hours and minutes, strikes a number of strokes equal to that number of hours and minutes; and it is observed that the number of minutes is less by 41 than the square of the number which the clock strikes at the last time of striking. The clock does not strike twelve during the time. How much is it too fast?

Ans. 3 hours and 23 minutes. (30) A and B engage to reap a field for £4 10s.; and as A alone could reap it in 9 days, they promised to complete it in 5 days. They found, however, that they were obliged to call in C, an inferior workman, to assist them for the last two days, in consequence of which В received 3s. 9d. less than he otherwise would have done. In what time could B or C alone reap the field ?

Ans. B could reap it in 15 days, C in 18. (31) The fore wheel of a carriage makes 6 revolutions more than the hind wheel in going 120 yards; but if the periphery of each wheel be increased 1 yard, it will make only 4 revolutions more than the hind wheel in the same space. Required the circumference of each.

Ans. 4 and 5. (32) The intensity of two lights, A and B, is as 7:17, and their distance apart 132 feet. Whereabouts between is the point of equal illumination?

(33) The loudness of a church bell is three times that of another. Now, supposing the strength of sound to be inversely as the square of the distance, at what place between the two will the bells be equally well heard.

(34) Supposing the mass of the earth to be 1 and that of the moon 0.017, their distance 240 thousand miles, and the force of attraction equal to the mass divided by the square of the distance; at what point between will a body be held in suspense, attracted toward neither ?

(35) The hold of a vessel partly full of water (which is uniformly increased by a leak) is furnished with two pumps, worked by A and B, of whom A takes three strokes to two of B's; but four of B's throw out as much water as five of A's. Now B works for the time in which A alone would have emptied the hold; A then pumps out the remainder, and the hold is cleared in 13 hours and 20 minutes. Had they worked together, the hold would have been emptied in 3 hours and 45 minutes, and A would have pumped out 100 gallons more than he did. Required the quantity of water in the hold at first, and the hourly influx of the leak.

(36) To divide two numbers, a and b, each into two parts, such that the product of one part of a by one part of b may be equal to a given number, p, and the product of the remaining parts of a and b equal to another given number, p'.

ab-(p'-p)+ V {ab-(p'-p)}: -4abp Ans. x=4

26 , ab+(p'—P) F V{ab-(p'-p) }2—4abp

26 _ab-(p*—P)+ V{ab(p'P)}? — 4abp

20 ab+(p-1) F V {ab-(p'-p) }? — 4abp

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(37) To find a number such that its square may be to the product of the

differences of that number, and two other given numbers, a and b, in the given ratio, p:9.

(a+b)p+v(ab)?p?+ tabpq

2(p-1) (38) There is a number consisting of two digits, which, when divided by the sum of its digits, gives a quotient greater by 2 than the first digit; but if the digits be inverted, and the resulting number be divided by a number greater by unity than the sum of the digits, the quotient shall be greater by 2 than the former quotient. What is the number?

Ans. 24. (39) A regiment of foot receives orders to send 216 men on gairison duty, each company sending the same number of men ; but before the detachment marched, three of the companies were sent on another service, and it was then found that each company that remained would have to send 12 men additional in order to make up the complement, 216. How many companies were in the regiment, and what number of men did each of the remaining companies send on garrison duty ?

Ans. There were 9 companies, and each of the remaining 6 sent 36 men.


FIRST DEGREE. 193. If we add to this trinomial, in order to complete the square of the first two terms, the term ip, and afterward subtract the same, so as not to change the quantity, it becomes

x+pr+ip-ip-9, which may be written thus :

(x+\pe-(ip?+9).............. (2) But the difference of the squares of two quantities being equal to the product of their sum and difference, the expression (2) is equal to the following:

(r+p+ Vipp +9)(r+p-Vip?+9)... (3) We perceive from this expression that the two factors of the first degree, which compose the trinomial of the second degree, are x minus each of the roots of the equation of the second degree, formed by putting this trinomial equal to zero.

Moreover, by equating (3) to zero, we perceive that the only way of satisfying the resulting equation is by making one or other of the factors of the first degree, of which it is composed, equal to zero. The first,

x+p+vipi+q=0, gives r=-\p-Vip+q; and the second,

x+ip- Vip +9=0, gives r=-pt Vip: +9. Hence there are but two values of x which will satisfy the general equation

22 +pr-q=0.


1o. Decompose the trinomial xo_72+10 into two factors of the first degree.

From the equation 22—7x+10=0 we find the roots r=5 and r=?. Hence

22—7x+10=(2—5)(r—2). 20. 3.22 – 5x – 2.

Equating this trinomial to zero, after dividing by 3, we obtain the equation x? _*- =0, the roots of which being r=2 and x=-!, we have

3x”—5x—2=3(x— 2)(x+1)=(x-2)(3.2+1). 39. 4+5 +3.

Ans. (x+1-1V13)(x+1+iV13). 4o. 4.x2 - 4x+1.

Ans. (2x— 1)?* 5?. 2-5+7.

Ans. (r->)+. 194. To complete the analysis of the 20 degree, it would be necessary to consider the case where the unknown quantities exceed the equations in number. The more simple is that when there is but one equation and two unknown quantities. If it be resolved with respect to one of the unknown quantities, y, for example, an expression is found generally containing r under a radical; so that, by giving to x any rational values whatever, irrational values would be found for y. It might be proposed to find rational values for x, for which the corresponding one of y should be rational also. But the difficulty of this problem, unless it be restricted to some very simple cases, is beyond mere elements. We add one or two here. For further information upon the subject, the student is referred to the Theory of Numbers, by Legendre, a separate and very elegant treatise, in one quarto volume.

