And this Method holds good in thofe other Equations, wherein the highest Powers are at, a, as, &c. As, for inftance, The fame may be done with all the reft, Care being taken to add, or fubtract, according as the Cafe requires. But all Quadratick Equations may be more easily refolved by compleating the Square, which is grounded upon the Confideration of raifing a Square from any Binomial, or Refidual Root. (See Sect. 5. Chap. 1.) Viz. if a + b be involved to a Square, it will be a a+2ba+bb; and if ab be fo involved, it will be a a2ba+bb. Whence it is easy to obferve, that a a+2ba=dc (Cafe 1.), and aa-2ba dc (Cafe 2.), are imperfect Squares, wanting only bb to make them compleat. And therefore it is, that if half the known Co-efficient be involved to the fecond Power, and the Square be added to both Sides of the Equation, the unknown Side will become a compleat Square. Thus Let laa+2ba=dc I But 2 Here half the Co-efficient 2 b is b, which being squared, bbb b Lis bb. I+23 aa+2ba+bb=dc+bb Cafe 1. ພ 2 Leta a· But 2 Again. 2 ba de Cafe 2. bbbb 1+ 33 aa-2ba+bb=dc+bb 3 ww2 4 a—b=√ dc+bb, &c. as before. But in Cafe 3. you must change the Signs of all the Terms in the Equation, Thus 12ba-aa de Cafe 3. I+12 a a 2 ba ——dc = Then 3 aa- And this Method of compleating the Square, holds true in those other Equations. Viz. 1a aaa+2baa=de Cafe 1. For 2 bbbb, as before. 4-b5aa=√ dc+bb: -b : 5 w ,2161a = √√dc+bb: -b, and fo on for the reft. Or let a 2 baaa=dc, as before, Cafe 1. And 1 + 2 4 I aw 2 b 3 +2baaa+bb = dc + b b a a a + b =✔ d c + b b 5 aaa=√ dc+bb: —b 5 w3 16 a = 3 √ √ dc + bb : — b, &c. . COROLLARY. Hence it is evident, that whatsoever Method is ufed in folving thefe (or indeed any other) Equations, the Result will fill be the fame, if the Work be true; as you may obferve from the Operations of this Section: for both thefe Methods here propofed, give the fame Theorems in their respective Cafes for the Value of (a). Thus Thus, when a a+2ba=dc, then The like Theorems may be eafily raised for the reft. If the known Co-efficients (of the second or lowest Term) be any fingle Quantity, as a a+badt, &c. then is b it's Half, and bb will be the Square of that Half; that is, bxbbb, and then the Work will ftand Thus Iaaba=dc ICO 2 aa+ba+4 b b = dc+÷bb 3—1264a= √ di +÷bb:-b, and fo for the reft. Note, C□ placed in the Margin against the second Step, fignifies that the imperfect Square a a+ba in the first Step, is there compleated, viz. in the fecond Step. Now by the Help of thefe Theorems, it will be eafy to calculate or find the Value of the unknown Quantity (a) in Numbers. EXAMPLE 1. Suppofe a a+ 2 ba=z. Let b 16, and z = 4644. But z+bb4644 +256=4900, and 4900=70. But every Adfected Equation; hath as many Roots (or rather Values of the unknown Quantity) either real or imaginary, as are the Dimenfions (viz. the Index) of it's highest Power; and therefore the Quantity a, in this Equation, hath another Value either Affirmative or Negative; which may be thus found. The given Equation is aa +32a4644, and it's Root a=54. Let thefe two Equations be made equal or or equated to o, viz. to Nothing. Thus, Thus, a a+32a-46440, and a-540. Then divide the given Equation by it's firft Root, and the Quotient will fhew the fecond Value of a. Thus, a-540) a a+ 32 a 46440 (a+86=0 aa -54 a +86a-4644 (0) which Hence the fecond Value of a is -- 86, or 86 feems impoffible, viz. that an Affirmative Quantity fhould be equal to a Negative Quantity; yet even by this fecond Value of a, and the fame Co-efficient, the true (or firft) Equation may be formed Thus, Let | 1a86 122 a a=7396, viz. -86x-86=+7396 I x 32 3 32a = 2752 2+34/aa +32a=4644, as at firft. Suppofe I EXAMPLE 2. a a-7a948,75, then per Theorem 2. I CO 2 аа - 7a+2948,75 +42961 2 w2 30a (or 3,5)=√961=31 3+3,514 a = 31+ 3,5 = 34,5 Again, for the fecond Value of a, let aa-7a-948,75=0, and a34,50. aa— Then a 34,5=0)aa-7a-948,75=0 (a+27,50. Confequently this fecond Value is a=27,5 which will form the original Equation, aa7a948,75 if it be ordered as the last was. EXAMPLE 3. Suppofe 36 a-aa 243, then per Theorem 3. a=18 √324-243, viz. half 36 fquared is 324, &c. that is, • = 18 — √ 81; but ✔81=9, therefore a = 18—9=9. Now this third Form is called an ambiguous Equation, because it hath two Affirmative Values of the unknown Quantity (a), both which may be found without fuch Divifion as was used before. before. For in this Cafe, a 18+ 81, viz. a or, a=18-9=9, as before. And both thefe equally true, as to forming the given Equation; 243 For if a 9, then a a=81, and 36 a -81243, therefore a 9. = 18+9=27, Values of a are viz. 36a-aa 324; but 324 = Again, if a 27, then will a a729, and 36a=972: But 972-729-243, confequently it may be, a=27. Now either of thefe Values of a may be found by Divifion, as those were in the other two Cafes, one of them being first found by the Theorem. Thus, let 36 a- a a 2430, and 9 — a=0 then 9a0) 36a-aa-2430 (a-27=0 Hence, if a270, then a 27, as before. Notwithstanding all Quadratick Equations of this third Form have two Affirmative Roots (as in this) yet but one of thofe Roots will give a true Anfwer to the Queftion, and that is to be chofen according to the Nature and Limits of the Queftion, as fhall be fhewed further on. SCHOLIU M. From the Work of the three laft Examples, it may be observed; that the Sum of both the Roots will always be equal to the Co-efficient of their respective Equations, with a contrary Sign. Thus. In Example 1. aa+32a=4644 In Example 2. Here a= 54} Add 86 |