Hence, a1 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 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. SCHOLIUM. From the Work of the three last 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 = 465 Here a= 54} Add 86 before. For in this Cafe, a 18+81, viz. a = 18+9=27, or, a 1899, as before. And both thefe Values of a are equally true, as to forming the given Equation; viz. 36 a — a a =243. For if a = 9, then a'a 81, and 36 a 324; but 324 -81243, therefore a = 9. Again, if a = 27, then will a a = 729, and 36 a = 972: But 972-729243, confequently it may be, a = 27. Now either of thefe Values of a may be found by Divifion, as thofe were in the other two Cafes, one of them being first found by the Theorem. Thus, let 36 a - All 2430, and 90 then 90) 36 a-aa-2430 (a 27 = 0 Hence it is evident, that if either of the Roots be found, the other may be eafily had without Divifion. If the Contents of this Section be well understood, it will be eafy to give a Numerical Solution to any Quadratick Equation, that happens to arife in refolving of Questions, &c. And as for giving a Geometrical Conftruction of them, I think it not proper in this Place; because I here fuppofe the Learner wholly ignorant of the first Principles of Geometry, therefore I fhall refer that Work to the next Part. CHAP. IX. Of Analyfis, or the Method of refolving Problems exemplified by Variety of Numerical Questions, N. B. HE ERE Iadvife the Learner to make Use always of the fame Viz. {And e reprefent a lefs Number } or other Quantity, aa + cez the Sum of their Squares. aaeex the Difference 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 Queftions. Question 1. Suppose s and d were given, and it were required .e.p.q z..and x by them to find a s=240 } Then and fuppofed=192 2 9—10/12/a a—ee = sd = x = 46080, x found. Queflion 2. Let s and p be given, to find the reft, That is { 1|ate=s=240⋅ } Quære a. evd. q. %. x. 2ae=p=5184 I2 3aa +2ae+ee=ss= 2 X 4 4 4ae=4p= 3-4 5ad-2 ae + ee=ss-4p=36864 57600 20736 4P 5 w2 6a6a-e=√ss—4p=d=192 1+6 72a=s+ √ss-4p ་ s+-4, hence a= 7 ·28a= =216 2 12 13 14 aa+eess⋅ ·2p=x=47232 1315 aa ees√ss=4p === 46080 |