we muft bring them by Trials to Equality, which in moft Cafes is eafily done, obferving their Difference, and the Nature of the Scheme or Figure. Before I give Examples, I will premife this following Lemma, which fhews the Ground and Demonftration of this Way of Conftruction. Let ABD be a Semicircle (See Fig. 1.) on the Dia- Fig. 1 meter AD, AB, and AZ, two Subtenfes drawn at Pleafure from the End of the Diameter A, from B and Z are drawn the infinite Lines BC and ZS perpendicular to AD, BC interfecting it in m, and ZS in n; from A, draw the Line AS, interfecting BC in R, and ZS in S, I fay, that AB2: AZ2 :: mR : NS. : 2 For by the Property of the Circle DA x mA = AB', and DA x nA = AZ', then DA × mA : DA × nA :: mA nA: mR: nS, that is, AB2: AZ :: Rm: nS. Multiply the Extremes and Mean together, and it will be AB2 × nS = AZ × mR; if therefore we suppose AB =b, nS=c; AZ = Square Root of the Multiplier of two Dimenfions (in a Cubic Equation reduced into the Form above directed) then will mR be equal to the other Multiplier of one Divifion, fo in the firft Form above (zz x y+p = b2 × c) if AZ z, then is mR=y+p; and in the fecond Form zz -- pz × y = b2 × c ; if AZ = √zz+pz, then is mRy; and if y=x, then xx in the given Equation. Suppose I would conftruct this Equation, x3-4x2 = 72, or x3-px2=n, I take 16, as a convenient, fquare Number (which I call bb) and therewith I divide 72; the Quotient is 43, which I call c[c, and bbc = bb 72] I deduce alfo the other Side of the Equation into two Multipliers (as above) and then it is zz x 4-p=72 16 × 4, which is the firft Form for Construction. I defcribe Y 2 Fig. 1. . I defcribe a Semicircle ABD (See Fig. 1.) of a convenient Bignefs from my Scale of equal Parts which here, for this Figure, is of 24 in an Inch, 10 of which Parts make an Unit or 1.) and having drawn the Diameter AD, I take 4 (Units or large Divifions) off the Scale, and draw the Chord AB4=b, from BI draw the Infinite BC, Perpendicular to AD, and interfecting it in m. I take 4 (c) off the Scale, and fet that Distance with the Compaffes from m to C, and through CI draw CS parallel to AD. For the firft Trial, I confider, that the Root must be bigger than 4 or p (elfe the negative Term—4×× would take more than xxx, and fo the given Quantity would be negative) therefore taking 4 (p) from the Scale; with the Centre A, and Radius Ap=4, I defcribe the little Arch Pp; and then (at a venture) draw the Chord Ax (≈) interfecting the Arch Pp in p; fo is App 4, and the Line pp, or x-4. From z I draw ZS perpendicular to CS, and interfecting it in S, and then the Line AS interfecting MC in r. So is Mr =y—4, or y—p (the other Multiplier in the Equation) which being greater than the Line pz (to which it should be equal, it fhews, that z was taken too little. After the fame Manner I try another 2, which the View of the Scheme will now direct me to limit, till I find AZ, which anfwers the Demand; for making AZ =z, then is PZ ( = ≈—4) = MR (=y-4) confequently Zyx, z taken from the Scale is equal to 6, the Root fought. The fame Conclufion would follow, if I had inverted the Order of proceeding, and had begun with Mr, and thereby found A≈ (in a firft Trial) for in this Cafe I must have taken a Line for y (by Guefs) and made_mr= y-p, and then having drawn AS interfecting mC in "', and CS in S. Alfo SZ parallel to BC, touching the Semicircle in ≈; I draw Az, which will be equal to x, fo is the Line px-p, which ought to be equal to mr ; but not being fo, another Trial must be made. EXAMPLE Let the Root a =24, or x3-12 which dividing 24, Equation into this In the Semicirc AB=b=3. BC CS parallel to Al where AP-p=3, 14. Pd is perpendi taken by Guess, Circles drawn wi CZ, &c. fo are DQ. &c. = √z ZD (for Inftance) %. PZ=x-p. √2 DP2+PQ2)=zz—pz+99; therefore DQ=√zz—pz+qq. Having found dQ= √ zz−pz+qq. I draw Ax (See and than loo in the former Example) Fig. 2. Fi if mr Az, then is Az the Root of the double Cube AB2: Az2 :: mr : nS = mc = 2AB; therefore (when Axmr). Fig. F producing it, make ACc [but if b be greater than c, I make AB=c, and AC=b] Through CI I draw ZCP, perpendicular to AD, and applying AZ = √zz±pz±99, fo is the intercepted Chord AY=y; and if y=z, then is the Root fought. Elfe Trial must be made either |