And consequently, +2.601676, -2.261806, and -.339870, are the three roots required. EXAMPLES FOR PRACTICE. 1. Given x2 + 9x = 30, to find the root of the equation, or Ans. x= 2.180849. the value of x. 2. Given x3 the value of x. 3. Given x3 the value of x. 4. Given x3 of x. 3x 2x = 5, to find the root of the equation, or Ans. x= 2.0945515 3, to find the root of the equation, or Ans. 2.103803 27x=36, to find the three roots or values Ans 5.765722, 4.320684, and 1.445038. 48x2 200, to find the root of the equaAns. 47.9128. 24, to find the root of the equation, Ans. 5.1622 5. Given x33. tion, or the value of x. 6. Given x3 22x or the value of x. OF BIQUADRATIC EQUATIONS. A biquadratic equation, as before observed, is one that rises to the fourth power, or which is of the general form · 4 x2 + ax3 + bx2 + cx + d = 0. The root of which may be determined by means of the following formula; substituting the numbers of the given equation, with their proper signs, in the places of the literal coefficients, a, b, c, d. RULE 1.* *-Find the value of z in the cubic equation z3 + Gac 1 1 b 12 b2 _ d) a) z b3 += (c2 + da2) * This method is that given by Simpson, p. 120 of his Algebra, which consists in supposing the given biquadratic to be formed by taking the difference of two complete squares, being the same in principle as that of Ferrari. Thus, let the proposed equation be of the form c1 + ax3 + bx2+ cx + d =0(1), having all its terms complete; and assume (x2+ax+p)2 (qx+r)2 = x4 + ax3 + bx2+cx+d. Then, if x2+ax+p and qx+r be actually involved, we shall have 24+ax3+2p x4+ax3+bx2+cx+d. x2+ ap And consequently, by equating the homologous terms, there will arise by one of the former rules; and let the root, thus determined, be denoted by r. Then find the two values of x in each of the following quadratic equations :— and they will be the four roots of the biquadratic required. p2 - d 2qr where, since the product of the first and last of the absolute terms of these equations is evidently equal to of the square of the second, we shall have 2p3+(aα2—b) p2 — 2dp — d (‡a2 — b) — 1 (a2pî— Qacp+c2). Or, by bringing the unknown quantities to the lefthand side, and the known to the right, and then dividing by 2, b p3- p2 + 1 ( ac — 4d) p = } (c2 + a2d) — § bd . . . . (2.) 2 From which last equation p can be determined by the rules before given for cubics. And since, from the preceding equations, it appears that q = √(2p+ a2 —b) and r ap-c or V (p2— d), it is evident that the several values of z can be obtained from the quantities thus found. For, because 24+ ax3 + bx2 + cxd, or its equal (x2+ax+p)2→→ (qx+r)2 = 0, it is plain that (x2 + ax + p)2 = (qx+r)2. And, therefore, by extracting the roots of each side of this equation, there will arise x2+ tax +p=qx+r; or x2 + (‡a — q) x = r—-p. Whence, by substituting the above values of p, q, and r, for their equals, and transposing the terms, we shall have x2+{‡a=√(2p+‡a2—b) } x+p‡√(p2—d) = 0, for the case where ap-c is positive; and x2+ av 2p+ta2—b) } x+p±√(p2− d) =0, for the case where apc is negative: which two quadratics give the four roots of the proposed equation. b And by putting p=z+ in the reducing equation (2), in order to 6' destroy its second term, the several steps of the investigation may be made to agree with the expressions given in the above rule. EXAMPLES. 1. Given the equation x1 10x+35x2 50x + 24 = 0, to find its roots. Here a= 10, b 35, c = ~ 50, and d 24. Whence, by substituting these numbers in the cubic equation, 8d), we shall have the following reduced equation, which, being resolved according to the rule before laid down for that purpose, gives But, by the rule for binomial surds, given in the former And if this number be substituted for r, 1 (1 1 10 for a, 35 for b,. + 2 (r − b ) } ) x = − (r + = b) + v ·a2+2(r they will become, after reducing them to their most simple terms, Whence the four roots of the given equation are 1, 2, 3, and 4. Or, when its second term is taken away, it will be of the form x2 + bx2 + cx + d = 0, to which it can always be reduced; and in that case, its solution may be obtained by the following rule: RULE 2.-Find the value of z in the cubic equation and let the root thus determined be denoted by r, Then find the two values of x in each of the following quadratic equations : and they will be the four roots of the biquadratic equation required.* * The method of solving biquadratic equations was first discovered by Louis Ferrari, a disciple of the celebrated Cardan before mentioned; but the above rule is derived from that given by Descartes, in his Geometry, published in 1637, the truth of which may be shown as follows: Let the given or proposed equation be x2 + ax2 + bx + c = 0, and conceive it to be produced by the multiplication of the two quadratics x2+px+q=0; and x2‡rx+s= = 0. Then, since these equations, as well as the given one, are each =0, there will arise, by taking their product, x4+(p+r) x3 +(s+q+pr) x2+(ps+qr)x+q s = x++ux2 + bx+c. And consequently, by equating the homologous terms of this last equation, we shall have the four following equations, p+r=0; s+q+pr = a; ps+gr = b; qs = c; Or, 1= P; s + q = a + p2; s q ; qs C. Whence, subtracting the square of the third of these from that of the second, and then changing the sides of the equation, we shall have a2+2ap2+p4 b2 p2 4qs, or 4c; or p6+2ap1+(a2 — 4c) p2 = b2. Where the value of p may be found by the rule before given for cubic equations. b Hence, also, since s+q=a+p2, and s addition and subtraction, there will arise, by Ρ b 2p 1 2 1 b s == a += p2+ ;q= a+= p2 where p being known, s and q are likewise known. And, consequently, by extracting the roots of the two assumed quad 2 2p |