Page images

that (a+b+6+ &c)2 = ƒ2, and (ab 4 ac + ic+ &c) = g. Thus we have ƒ2 = (a2 + b2 + c2 + &c) + 2g; and consequently a + b + c2 + &c = ƒ3 — 2g.

1= -

3dly. The sum of the cubes of the roots, is equal to 3 times the rectangle of the coefficient of the second and third terms, made less by the cube of the coefficient of the second term, and 3 times the coefficient of the fourth term: viz, a3 + b3 + c3 + &c - ƒ3 + 3ƒg — 3h. For we shall by actual involution, have (a + b + c + &c)3 = a3 + b3 + c3 + &c + 3(a + b + c) x (ab + ac + bc) — 3abc. But (a+b+c+&c)3 ~ ƒ3, (a + b + c + &c) x (ab + ac + bc + &ç)=−ƒg, abch. Hence therefore, − ƒ3 = a3 + b3 + c3 + &c 3fg3h; and consequently, a3 + b3 + c3 + &c = −ƒ3 + 3fg3h. And so on, for other powers of the roots..



In Every Equation, which contains only Real Roots:

I. If all the roots are positive, the terms of the equation will be and alternately.

II. If all the roots are negative, all the terms will have the sign +.

III. If the roots are partly positive, Dartly negative, there will be as many positive roots as there are variations of signs, and as many negative roots as there are permanencies of signs; these variations and permanencies being observed from one term to the following through the whole extent of the equation.

In all these, either the equations are complete in their terms, or they are made so.

The first part of this theorem is evident from the examination of equation A; and the second from equation B.

To demonstrate the third, we revert to the equation c (art. 3), which has two positive roots, and one negative. It may happen that either c > a + b, or c < a + b.

In the first case, the second term is positive, and the third is negative; because, having c> a+b, we shall have ac + be > (a + b) > ab. And, as the last term is positive, we see that from the first to the second there is a permanence of signs; from the second to the third a variation of signs; and from the third to the fourth another variation of signs. Thus there are two variations and one permanence of signs; that is, as many variations as there are positive roots, and as many permanencies as there are negative roots.

In the second case, the second term of the equation is negative, and the third may be either positive or negative. If


that term is positive, there will be from the first to the second a variation of signs; from the second to the third another variation; from the third to the fourth a permanence; making in all two variations and one permanence of signs. If the third term be negative; there will be one variation of signs from the first to the second; one permanence from the second to the third; and one variation from the third to the fourth: thus making again two variations and one permanence. The number of variations of signs therefore, in this case as well as in the former, is the same as that of the positive roots; and the number of permanencies, the same as that of the negative



Corol. Whence it follows, that if it be known, by any means whatever, that an equation contains only real roots, it is also known how many of them are positive, and how many negative. Suppose, for example, it be known that, in the equation .rs +3.x4 1200, all the roots are real: it may immediately be concluded that there are three positive and two negative roots. In fact this equation has the three positive roots = 1, x = 2, x = 3 ; and two negative roots, x =

- 27x2 + 166x

4, x =

[ocr errors]


If the equation were incomplete, the absent terms must be supplied by adopting cyphers for coefficients, and those terms must be marked with the ambiguous sign. Thus, if the equation were 205

20x3 + 30x2 + 19x

30 = 0,

all the roots being real, and the second term wanting. It must be written thus:

[ocr errors][merged small][merged small][merged small][merged small]

Then it will be seen that, whether the second term be positive or negative, there will be 3 variations and 2 permanencies of signs and consequently the equation has 3 positive and 2 negative roots. The roots in fact are, 1, 2, 3, −1, −5.


This rule only obtains with regard to equations whose roots are real. If, for example, it were inferred that, because the equation x2 + 2x + 50 had two permanencies of signs, it had two negative roots, the conclusion would be erroneous; for both the roots of this equation are imaginary.


Every Equation may be Transformed into Another whose Roots shall be Greater or Less by a Given Quantity. In any equation whatever, of which is unknown, (the equations A, B, c, for example) make x = z+m, z being a new unknown quantity, m any given quantitý, positive or negative:

N 2

negative: then substituting, instead of r and its powers, their values resulting from the hypothesis that x=z+m; so shall there arise an equation, whose roots shall be greater or less than the roots of the primitive equation, by the assumed quantity m.

