But by a Numerical Operation, the faid Notification, as well as the Method of the Procefs of each Theorem, will be further illuftrated. Therefore, 1. Suppose x2 = 2 N, feek x by the firft General Theorem. N-N2 Then n = = T, and take m= 1. 2m Therefore I+T (=.5)=1.5=m the 2d. ·.· 1.5-T (=—.088)= 1.417m the 3d. 1.417-T(.002783)=1.414217m the 4th. 1.414217-T (= .000003437622) 414213562378m the 5th = x. = I. 2. Suppose xx=2= N. feek by the fecond General Theorem. Then x = N+N2 T, and take m=1, as before. 2m Therefore T = 1.5= m the 2d. T= 1.416m the 3d. T= 1.414215=m the 4th. By which it is evident, Firft, that both Theorems amount to the fame Thing, the Difference being only in the laft Figure, which would be corrected the next Operation. Secondly, that will proceed into an Infinite Series, if a Surd. Thirdly, that each Operation gives double the Number of Figures of the last. 3. Suppofe x4=28398241=N. feek x by Theorem 1. N-m4 Then n = 3m3 =T, and take m = 10. Therefore 10-T (=—3)=7=m the 2d. 7+T (=+·4)=7.4m the 3d. 7—T (=—.1)=7.3=x, the true Biqua dratic Root fought. 4. Suppose x42839.8241 N. as before, feek x by the fecond Theorem. T Then Then x = N-3m4 =T, and take m=5. 4m3 Therefore T=5.6m the 2d. Root fought. T= 8.2 m the 3d. T= 7.4m the 4th. T= 7.3m the 5th.x to the true From which two laft Examples it appears, Firft, that either Theorem will find the true Root, if it have one. Secondly, that it matters not, whether m be taken above or below the Root, or how far from it. 5. Suppose *2+587x= x2+587x=987459, or xx+px = N. Seek by Theorem the firft (i. e.) n = N-m2-pm 2m-+-p = T, because of the Punctations we are to take. 1. Operation xx+5x=98 Land suppose m=8. 2. 3. xx+58x=9874. xx+587x=987459 Therefore 8-T(=—.2)=78=m the 2d. 78—T(=—3.4)=746=m the 3d. 746—T(=—3·34)=742.66—m the 4th. 742.66—T(——.012689)=742.647311=x sought. Again. 6. Suppose xx-20x53482, or xx-x N. Seek by the second General Theorem. N+m2 Then x = =T, and take m➡250. 2m-p Therefore T=241=m the 2d. T = 241.4m the 3d. T= 241.475=m the 4th. T=241.477860m the 5th. =* fought. From these two laft it is plain; Firft, that there is no abfolute Neceffity for Punctation. Secondly, Secondly, That Punctation does nevertheless shorten the Work where it can be done. But I hope I have faid enough to make the whole Matter, as well as the Manner of proceeding plain and eafy to the meaneft Capacity; and though I have given Numerical Examples, no farther than an affected Quadratic, yet it is the fame to any Degree of Power, or Affectation whatfoever, Regard being had to its proper and particular Theorem deduc'd from either of the General Ones. I fhall therefore now proceed to Roots in General; but first I will fhew by an Example how the Cube may be compleated as a Quadratic. A new Way of Compleating the Cube. Uppose x3+12x2+48x=152. Here you fee it is a perfect Cubic Equation: Now the fame may be compleated thus. I observed the Canon for the Cube Root (x2+3bx2+ 362x+63) and I found the third of the Co-efficient (or 3b) cub'd, is always the fourth Term of a regular Cube, which I tried in this Cafe, calling 12 = 3b, and 48=362, the Equation then will ftand thus, 3+3bx2+362x = 152; then adding 64 the Cube of 6, a third of the Co-efficient 36, to both Sides of the Equation, and we shall have x2+3bx2+3b2x+64 = (152-+64) 216, and extracting the Root x+4=216) 6. and x = 2. Equation, OPERATI o N. Ix3 +12x2+48x=152 per Canon, 2x2+3bx2+3b2x=152 2 CO 3 w 3 x3+3bx2+3b2x+64 = 216. 4x+4=2166 345 An Univerfal Solution of Cubic and Biquadratic Equations, analytically. §. 1. Of the Univerfal Cubic Equation. That the Arithmetical Calculus may appear the eafier, and fitter for Operation, put the Cube Root of the irrational Binomial r+√r2-q3 to be m+vn, the three Roots of the fame Equation will be xp+2m, and x=p¬m±√—3n. Therefore in any given Cubic Equation, we must compare p, q, r, &c. for thefe being known, all the Roots of the Equation become known. Let the Root be fought of this Cubic Equation, viz. §. I. x3 = 2x2+3x+40 Hence 3 or p =—-—. Secondly, 39—(360°) — — — — 3. or 97 ~ 23 Hence it will be according to the Prescript, 3p=2, 2 4 = 13 9 §. 2. In this Equation x3=12x2-41x+42: In the firft Place it will be 3p=12, or p = 12 =4. Secondly, 3 (p2—39 × p) 36=42, or r=3. and from thence r2-q3 3+√ 100 27 100 27 But the Cube Root of the Binomial Surd, (=r+ √√r2 —q3) is to be extracted according to the fought Methods of Arithmetic, and is −1+√−1, (=m+√n,) and therefore the Root 4 3 ·(p+2m=4—2=) 2; or likewife x=(p—m±√—zz =4+1 ±(√4) 2 =) 7 or 3. Again, the other Root (=m+vn) and therefore the Root x=(p+2m = 4+ 3=)7; and likewise x = (p—m±√−3n=4—3 ± The third Root of the fame Binomial, viz. 3+√ |