3b+ √\bb—p3]3: whence y() is given ལ ; and consequently s ( = x + y) :: which is, evidently, the true root of the equation s3-3ps=b. From whence the root of the equation x3+ax=b, wherein the second term is positive, will be given, by writing x for s, and a for-p; whence x is found In like manner, if things be supposed as above, and there be, now, given z5+y3=b; then, by the problem there referred to, we likewise have s—5ps3+5p2s=b. But the p for its equal y, becomes first equation, by substituting + b: whence i b≈5 = p3, z5 = ab + √4bb—p3, and ≈ — §b+√‡bb—p3]} ; and consequently s (=z+y=≈+2)=3b+√¥bb—p3]} + p 4b+4bb-ps } } = the true root of the equation $5-5ps3+5p2s=b. Which a by substituting æ for s, and for p, gives x fb + √ 4bb + zas root of the equation x+ax+ { u2 x = b. Generally, supposing z" + y2 = b, or z2 + 3b + √ Abb—p; therefore s (x + y) =≈+ P žb + √‡bb + p′′]÷; which is n the true root of the equations"- npsn 2 + n. This equation, by writing x for s, and for— p, becomes a + ax^2 + a n theorems are included, with innumerable others of the same kind; but as every one of them, except the first, requires a particular relation of the coefficients, seldom oc curring in the resolution of problems, I shall take no further notice of them here, but proceed to The Resolution of biquadratic equations, according to Here the second term is to be destroyed as in the solution of cubics; which being done, the given equation will be reduced to this form, x4 + ax2 + bx + c = 0; wherein a, b, and c, may represent any quantities whatever, positive, or negative. Assume a2+ px + q × x2 + rx + 8 = x2 + ax2 + bx + c; or, which is the same, let the biquadratic be considered, as produced by the multiplication of the two quadratics x + px + q=0, and x2 + rx + 8 = 0: then, these last being actually multiplied into each other, we shall have x + ax2 + bx + pr by equating the homologous terms (in order to deter mine the value of the assumed coefficients, p, q, r, and s,) 0, s + q + pr = a, ps + qr the first of which r = we have p + r = c; from = and qs third s q= b - P = b, p; from · pr) = a + p2; and from the a — pr) Now, by subtracting the square of the last of these from that of the precedent, we have bb 4qs = a2 + 2ap2 + ppp that is, 4c = a2 + 2ap2 + a2 bb pp (because qs = c); and therefore p + 2ap } p2 = b2; from which p will be determined, as in example the second, of the solution of cubics. b Whence s (= a + 1p2 + 7) and q (≈ 1a + §p2 — 2p 2p are also known. And, by extracting the roots of the two assumed quadratics x2 + px + q = 0, and x2 + rx + s = 0, we have x, in the one, — — q; and, in the other, 4 four roots of the biquadratic, x+ax2 + bx + c = EXAMPLE. Let the equation propounded be y1—4y3 — 8y + 32 0; then, to take away the second term thereof, let x* x + 1 = y; whence, by substitution, 4* 6x216x+21= 0; which being compared with the general equation, 4*+ ax2 + bx + c = 0, we here have a = 6, b 16, and c = 21; and conse + a2 quently po-12p4—48p2 (= p° + 2apa } p2) = 256 (= b2.) Now, to destroy the second term of this last equation also, make+4= p2; and then, this value being substituted, you will have z3 - 96% 576; whence, by the method above explained, z will be found ( = 288 + 28872 3213 2 3; which are the four roots of the equation x-6x216x+21; to each of which let unity be added, and you will have 4, 2, 1+ 1 + √ — 3, and 1 ✔ 3, for the four roots of the equation proposed; whereof the two last are impossible. And that these roots are truly assigned, may be easily proved by multiplying the equations, y-40, y—2 0, y + 1 - ✔ 3 0, and y + 1 + √ 3= 0, thus arising, continually together; for, from thence, the very equation given will be produced. = The resolution of biquadratics by another method. In the method of Des Cartes, above explained, all biquadratic equations are supposed to be generated from the multiplication of two quadratic ones: but, according to the way which I am now going to lay down, every such equation is conceived to arise by taking the difference of two complete squares. 2 - Here, the general equation + ax3 + bx2 + cx + d = 0 being proposed, we are to assume x2 + дax + A]2— Bx + C12 = x2 + ax3 + bx2 + cx +d; in which A, B, and C, represent unknown quantities, to be determined. Then, x2 + ax + A, and Bx + C being actually involved, we shall have cx+d: from whence, by equating the homologous terms, will be given, 1. 2A+ a2-B2b, or, 2A + 4a2 — b = B2; 3. A2- C2 = d, or, A2 -d = 2BC; = C2. Let now the first and last of these equations be multiplied together, and the product will, evidently, be |