Corol. The principal use of this transformation is, to take away any term out of an equation. Thus, to transform an equation into one which shall want the second term, let m be so assumed that nm a = 0, or m = n being the index of the highest power of the unknown quantity, and a the coefficient of the second term of the equation, with its sign changed then, if the roots of the transformed equation can be found, the roots of the original equation may also be found, because x = +


[ocr errors]


Every Equation may be Transformed into Another, whose Roots shall be Equal to the Roots of the First Multiplied or Divided by a Given Quantity.



1. Let the equation be 23+ az2 + bx + c = 0: if we put fzx, or z = the transformed equation will be 3 + fax2 +fbx +ƒ3c = 0, of which the roots are the respective products of the roots of the primitive equation multiplied into the quantity f.




[ocr errors]


By means of this transformation, an equation with fractional quantities, may be changed into another which shall be free from them. Suppose the equation were z3 + + d + 1/ = 0: multiplying the whole by the product of the denominators, there would arise ghkz3+hkaz2 + gkbz + ghd = 0: then assuming ghkz = x, or z = the transformed equa. would be x3+hkax + g2k2hbx‍+g3k3h3d=0. The same transformation may be adopted, to exterminate the radical quantities which affect certain terms of an equation. Thus, let there be given the equation z3 + az2 √ k+ bz+c√k: make z✔k r; then will the transformed equation be 3+ akx2 + bkx + ck2 0, in which there are no radical quantities.



2. Take, for one more example, the equation z3 + az2 + bx + c = 0. Make = x; then will the equation be




transformed to "+++= 0, in which the roots



are equal to the quotients of those of the primitive equations divided by f.

It is obvious that, by analogous methods, an equation may be transformed into another, the roots of which shall be to those of the proposed equation, in any required ratio. But the subject need not be enlarged on here. The preceding succinct view will suffice for the usual purposes, so far as relates to the nature and chief properties of equations. We shall therefore conclude this chapter with a summary of the most useful rules for the solution of equations of different degrees, besides those already given in the first volume.

I. Rules for the Solution of Quadratics by Tables of Sines and Tangents.

1. If the equation be of the form x2 + px = 9:


Make tan A= ✔g; then will the two roots be,

x = +tan

[ocr errors]
[merged small][merged small][ocr errors]

x =

[merged small][ocr errors][merged small][merged small]


q: then will

[ocr errors]

xcot A√9.

3. For quadratics of the form x2 + px = — q.

Make sin A =


√9: then will

[ocr errors][ocr errors]

cot A√9.

x = - tan Ag 4. For quadratics of the

Make sin A =

x = +tan A√9 In the last two cases, if



form 2 — px = — q•

✔g: then will


[ocr errors]
[ocr errors][merged small]

and consequently the values of x.

The logarithmic application of these formulæ is very simple. Thus, in case 1st. Find A by making

[blocks in formation]

Then log x = S+ log tan A +
-(log cot A+ log q

Note. This method of solving quadratics, is chiefly of use when the quantities p and q are large integers, or complex fractions.

II. Rules for the Solution of Cubic Equations by tables of Sines, Tangents, and Secants.

1. For cubics of the forin x3 + px ±q=0.

[merged small][merged small][ocr errors][ocr errors][ocr errors][ocr errors][merged small][merged small][merged small][merged small]
[ocr errors][ocr errors]

demde of sin 3 should exceed unity, s would , and the equation would fall in what is called Paucible case of cubics. In that case we must make

[ocr errors]

2005A= 2p: and then the three roots would be x = sin A. 2√ √ P.

x=sin (60° - A). 2√ p.

x = ± sin (60° + a) . 2 √ {p.

If the value of sin в were 1, we should have в = 90°, tar A = 1; therefore A = 45°, and x = 2√÷p. But this would not be the only root. The second solution would give 1: therefore a = ; and then

cosec 3A



[merged small][ocr errors][merged small][ocr errors][ocr errors][subsumed]

Here it is obvious that the first two roots are equal, that their sum is equal to the third with a contrary sign, and that this third is the one which is produced from the first solution*.

In these solutions, the double signs in the value of x, relate to the double signs in the value of q.

N. B. Cardan's Rule for the solution of Cubics is given in the first volume of this course.

The tables of sines, tangents, &c, besides their use in trigonometry, and in the solution of the equations, are also very useful in finding the value of algebraic expressions where extraction of roots would be otherwise required. Thus, if a and b be any two quantities, of which a is the greater. Find x, *,

[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]


log 2+ log cos y
log 2+ log sin ly.

log (a+b)" ➡ · [log a + log cos t + log tan 45° ±‡t)].


The first three of these formule will often be useful, when two sides of a right-angled triangle are given, to find the third.

III. Solution

« PreviousContinue »