I+33aa2ba + b b = dc+bb 2 3 w2 | 4|ab=√ 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. =-dc I±2aa2ba = ~ Then 3 aa-2 b a + b b = bb-dc, &c. And this Method of compleating the Square, holds true in those other Equations. Viz. Ia aaa+2baa=de Cafe 1. For 2 bbbb, as before. 2 ·⋅4—65aa = √ dc + Bb: b la = √ : √dc+bb:b, and fo on for the reft. Or let 146+2 baaa=dc, as before, Cafe 1. And 2 bbb b 1+2 a +2baaa+bb = dc + b b 3 4 a a a + b = √ dc + b b 4—65 aaa = √ dc+bb: -b 4 3 5 w 6α= : 3 √ : √dc+bb: — b, &c. COROLLARY. Hence it is evident, that whatsoever Method is ufed in folving thefe (or indeed any other) Equations, the Refult will fill be the fame, if the IVork 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+ 2 bade, then The like Theorems may be easily raised for the rest. If the known Co-efficients (of the fecond or loweft Term) be any fingle Quantity, as a a+bade, &c. then is and bb will be the Square of that Half; that is, and then the Work will ftand ICD 2αa+ba+, ÷ b b = dc + 4 b b 2 w2 3a + b = √ dc + ÷ bb b x b it's Half, 1 b = 1 bb, 364a= √ de + 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 firft 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. Suppose 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 Dimensions (viz. the Index) of it's higheft 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 these 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 (af 86=0 Hence the fecond Value of a is 86, or 86a which feems impoffible, viz. that an Affirmative Quantity should 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 ||a86 12 2 aa=+7396, viz. - 86 x 867396 I x 323 32a = 2752 2+3/4/aa+32a=4644, as at firft. EXAMPLE 2. Suppofe | aa7a948,75, then per Theorem 2. 2 2 e 3a (or 3,5) = 961 = 31 w 3+3,5141a = 31+ 3,5 = 34,5 Again, for the fecond Value of a, let aa-7a-948,75=0, and a 34,50. Then a-34,5=0)aa-7a-948,75=0 (a+27,5=0. Confequently this fecond Value is a=27,5 which will form the original Équation, a a-7a948,75 if it be ordered as the laft was. EXAMPLE 3. Suppofe 36 a-a a 243, then per Theorem 3. a=18 324-243, viz. half 36 fquared is 324, &c. that is, a=18 / 81; but 819, therefore a 18-99. 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 or, a=18-9=9, as before. And both thefe Values of a are equally true, as to forming the given Equation; viz. 36a-aa 243. For if a=9, then a a=81, and 36a=324; but 324 81243, therefore a = 9. 18+ 81, viz. a=18+9=27, Again, if a 27, then will aa=729, and 360=972: But 972-729=243, confequently it may be, a=27. Now either of these 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- -aa. 2430, and 9-4=0 then 9-a=0) 36a-aa-2430 (a-27=0 Hence, if a 270, 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 Answer to the Queftion, and that is to be chosen according to the Nature and Limits of the Question, as shall 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. à a +32a=4644 Here a= 24=32 In the last Example aa-7a=948,75 And a=24,5 Add 2a=+7 36a- -aa=243 Which was changed into a a-36a=243 Here a 9 9} And a275 2a=36. Dd Add Hence Hence it is evident, that if either of the Roots be found, the other may be eafily had without Divifions. If the Contents of this Section be well underftood, it will be eafy to give a Numerical Solution to any Quadratick Equation, that happens to arife in refolving of Queftions, &c. And as for giving a Geometrical Conftruction of them, I think it not proper in this Place; becaufe I here fuppofe the Learner wholly ignorant of the firft Principles of Geometry, therefore I fhall refer that Work to the next Part. CHAP. IX. Of Analysis, or the Method of refolving Problems exemplified by Variety of Numerical Duestions. N. B. HERE I advise the Learner to make Ufe always of the fame Letters to reprefent the fame Data in all Questions. If a reprefent any Number Viz. {And a reprefent a lefs Number e rates their Sum... } or or other Quantity, aa+eez the Sum of their Squares. Any two of the fix (s, d, p, q, z, x) being given, thence to find the reft; which admits of fifteen Variations, or Questions. Question 1. Suppofe s and d were given, and it were required by them to find a. e. p. q.z. and x |