Chap. II. Cubick, by Cardan's Method. 187 Value of a, in the faid Theorem, will become = 2 of Vs: and (by Chap. I. Part XV.) - I Confequently one of the Values of a, in the propos'd E quation, is (because r8, and s = √ 4.631089 243 X 2 .481201.) &c. = 4 &c. = 4+ .003341 &c. .000009 &c. &c. = 4.003332 &c. b 2 CHAP. CHA P. III. A Rule of Des Cartes for diffolving a BIQUADRATICK, into two QUADRATICKS. A NY Biquadratick Equation, which hath the 2d Term, being propos'd to be diffolv'd into two Quadraticks, muft have its 2d Term first deftroyed, (by Chap. 1. Prop. I. Schol. 1.) and Let the Equation thence produc'd, be fuppopos'd equal to x4*+qx2 + rx +so; in which Aquation x is fuppos'd to have the Sign + prefix'd to it; but the Values of q,ror's, may either be Affirmative or Negative. Let us fuppofe this Equation to be produc'd by the two following Quadraticksx+ex+fo, and x1-ex 2 go; in which Equations x is understood to have the Sign+prefix'd to it; and to the end that the second Term fhou'd be wanting in their Product, ex must have the Sign +in one Equation, and in the other, and the Values of ƒ and g are to be determin'd in the following manner; then f x2+q x2+rx+s=x1 +f, e x +eg +8 And by Equating their respective Terms, we have q=f÷ g-e2, r=8-ƒxe, and s = fg; wherefore q+e= ƒ+8=8-ƒ; consequently =g—ƒ; =f, and the Product of the two last Steps is pofe ey, and then you'll reduce the foregoing Equation to to this Cubick one, ý3 +2qy2+11= r2, whose se y 3 4 s. cond Term may be deftroyed, and the Root of the Equati on thus had, extracted by the foregoing Chap. and confequently the Value of y found; as alfo of ey; as alfo of f 2 and laftly of g⇒ 2 2 -. And by extracting the Roots of the two Quadraticks x + ex+ f = o, and x2 — ex-go, you'll have the four Roots of the Biquadratick x 4 4 4 If a 2 a3 — a 20, and the Values of a be required. 4 First, In order to deftroy the fecond Term, fuppofe a = x+, then the foregoing Equation will become x**4x2 -2x+4=0, in which-=q, -2=r, and 2 2 = s; wherefore y3 + 2 q y 2 + 11 y: — r r ( = 0 ) =y3 45 -3-3-4=0: And likewife to deftroy the fecond Term of this Equation, fuppofe y=v+1, then it will become v3 συ 9=0: Whence will be found, 4 + √ 2018 44 - √ 121 28 +1=2+1=3: And v+y= 3 + 4. Now y being thus found, its Square Root-2 is = e, 2 2 e r -+ e ; that is,+4+1, or 11⁄2 = ƒ, 2 , org: Wherefore the Equations x * 2 +ex+fc, and x-ex-go, are equal to x + 2 x 10, and 2-o refpectively; whence x ' * = −1 ±√ − 1, and x = 1±√={}: But a is (by Suppofition) = x+; confequently the 4 Values of a in the propos'd Equation, are + √ − 1 − 1 − √3/20 It and Note. One D. T. In A&t. Erudit. Lipf. published a Method of Solving Equations by destroying all the middle Terms; which, beal caufe very tedious, and as to what relates to Cubicks, lefs Practicable in every Cafe than Cardan's, Ifhall here omit. CHA P. IV. The Refolution of EQUATIONS by Stephinus's Method. THE Roots of a given Equation are by this Method found out by (fometimes frequent) Tryals': Thus; J Example I. 2 == Suppofe the given Equation to be a3 - 9 a 2 + 26 a —— 24, 'tis required to find the Values of a.. 3 First, I fuppofe a=1, and working according to the Æquation, find that a 3 (1) — 9 a2 (— 9) + 26 a (+26) 18; but it ought to be 245; wherefore I conclude a is I. I try again, and fuppofe and fuppofe a = 2, then will a3 (+8) 9 a2 (— 36) + 26 a (+52) = 24, which anfwers my Defire, and gives me one real Value of a: after which I may 'divide the Equation a39 a a + 26 a — 180, by a 20, which will bring it down to the Quadratick: may proceed further in the fame Method, and find alfo, that 3 and 4 are the two other real Roots." Example II. Or I If this irregular Equation was propos'd (where also the abfolute Number is a Fraction) x+5x=184638.6801. I can difcover at firft Sight almoft, that a must be at least 10, and trying with 10, I find it too little; but trying with 100 I find that by much too great; proceeding again I find 30 too too much; Itry with 20, and find it by fomethiug too fmall, but 21 I find too big; wherefore I know that x must be = 20, with fome Decimal Fraction annex'd; and at laft I difcover 20.7 to be the very Root fought, or at least one of the Roots of the given Equation. LEMMA to Chap. V. How to find all the Divifors of a given Number or Quan tity. 1 RULE. Divide the given Quantity or Number by the leaft of its Divifors that is 1, and the Quotient by the leaft of its Divisors that is, and fo on till you have 1 for a Quotient, and you'll have all the prime Divifors of the given Number or Quantity; then multiply each two, three, four, c. of the prime Divifors into themselves continually, and the feveral Products are the compound Divisors. Examples. I. So if all the Divifors of 60 were required. First divide it by 2, and the Quotient 30 by 2, and the Quotient 15 by 3, and the Quotient 5 by 5: then the prime Divifors are 1, 2, 2, 3, 5; and the Compound ones produc'd by each two are 4, 6, 10, 15; by each three are 12, 20, 30; by each four (or by all) 60. II. Again, If all the Divifors of 21 a b b were required. Divide it by 3, and the Quotient 7 abb by 7, and the Quotient abb by a, and the Quotient b b by b, and the b by b: then the prime Divifors are 1, 3, 7, a, b, b; and the Compound ones produc'd by each two are 21, 3 a, 3 b, 7 a, 7b, ab, bb; by each three, 21 a, 21 b, 3 ab, 3 bb, 7 ab, 7b2, abb; by each four, 21 ab, 21 bb, 3 abb, 7 abb; and by all five, 21 abb. III. In like manner all the Divifors of 2 abb are 1, 2, a, bb 3 ac; 2a, 2b b — 6 ac, abb 2 abb - 6 a ac. 6aac 3 a 2 CHAP. |
