Example. - If a4 — 2 a3 — a+2=0, and the Values of a be required. * First, in order to destroy the fecond Term, fuppofe a = +; then the foregoing Equation will become + —— § x2 - 2x+2123=0, in which is = q, — 2 = r, and =s; wherefore y3 +2qy2 +¶¶ y — rr (= 0) = y3 - 2 +99 -45 3y2-3y40: And likewife to deftroy the fecond Term of this Equation, fuppofe y=v+1, then it will become v3 6V 9 o: Whence v will be found, by the laft Chap. 42 + √20 4-8 : + 3 / 4 3 —√124: = √8 +1=2+1=3: And v+1=y=3+1= 4. Now y being thus found, its Square-Root 2 ise, : 2 org: Wherefore the Equations x2 + ex+go, are equal to x2 + 2 x ex+f: =0, and x2 +14=0, and x2-2x+4= o refpectively: Whence x = −1 ±√ −4, and x = 1±√ = {1}}: But a I is, by Suppofition, x+; confequently the four Values of a in the propos'd Equation are +4, — 3 — √, 2 and i. 3 4 - M. Tschirnhaus, in A&t. Erud. Lipf. publifh'd a Method of folving Equations by destroying all the intermediate Terms, which I will not infert here, because it is very tedious; and, as to what relates to Cubicks, less praEticable, in every Cafe, than Cardan's. CHAP. CHA P. IV. The Solution of Equations by Stevinus's Method. TH HE Roots of Equations are, by this Method, found out by (fometimes frequent) Trials: Thus, Example I. Suppofe the propos'd Equation to be a3-9 aa+26a=== 24; 'tis required to find the Values of a therein. First, I fuppofe a=1; and, working according to the Equation, find that a3 (1) — 9 a a (−9) + 26 a(+26) 18; but it ought to be 24; wherefore I conclude a is CI. I try again, and fuppofe a=2; then will a3 (+8) — 9 aa (36) + 26 a(+52)=24, which anfwers my Defire, and gives me one real Value of a: After which I may divide the Equation a39aa26a240 by a -2, which will bring it down to a Quadratick; or I may proceed further in the fame Method, and find also that 3 and 4 are the two other real Roots. Example II. If this irregular Equation was propos'd (where alfo the abfolute Number is a Fraction) 4+5x=184638.6801. I can difcover at firft fight almoft, that a must 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 much; I try with 20, and find it by fomething too fmall; but 21 I find too big: Wherefore I know that x must be 20 with fome Fraction annex'd; and at laft I discover 20.7 to be the very Root fought; or, at leaft, one of the Roots of the propos'd Equation. LEMMA to CHAP. V. How to find all the Divifors of a propos'd Number or Quantity. RULE. Divide the propos'd Quantity or Number by the leaft of its Divifors that is 1, and the Quotient by the leaft 1, and fo on 'till you have I leaft of its Divifors that is for a Quotient, and you'll have all the prime Divifors of the propos'd Number or Quantity: Then multiply each two, three, four, &c. of the prime Divifors into themselves continually, and the several Products are the compound Divisors. Examples. I. So, if all the Divifors of 60 were required. First, di vide 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 (always omitting Unity) 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 abb were required. Divide it by 3, and the Quotient 7 abb by 7, and the Quotient abb by a, and the Quotient bb by b, and the Quotient 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, 3b, 7 a, 7b, ab, bb; by each three 214, 216, 3 ab, 3 bb, 7ab, 7bb, abb; by each four 21 ab, 21 bb, 3 a bb, 7abb, and by all five 21 abb. III. In like manner all the Divifors of 2 abb-6 a acare I, 2, a, bb 3 ac; 2a, 2 bb. -6ac, abb-3aac; 2abb-6 aac, CHAP. V. John Kerfey's Method of finding the Roots of fome Adfected, Cubick, Biquadzatick,&c. Equations. FIRST prepare the propos'd Equation thus ; viz. 1. If the Coefficient of the highest Power of the unknown Root be greater than 1, divide the whole Equation by that Coefficient; and then 2. If any of the Terms be Fractions, multiply the unknown Root by fuch a Number as will give an Integer Product, and the Coefficient of its higheft Power 1 (by Propof. II. Chap. 1.) And the Root of the Equation (whether it be at first, or by these Directions thus