$74. By comparing the fecond and fourth equations of the laft article, you may obtain another value of k. For n=q—1—mk=; fo that (m being equal to p— k) — = q— 1 —pk+k2, and k2 — pk+q — 1 —=0. Which gives S k = p ± √ √ = p2 − q +1+. So that the = √ √ - 9+1+ quadratic divifor required becomes p = √ = p2 − q + 1 + 1/ This divifor must be tried when = 7 and at the fame time /=, the former expreffion not ferving in that cafe. By this formula, divifors may be found whose fecond terms may be irrational. How the divifors of higher equations may be found, when they have any, may be understood from what has been faid of thofe of four di menfions. SUPPLE→ SUPPLEMENT to CHAP. VII. Of the Reduction of Equations by Surd divifors. A N equation of four, fix, or more dimen fions, although it may admit of no rational divifor, may have one that is irrational. As the biquadratic + px3 + qx2 + rx + s = 0, which we fuppofe to be irreducible by any rational divifor, may yet, by adding a square k2x2 + 2klx + l2 multiplied into fome quantity n, be compleated into a fquare x2+ 1⁄2 px + Q In which cafe we fhall have x2 + 2p + 2 = √ n xkx+, and x is found by the resolution of an affected quadratic equation. 2 To reduce a biquadratic equation in this mane ner, we have the following RULE. 2 *If the biquadratic is x+px3+qx2+rx+5=0, where p, q, r, s, represent the given coefficie ents under their proper figns, put q- p2=a, r — — ap = ß, sa. And for n take fome integer common divifor of ẞ and 25, that is not a fquare number, and which, if either p or r is an odd number, must be odd, and, di* Arithmet. Univers. pag. 264. P 2 vided vided by 4, leave the remainder unity. Write β likewife for k fome divifor of if p is an n even number, or the half of an odd divifor if p is odd, or o if ß = 0. Subtract β nk from pk, and let the remainder be 1. For a+nk2 2 and try if, dividing Q 2 put EXAMPLE I. Let the equation propofed be x++ 12x — 17 =0, and because po, qo, r=12, 5=-17, we shall have a=0, ß=12, (=-17. And ß and 25, that is 12 and -34, having only 2 for a common divifor, it must be n = 2. Again, =6, whofe divisors 1, 2, 3, 6, are to be fuc n and 6 are substituted for k, Qbecomes 4 and 36, and I and 2-s being an odd number, is not divifible by n (2). Wherefore 2 and 6 are to be set aside. But when 1 and 3 are written for k, 2 is 1 or 9, and 2-s is 18 or 98 respectively; which numbers can be divided by 2, and the roots of the quotients extracted, being ± 3 and +7; but only one of them, viz. -3, coincides with 7. I put therefore k 1, 1=-3, 2=1, and adding to both fides of the equation nk2x2 + 2nklx +n, that is, 2x2 — 12x + 18, there refults x4+ 2x2 + 1 = 2x2-12x+18, and extracting the root of each, x2+1=±√/2xx-3. And again, extracting the root of this laft, the four values of x, according to the varieties in the figns, are being the roots of x++ 12x — 17 = 0, the equation at first proposed. EXAMPLE II. -II Let the equation be x-6x3-58x2-114x -110, and writing-6, 58, -114, -11 for p, q, r, s, respectively, we have — 67 = α, 3156, and 1133. The numbers Band 25, that is one common divifor 3, that is n=3. And the B 3k, and dividing or 105, by it get n 35, and this fubtracted from the quotient be equal to . But pk3x3, leaves 26, whofe half, 13, ought a+nk2 —-67 +27, that is, 2 or 2 -20 is equal to 2; and 2-s=411, which is indeed divisible by n=3; but the root of the quotient 137 cannot be extracted. Therefore I reject the divifor 3, and try with 5k; 105, the quotient is by which dividing β - n 21, and this taken from pk-3X5, leaves 6=27. At the fame time, Q(= a+nik2) - 67+75 2 2 divifible by n, and the root of the quotient 9, that is, 3, coincides with . Whence I conclude that putting =3, k=5, 2=4, n=3, adding to both fides of the equation the quantity nk2x2 + 2nklx + n2, that is, 75*2 + 90x + 27, and extracting the roots, it will be x2 + 1⁄2 px + 2 = √2 × 4x + ↳, or In like manner in the equation 4 gx3 f 15x2-27x+9=0 writing-9. +15,— 27, +9, for |