Alfo; 2. If a3-pa-q; then (p being b, and

q=c)

Ex. 1. Cafe 1.

If i3 3 ii9i13, and one of the Values of i be required.

First, to deftroy the fecond Term, I fuppofe a-1=i, then the foregoing Equation will become a3-6 a = 20: And, by the first particular Theorem, a 3:10--100-18:

2

2:10-1100-8:20. 39 2 3 0 4 8 4 5 4 1 3 +

Tho' the foregoing and Cardan's Theorems are equally true, yet, as to the Practick Part, inafmuch as Divifion is eafier than the Extraction of the Cube-Root, the former

are preferable, except only when the Value of

CC b3
--|-· is
27

4

o In that refpect they were, by fome, deem'd impra&ticable and falfe; but you may fee by the laft Example that those are not fo: 'Tis true the Method of applying them to Practice, in Cafes of this Nature, is not perfect; it being only by Trials I came to difcover that 2-3 is the Cube-Root of 10-24 3.

But fince the Extraction of the Cube-Root of 10

243 is more difficult than finding the Values of a in the Equation a3 21 a 20, I will now fhew how by knowing the latter you may perform the former. Thus Suppofe the Cube-Root of 10-243 to be

y (that the first Member is alwaysa is appa rent from Cardan's Theorem); then a3 \- 3 aa √ —y + 3 ax―y (— 3 ay)—y√-y=-10+√243: And the rational Part of the first Part of this Equation must be equal to the Rational Part of the latter; viz. a3-ayis -10; that is (knowing one of the Values of a to be 4)8-6y -10; therefore y3: And confequently a√ÿ 2-3 is the Cube-Root of 10-243.

=一

Or, fince one of the Values of a is likewife 1; therefore the Cube-Root of 10-243 will be alfo found (by this Method) to be (foray is

27

4

5;

ax

the

Or laftly, fince one of the Values of a is alfo · Cube-Root of 10--√243 will, in like manner, be found 2+ √-4.

-

I will now proceed to the Discovery of that general Theorem from which Cardan's are deducible; and then fhew how Equations, in which the Value of cc+b3 is negative, may be thereby folv'd.

I

Having proceeded as far as the 16th Step inclufive of the foregoing Operation, you'll from the 7th and 15th find for See the following Page.

the

*O

From the 3 and 19th Steps Cardan's Theorems may be

eafily deduc'd.

b3

Example in which the Value of 4 cc + b3 iso.

If a3-12.03a=16;

2

I

that is, if a3-paq, and that qq — — p3 is 0;

"Tis required to find one of the

the Help of Cardan's Method.

27

Values of a proxime by

The particular Theorem for evolving this Equation deducible from the 19th Step, is

4

$3

P3: — √—s; then the

27

Values of a in the faid Theorem will become 3:r+ √ ~s:+3:r−√-s: And (by Part XV. Chap. 1.) 3:r±√−s:=r3}+{xr3¬3×±√-s+x

I

therefore a (= 3√ir + √−s:+V : r — √ − s :) = 2 X

I

2055

243 × r

&c. fine fine: Confequently one of the Values of

a in the propos'd Equation is (becaufer is equal to 8, and/

sequal to—.481201) = 2 × 83 (= 2 × 2)+

4.631089

243×2

I I

497664

.962402

9×25 Sic. = 4

+.003341 &c.—.000009 &c, &c. = 4.003 3 3 2, &c.

Having thus long dwell'd upon this Subject, and render'd it tedious to the Learner, and Prefs, I am forry I must leave it imperfect as it is: For could the Canon in Page 182 be as effectually applied to Cafes wherein 63+cciso (in which Cafes all the Roots in all original Cubicks are always real) as it could to the Cafes wherein 3+4cc is

27

4

o, we should have a perfect Method for extracting the Roots of all forts of Cubick Equations; and confequently of all Biquadraticks, by the following Chap. But the above Method of Series (tho' it be the belt I could deduce from the foregoing Theorems for to apply to the Cafes herein first mention'd) is, in moft Examples, very prolix and troublefom: And therefore I must refer fuch Equations to be evolved by the numeral Exegefis, the converging Series, or by the Trifection of an Angle.

*See Prob. 9. chap. 3. Part II. Book II.

CHA P. III:

A Rule of des Cartes's for diffolving a Biguadratick into two Quadzaticks.

NY Biquadratick Equation, which hath the fecond

A Term, being propos'd to be diffolv'd into two Quadra

ticks, must have the fecond Term first deftroy'd (by Chap. 1

* 0 2

Prop.

2

Prop. 1. Sch. 1.); and let the Equation thence produc'd be fuppos'd equal to x + x + q x x +rx+s=0; in which Equation is fuppos'd to have the Sign + prefix'd to it; But the Values of q, r and s may either of them be affirmative or negative. Let us fuppofe this Equation to be produc'd by the two Quadraticks x2+ex+ƒ=o, and x2 cx+go; in which Equations 2 is underflood to have the Sign + prefix'd to it: And, to the End that the fecond Term should be wanting in their Product, ex must have the Sign in one Equation, and in the other, and the Values of ƒ and g are to be determin'd in the following man

+f

ner: Then x4qx2 + rx+s= x+x― e ex

x+fg: And, by equating their respective Terms, we have q=fts—ee, r:g-f: xe, and s=fg; wherefore

r

q+ce=ƒ+8,2=8—ƒ; consequently

e

f, and the Products of the two laft

rr

ee

4

=ƒg=s; which Equa

tion, when reduc'd, gives e629e4

suppose e2y, and then you'll reduce the foregoing Equa

tion to this Cubick one y3 +2 qy'

cond Term may be deftroy'd, and the Root of the Equation thus had extracted, by the foregoing Chap. and confequently one of the Values of y found, as alfo of e = √ y;

And, by extracting the Roots of the two Quadraticks, x2+ exƒ0 and ∞2

o, you will have the four Roots of the Biquadratick ++qx2 + rx+s=0'; to wit, x = -10√4ee-f:, and x: