s+nl2 = 2; and 2nkl must be even. And r2nkl (an odd number) = p2 an even number, which is abfurd.

cr

2. Let N represent any number in general, I an odd number; then I fay, " every odd number is a multiple of four, more or less unity," that is, I = 4N± 1. "The fquare of an odd number is 4N-+ 1," (that is fome multiple of 4, more unity;) and "if from fuch a square there be taken any multiple of 4, the remainder, if greater than unity, will be 4N+1."

Hence it follows that n = 4N+1. For feeing nl2 = 2'—s; becaufel and Qare the halves' of odd numbers, we have, according to the pre1-45, or without the

fent notation,

n x 12

4

x

4.

common denominator n × I2 = I2 — 4s, that is, n × 4N + 1 = 4N + 1, and confequently, n = 4N+1. For it is not 4N-1 but 4N+I that can give the product 4 N + 1. *

In the former Editions, there were here inferted two Rules for the Cafe of Bo: which, though true, Mr. Tho mas Simpfon has, in his Mifcellaneous Tracts, published 1757, fhewn to be unneceffary. In this, therefore, they are omitted.

It is only to be regretted that Mr. Simplon fhould, through inattention, have placed this inaccuracy, not to the account of the Editor, as he ought to have done, but to that of Mr. MACLAURIN. The whole explanation of Sir Ifaac's Method of Reducing Equations by means of Surd Divifors, is (pag. 213.) profeffedly a Supplement; as is likewife the Addition to Chap. 14. Part I.

And the

In like manner the other limitations may be determined: and what has been faid may lead to the invention or demonstration of fimilar Rules for the higher equations of even dimensions, if any one pleases to take the trouble.

CHA P. VIII.

OF THE RESOLUTION OF EQUATIONS BY CARDAN'S RULE, AND OTHERS OF THAT KIND.

$75. WE

E now proceed to fhew how an expreffion of the root of an equation can be obtained that shall involve only known quantities. In Chap. 11. Part. I. we fhewed how to refolve fimple equations; and in Chap. 13. we fhewed how to refolve any quadratic equation, by adding to the fide of the equation that involves the unknown quantity, what was neceffary to make it a complete square, and then extracting the fquare root on both fides. In § 27. of this Part, we gave another method

Editor thought he had, in his Preface, fufficiently intimated that a few such infertions had been made, and the reason why: though he cannot recollect any others worth mentioning; if it is not $123, 124, of Part II.

of

of refolving quadratic equations, by taking away the second term: where it appeared that if x2 —px + q = 0, x = {p ± √ ;p2 — q.

$76. The fecond term can be taken away out of any cubic equation, by § 25; so that they all may be reduced to this form, x3 + qx + r = 0.

Let us fuppofe that x=a+b; and x2 + qx+r ≈ a3 + 3à3b + 3ab2 + b3 + qx + r = a3 + zab x a + b + b2 + qx+r=a3 + zabx + b2 + qx+r (by fuppofing 3ab = − q) = a3 + b3 + r = ∞

3

2741 and confes

27a"

+ r=o; or, a® + ra3 =

Suppose a3 =z, and you have x2 + rz =

which is a quadratic whofe refolution gives

3

q
r2+ = a3,
27

and a =

hence we have three expreffions for x, viz. 1o. x = p + 2m,

2°.x=p- m + √=3n,

3°•x=p

and thefe give the three roots of the proposed

cubic equation.

EXAMPLE

I.

§ 80. Let it be required to find the roots of the equation - 12x2 + 41% — 42 = 0. Comparing the coefficients of this equation with thofe of the general equation

2°. 3p2 — 39 (= 48 — 39) = 41 . . . . q =

3°• 3pq-p3-2r (=-36—2r)=~42.. r = 3;

343

and consequently, r2 —— q3 — 9 — =

and r + √r31—q2 = 3 + √

27

100

27

cube root of this binomial is found to be

these substituted for m +✔n give the fame values for x, as are already found.

In the equation 3 + 15x2 + 84x — 100 = 0,

you

find

-

P = 5,9 = 3, r = 135, and r + √ r2−q3 = 135 + ✓18252, whose cube root is 3+12; fo that x (=p + 2m) = −5+6=1. The other two values of x, viz. — 8 + √−36, 8-36, are impoffible.

After the fame manner, you will find that the roots of the equation x+x-166x+6600, 15,75. The Rule by which we may discover if any of the roots of an equation are impoffible, fhall be demonstrated afterwards.

are

$82. The roots of biquadratic equations may be found by reducing them to cubes, thus: Let the fecond term be taken away by the Rule given in Chap. 3. And let the equation

that refults be

** * + qx2 + rx.+ s = 0.

And let us fuppofe this biquadratic to be the product of these two quadratic equations,

* +