A Problem, producing a Simple, Onadratick B N NI C A B D גיי F Having given the Square A D, and a ftraight Line BN, you are to produce the Side AC to E; fo that EF E drawn from E towards B fhall be equal to BN. It will be evident, if you imagine a Semi-circle to pafs thro' the Points B and E, that HG the most commodious Way will be to find the Line DG, that you may have the Diameter BG, upon which, having afterwards defcrib'd a Semi-circle, there will be need of no other Operation to fatisfy the Problem, than to produce the Side A Ĉ, till it meets the prescrib'd Periphery. Sclution. 2 NBC FE 4. And 4BF = y. Then 5BG=b+x. And 6 BE = 6 BE=y+c. The As BFD, GEH, and BEG are fimilar; wherefore 7b (BD) y (BF)::b(EH) ·y (EG). 8b+x(BG)·y(EG)::y+c(BE)·b(EH) 9 b b + bx = yy+yc. 9bb 47. 1 Eucl. El. 10 b b + 2b x + xx (BGq) = yy (EGq) +y+2cy + cc (BEq). 10-9.11 6x+xx=yy + cg+cc. II .cc. 126 x + xx cc = yy + cy. 12=9.136x+xx — cc (=yy+cy)=bb+bx. 13, 14 x x = bb + cc. 14 w. 2. 15 x = √ :bb + cc:= 5. Conftruction. Having produc'd the Side of the Square B A to N, fo that BN fhall be cqual to the given straight Line BN; then, fince BD is, and BNC, the Hypothenufe DN will be =√:66+cc:=x. Having therefore made DGDN, and and defcrib'd a Semi-circle upon the whole Line BG, if AC be prolong'd until it meet the Periphery in E, you'll have done that which was required. 47. 1 Eucl. El 56 7+ a a a=DC= ? 48. b BD BD .a. |ba = BD q ppaa BDq. 716α = pp-aa. 8a a + ba=pp. Cafe 1. of adf. Quadrat, Comp. 9a a + b a + b b = p p + b b. a + b = √ : pp + bb : Make BM=b, and erect the BC, which make p On Mas a Center with the Radius MC, defcribe a Semi-circle NCL, interfecting the ftraight Line BM produc'd both ways in the Points N and L. 4 Conftruction. C : b = 27 or-75. I fay, that the ftraight Line NB is the affirmative Root, and LB the negative, of the Equation a a+bapp. For feeing BM is half the Difference of the ftraight Lines NB and BL; if NB be put for a, then BL will be a+b; and therefore, fince NBBC:: BC BL (per 13. 6 Euck. El.), i. e. app a + b, a a + bapp, as before. In like Manner, if LB bea, and confequently LB-a, NB will be ab; And (by 13. 6 Eucl. El.) NB x LB is BCq; that is, aa+bapp: Wherefore the Conftruction is right. Let I d AE 15. 4.6 Eucl. El. 47. 1 Eucl. El. 6 x 2. 7-8. 5 d+e(AB)p(BD): : a(AC). e = EB. 6de + ee pa. 7 d d + 2 de + ee+ee( ABq+BCq) =aa=dd + 2 de + zee. 82de + 2ee2 pa.. 9 ddaa 2 pa. Cafe 2. 9+ pp. 10 dd +ppaa-2 pa + pp. 11+p. 12V: dd + pp: +p=a = 75 or 3. Points N and L. I fay, that the ftraight Line LB is the affirmative, and NB the negative Root of the Equation aa 2 pa=dd. For feeing BM is half the Difference of the ftraight Lines LB and BN, if L B be put for a, then BN will be a 2p; And (by 13. 6 Eucl. Elem.) LB x BN=BCq; that is, 2pa= d d, as above. a a In like Manner, if NB be- a, then LB 21 d, and NB x LBBC q is 2 pa + a add; wherefore the Conftruction is right. IO W. 2.II II. 12 4 a = b = + √ : 4 b b − p p : pp: a = b + √ b b - pp: 48, or 27; = b : { but, because of the Limitation in the Problem, a 48. = Conftrution, the LBD to the Diameter A C. I fay, that both AD and DC are affirmative Roots of the Equation a aba =—pp. For A C or being their Sum [See Pag. ], if A D be puta, DC will be ha; or if DC be a, AD will be ha; whence, in both Cafes, baaa, or the Rectangle of AD x DC, will be equal to pp, or the Square of DB (per 13. 6 Eucl. Elem.), which was to be done. = This Equation fometimes becomes impoffible, viz. when p is fo great, as that the || BF does neither cut nor touch the Circle ABC; that is, when p is greater thanh: Forp ought to be a Geometrical Mean Proportional between the Parts (AD and DC) of k, and confequently not greater than an Arithmetical Mean, or b; Nor are they equal, except in the Cafe of Contact; where likewife a and p become equal. All fimple Equations may be conftructed by ftraight Lines only. See the four firft Problems of this Part II.] But Plane, or Quadratick Equations require, befides ftraight Lines, the Circle, or fome other Curve of the Conick-Sections to conftruct them. [See the four laft Problems.] In the three laft Problems you have Methods of conftructing all original adfected Quadratick Equations. And as for thofe of higher Degrees, I will not infert here the Methods of conftructing them; for, whenever 'tis required that any Thing in Geometry fhould be accurately determin'd, a Mathemati cian must not undertake to do it by Rule and Compass, becaufe of the Defect of Inftruments, and of our Senfes, whereby the Interfections of Lines imperfectly drawn are yet more imperfect: But he will give a Solution as near the Truth as you please by an Arithmetical Calculation, according to an Equation determining the Nature of the Problem. Moreover, Cubick and Biquadratick Equations are not to be conftructed by ftraight Lines and Circles; 'tis true, they may, by the additional Help of a Parabola, which, indeed, from the |