THE METHOD OF DIVISORS. m--2 561. Since the last term (v) of the equation (a)= 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 x3—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 |