PART X. The Solution of Adfected, Quadzatick, Æquations. ALI Adfected Quadratick Equations, do (as hath been already fhewn) fall under the Confideration of thefe four Forms or Cafes, viz. When there happens to be more Terms in one of these kind of Equations than two, and the highest Power of the unknown Quantity is Multiplyed into fome known Coefficient, you must reduce them by Divifion (as in Part VII. Se&t. 4) and for the Fractional Quantities that may arise by those Divifions, Substitute another whole Quantity. For Inftance, Let baa + caa — da — ca— de ↓ cb. f Then by Dividing by bc, you'll have de cb Then will aa-ka, be the new Æquation equal to the other, and now fitted for a Solution. But here we will leave it, and return to our four foregoing Cafes; In order to folve each of which, it will be requifite to premife this Lemma. Half (d) the Difference of any two Quantities (and whereof is the greater) Added to half (s) their Sum, is equal to the Greater; but Subftracted from half their Sum, is equal to the Leffer of them. * See Page 114. * It hath been already prov'd, that in each of the four foregoing Forms or Cafes of Adfected Quadratick Equations, the Sum (s) and Rectangle (r) of the two fought Roots, are given to find their Difference, and then the Roots themselves feverally. Now (by the foregoing Lemma) And 2 greater of the fought Roots. Ld - leffer of them. 2 And by Multiplying them by one another, you'll have SS 1 dd = r. 4 Then by Multiplying the last Æquation by 4, you'll have $5 dd 4. That is, by Tranfpofition ss And by Evolutions And by our Lemma 4rdd. 4rd. greater of the fought Roots. Wherefore the Canon for finding the two Values of a in this ift Cale, viz. is aa + pa bo is (because, as we have already prov'd, s p, and r= b'in this 1st Case) — 1 p + √ pp + 4b Greater Lefs or Negative Value of a. S is Alfo the Canon for finding the two Values of a in this 2d Cafe, viz, aa ho is (becaufe s Da λä+qα=6o pons = 2 huice, Z X X - 4 X + 4 + 1 X - 8÷60 Ft 4 lwn XX=64x=8 unca-6 36+24= Go } in this 2d Case) ≤ p + No pp + 4b Greater Lefs or Negative S Value of 4. Again, the Canon for finding the two Values of a in this 3d Form or Cafe, viz. aa rb in this 3d Čase), 2 qaho is (becauses isq and which Values are Affirmative if qq4h; and for that Reason this Equation is call'd Ambiguous; but if gg 4b, then the faid Values are Imaginary: for, in this latter Cafe, the SquareRoot of a Negative Quantity is to be Extracted in order to find the Value of a. But a Negative Quantity hath no Square Root; for whether the Root be Affirmative or Negative, its Square will be Affirmative; wherefore fuch impoffible Equations are said to have Imaginary Roots." And the Canon for finding the Values of a in this 4th Cafe viz. aa + qa+b o is (because s 9 and rb in this 4th - Greater Lefs Value of a, both of which Values are either Negative or Imaginary, and therefore this laft Cafe is not taken notice of by many Algebraifts. And I (for Brevities fake) will infert it no more. Tho' the foregoing Method of refolving Adfected Quadratick Equations, is what naturally follows from Mr. Harriot's Method of Compounding them; yet it is not the fame with his, which is a peculiar way of his own: and that is, after Tranfpofing.h, by Adding the Square of half the Coefficient, to each part of the Equation, he thereby makes the unknown Part a compleat Square in Species. Thus by Tranfpofing the known or given Quantity in the three first foregoing Forms or Cafes, they will become 1. aapab 2. aa pa = b 3. aaqah Refpectively. And thefe Mr. Harriot calls Canonical Equations. Now 'tis manifeft that if you Add the Square of half the Coefficient to both fides of each Equation, the firft Part of it will thereby become a perfect Square, by which Means the Value of the fought Root a will be made known; thus 1. In the 1st Form where aa pab, Add the Square of ptowitpp to each Part, and the Sums are aa pa pp = b + ÷ pp. raa-4a=12 4 X X + 4 x + 4 + 4X - S = 12-XX- 4 A Then by Extracting the Square-Root of each Part, x.se = ÷ p± √√pp, as before. -- 2. In the 2d Form, where aa pah. By Compleating or Adding the Square of to each Part, aa papp = b + 1⁄2 pp. Then by Extracting the Square-Root of each Part, 2 ga= b. And by Transposition 4 = p± √√b+pp, as before. 3. In the 3d Cafe or Form, where aa Compleat the Square, by Adding the Square of half the Coefficient-4, that is qq to each Part, and you'll have aa - qa+÷/gq = b + q¶• 4 Then by Extracting the Square-Root of each Part 2.3.4= ÷ Now by the help of these Theorems it will be eafy to Calcu late or find the Value of the unknown Quantity (a) in Numbers. Example 1. Cafe 1. Suppofe aa+a6; Then p= I, and b= 6; wherefore a = 1 + 2 1/2 fay, the two Values of a are + 2 and 3. And either of these Roots being known, the other may be found by Substracting the known one, from the Coefficient of the 2d Term with a contrary Sign, as is evident from the Compofition of adfected Quadra ticks. So, if I know that - 2 is one of the Roots of the above Equation a+a6, I can have the other 3 by Substracting 2 from the Coefficient of the 2d Term with a contrary Sign, that is from 1: or if I had known that 3 is one of the Roots (or Values of a) in the faid Equation, I can find the other2, by Subftracting 3 from I: or if one of the the Roots be known, the other may be had by reducing all the Terms in the given Equation to one Side, and Aquating them to o, and then Dividing them by a given (or known) Quantity c thus ·20) aa a-6=0 (a + 3 = 0 The truth of this is manifeft from the Compofition of Original Adfected Quadraticks. And tho' it may seem wonderful that the Value of a should be -3 in the foregoing Equation; yet by that Value the true (or firft) Equation may be found thus. 3 i22aa9, viz: = - 3 X-3 I + 23 aa+a=9−3=6, as at first. Example 2. Cafe 2. Suppose aa74 948. 75. Then-p-7 and b 948. 75. And by Canon (or Theorem) 2... = 42±√948.75 += 2 + √961 = 3 ± 31= 34.52 ·27.55' Now, either of thofe Values of a being known, the other may be had by Subtraction or Divifion, as in Example 1. |