Where e is the coefficient of x in both equa

*

tions, but affected with contrary signs; because when the fecond term is wanting in an equation, the fum of the affirmative roots must be equal to the fum of the negative.

Compare now the propofed equation with the above product, and the refpective terms put equal to each other will give f+8 — e2 = 9, eg-ef=r, fg = s. Whence it follows, that ƒ + 8 = q + e2, and gƒ=; and confeg

quently ƒ + g + g − ƒ (= 28) = q + ë2 +

ƒ × 8 (= s ) = ¦ × 92 + 29e2 + e* —

; and multiplying by 4e2, and ranging the terms, you have this equation,

es + 29e+ + q2 — 45 × e2 — q2 — 0. Suppofey, and it becomes y3 + 2qy2+

q* — 45 × y—r2 =0, a cubic equation whofe roots are to be discovered by the preceding articles. Then the values of y being found, their square root will give e (fince y = e); and having e, you will find ƒ and g from the equations

tracting the roots of the equations x2 + ex +ƒ=o‚' x2 —ex + g = o, you will find the four roots of the biquadratic **+qx2+rx+s=0; for either x=−↓e±√že2 −ƒ, or, x = + Le±√že2 — g•

§ 83.

Or if you want to find the roots of the biquadratic without taking away the fecond. term; suppose it to be of this form,

2r

and x=p+a±√√√ p2 + q — a2 +

a2 is equal to the root of the cubic,

-p2 + 2pr

-9

The demonftration is deduced from the laft

article, as the 78th is from the preceding.

CHA P. IX.

OF THE METHODS BY WHICH YOU MAY APPROXIMATE TO THE ROOTS OF NU MERAL EQUATIONS BY THEIR LIMITS.

$ 84.

WH

HEN any equation is propofed to be refolved, firft find the limits of the roots (by Chap. 5.) as for example, if the roots of the equation x2 16x + 55=0 are required, you find the limits are o, 8, and 17, by § 48; that is, the leaft root is between o and 8, and the greatest between 8 and 17.

In order to find the first of the roots, I confider that if I fubftitute o for x in 16x + 55, x2 the refult is pofitive, viz. + 55, and confequently any number betwixt o and 8 that gives a pofitive refult must be less than the leaft root, and any number that gives a negative refult nuft be greater. Since o and 8 are the limits, I try 4, that is, the mean betwixt them, and fuppofing x = 4, x2- 16x + 55 = 16 ~64+55=7, from which I conclude that the root is greater than 4. So that now we have the root limited between 4 and 8. Therefore I next try 6, and fubftituting it for x we find x2- 16x + 55 = 36 -96 +55=5; which refult being negative, I conclude that 6 is greater than the root required, which therefore is limited now between 4

and

and 6. And fubftituting 5, the mean between them, in place of x, I find x2 16x + 55 =2580 + 550; and confequently 5 is the leaft root of the equation. After the fame manner you will difcover 11 to be the greatest root of that equation.

$85. Thus by diminishing the greater, or increafing the leffer limit, you may discover the true root when it is a commenfurable quantity. But by proceeding after this manner, when you have two limits, the one greater than the root, the other leffer, that differ from one another but by unit, then you may conclude the root is incommenfurable.

We may however, by continuing the operation in fractions, approximate to it. As if the equation propofed is x2-6x+7= 0, if we fuppofe x 2, the refult is 412 + 7 = — 1, = which being negative, and the fuppofition of xo giving a pofitive refult, it follows that the root is betwixt o and 2. Next we fuppofe x=1; whence x2-6x+7= 1−6 +7=+ 2, which being positive, we infer the root is betwixt 1 and 2, and confequently incommenfurable. In order to approximate to it, we fuppofe

4

x = 1';, and find x2 — 6x + 7 = 2; − 9 + 7 = 2 ; and this refult being pofitive, we infer the root must be betwixt 2 and 14. And therefore we

49

try 14, and find x2-6x+7=

8

7

= 31% — 10,1% +7 = ——, which is negative;

16'

fo that we conclude the root to be betwixt

and 1. And therefore we try next 1, which giving also a negative refult, we conclude the root is betwixt 1 (or 1) and 13. We try therefore 1, and the refult being pofitive, we conclude that the root must be betwixt 1 and 11%, and therefore is nearly 12.

9

$86. Or you may approximate more easily by transforming the equation proposed into another, whofe roots fhall be equal to 10, 100, or 1000 times the roots of the former (by § 29.) and taking the limits greater in the fame proportion. This transformation is eafy; for you are only to multiply the fecond term by 10, 100, or 1000, the third term by their squares, the fourth by their cubes, &c. The equation of the laft example is thus transformed into x2- 600x + 7c000 = 0, whofe roots are 100 times the root of the propofed equation, and whofe limits are 100 and 200. Proceeding as before, we try 150, and find x-600x + 70000 = 22500 90000 + 70000 = 2500, fo that 150 is lefs than the root. You next try 175, which giving a negative refult muft be greater than the root and thus proceeding you find the root to be betwixt 158 and 159: from which