prepar'd, call a. Then 3, Reduce all the Terms of this Equation to one side of it, and the other Side will be o. Then find all the Divifors of the abfolute Number in the Equation fo reduc'd, and try whether any of those Divifors connected to the unknown Root a, by- or+will divide the total Sum of the reduc'd Equation without leaving a Remainder: For when fuch Divifion fucceeds, either the known Part of the Refidual, or Binominal Divifor, with a contrary Sign, is the defired Value of the Root a, or, at least, the Quotients give an Equation whofe firft Term hath fewer Dimenfions by 1, than the Equation divided: And if this Equation contains three or more Dimenfions, let it be examin'd by Divifion, as before; and fo on. By which Divifions the Roots of the propos'd Equation may be fometimes made known, or the Equation may be reduc'd to a Quadratick one; and then the fought Root will be found by the Canons given for folving Adfected Quadraticks. Examples: I. If a3-15aa74a-1200; what are the Values of a? The Divifors of 120, the abfolute Number, are 1, 2, 4, 8, 3, 6, &c. Then I try whether a−1, a+1, a—2, a+2, a-3, or a+3 will divide the Equation without leaving a Remainder But, finding that neither of them will do, I try next with a 4, which will exactly do; and therefore 4 is one affirmative Value of the Root a: And the Quotients being a a- 11a300, the other two Roots will be found to be 5 and 6, by dividing a a-11a+30=0 by a or by a-6; or by the Canon given for folving the third Cafe of Adfected Quadratick Equations. a3 5, II. Let it be required to find the Roots of this Equation ·36aa2a-720. The abfolute Number 72 can be exactly divided by 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72; wherefore the propos'd Equation is to be divided by a-and+1, a—and 2, &c. to find fuch of them as will exactly do it: But, fince here are a great many Divifors, and that (by the Compofition of Equations) there can be, at moft, but three fuch Divifors, which will exactly divide the propos'd Equation; you you may try with a great many of thofe Divifors before you find any of the three fought: Wherefore, * Mr. Waeffenar's to fave your felf a great deal of this Trouble, Method. transform the propos'd Equation into another, each of whofe Roots fhall be more or less than those of the propos'd one by a given Number, 1 is generally the most convenient. Suppofe therefore a=x-1, then the above Equation will become x3-39xx-77x-11=c, an Equation whofe Roots are each by I more than those of the propos'd one: And the Divifors of its laft Term are 1, 3, 37, and III. But, fince a=x- I, it is evident that if any of the Values of x be 1, 3, 37, III, or I, 3, 37, III, (as it, or they, muft be, if it has any rational one) that, or thofe of a will be o, 2, 36, 110, or 2, — 4, —38, 112: But, by what was before faid, all the Rational Values of a are inferted among the following ones, viz. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72, or -I, -2, 3, 4, &c. Confequently, if a has any Rational Value in the propos'd Equation, it, or they must be 2, 36, or 2,4: Wherefore you need now try to divide the Equation propos'd only by a 2, a 36, a-1-2, and a-1-4: Wherefore I try first to divide it by a-2; but that not fucceeding, I try next to divide it by a 36, which exactly docs, the Quotient being a220, an Equation wherein a is ---- 2, and -2; and confequently the three Roots or Values of a in the propos'd Equation are 36, 2, and CHAP. VI. 2. The Solution of Adfected Equations by Sir Ifaac Newton's Method. T HERE is an univerfal Method of extracting Roots, either in Numbers or Symbols, invented by Sir Ifaac Newton, which you may find in Pages 381, 382, and 383 of Dr. Wallis's Algebra; which is to this effect. First, Find the first or greatest Member of the Root fought y; and, if it not bey, let that Member --p be fuppos'd y; then, having fubftituted this Binomial and its refpe&tive Powers for y, and its Powers in the Equation, collect its feveral Terms into one Sum: Then find the second Member * P |