INDETERMINATE ANALYSIS OF THE SECOND DEGREE. Resolution in whole numbers of an equation of the second degree, with tuo unknown quantities, which contains but the first power of one of the unknowns.

195. The questions of indeterminate analysis, which depend upon equations of a degree superior to the first, go beyond the limits which we have imposed on ourselves in the present work; but when an equation of the second degree contains the second power of but one of the unknown quantities, the solutions of this equation in whole numbers may be regarded as a question of indeterminate analysis of the first degree.

Equations of the second degree in two unknown quantities, which do not contain the second power of one of these, are represented by the equation

mxy+nza +px+y=r ............ (1) Resolving this equation with respect to y, we find - nxo - pr+r

.............. (2) We deduce from it, by performing the division,

n, nq-mp, mer+mpq-nq*

y=-mt mt m (mrta) ; which gives

moy=mnr+nq-mp+mto....... (3) putting to abridge mor+mpq-nq=N.

N In order that x and y should be whole numbers, it is necessary that -

wa mrt should be a whole number ; we must, therefore, calculate all the divisors of

* This presents a case of what are called equal roots.

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the number N, and put mx+9 equal to each of these divisors successively, taken with the sign + and with the sign -. If the equations thus obtained furnish for x a certain number of entire values, these values are to be substituted in equation (3); and it is necessary, moreover, in order that y may be a whole number, that the second number which becomes a known quantity should be divisible by m.

It is evident that the member of entire solutions will be very limited, and that there may not be even one. If this method be applied to each of the following equations,

2xy-3x2 + y=1
5xy=2x +3y+18

| xy + r=2x+3y + 29,
considering only the positive solutions, we find
For the first equation ......

sx=0, y=1

1=3, y=4.

r=1, y=10 For the second equation ....3r=3, y=2

r=7, y=1. For the third equation .....3

Sx=4, y=21

''x=5, y=7. If the remainder, after the division of — nx? —px+r by mr+q, should be zero, equation (1) would be of the form (mr+9)(ar+by+c)=0; and we should have all the solutions of this equation by resolving separately the two equations mx+q=0, ax+by+c=0.

The method which has just been explained is applicable only in case m is not zero. Let m=0; equation (1) gives

nr +pry=

............. (4) Suppose that one value of r=a (a being a whole number) gives an entire value for y. If we place x=a+qt, t being any entire number whatever, we find

na?+pa y= -

- -(2nat + nqt? +pt);

9 by hypothesis, na’+pa-r is divisible by q; the value of y, corresponding to x=a+gt, will be then a whole number. As this conclusion is true, whatever be the sign of t, it follows that, if the equation admits of entire solutions, they will be found to be such as answer to a value of x between 0 and q. Consequently, to obtain all the solutions in whole numbers, it will be sufficient to substitute for x in the equation the numbers 0, 1, 2, 3, ....(-1, and each solution in whole numbers corresponding to one of these numbers will furnish an infinite number of others.

Equation (4), in which the object is to find values of x which render the polynomial nx? +pr-ra multiple of the given number 9, M. Gauss calls congruence of the second degree ; so, also, the equation ax+by=c, in which we seek to render ax-c a multiple of b, is a congruence of the first degree.

Further matter on the subject of indeterminate analysis will be given in connection with the theory of numbers, for which see a subsequent part of the work.

MAXIMA AND MINIMA. 196. When a quantity which is capable of changing its value attains such a value that, after having been increasing, it begins to decrease, or, having been decreasing, it begins to increase, in the first case it is called a maximum, and in the second a minimum. The same quantity may have several maximum or minimum values.


ro - 2r+2 To find what value of x will render the fraction - - a maximum or

2x-2 minimum. Equating the given function of x to z, we have

ro — 2x+2_

P=z.: x=2+1+V2-1.

23-2 We perceive at once that by making z=+1 we have x=2, and that the values of 2, a little less than 1, render x imaginary ; hence the given expression has a minimum value 1 corresponding to x=2.

In a similar manner, making z=-1, we have r=0; and a negative value of z, a little smaller than 1, would render x imaginary. But in algebra, negative quantities, which, without regard to the sign, go on increasing, ought to be regarded, when the sign is prefixed, as decreasing ; we may, therefore, say that a value of z, a little greater than -1, renders x imaginary, then z=-1 is a maximum corresponding to r=0.

As the subject of maxima and minima is generally treated by the aid of the differential calculus, we shall not dwell further upon it here, though it furnishes one of the applications of equations of the second degree.

THE MODULUS OF IMAGINARY QUANTITIES. 197. We have seen (191) in the equation of the second degree

2 +pr+9=1, that when q is positive, and greater than the roots are imaginary. Replace

Áp by — a, to avoid fractions ; and to express that q> putq=a? +62; the equation will become

22 - 2arta'+b=0; and, by the formula for the solution of equations of the second degree,

r=ał 62,


x=a+bv-1........ (1) The absolute value of the square root of the positive quantity a'+b is called the modulus of the imaginary expression (1). For example, the modulus of 3–4V-1 would be V9+16, or 5.

Two quantities, such as a+b V-1 and a-bv-1, which differ from one another only in the sign of the imaginary part, are called conjugates of each other. Two conjugate quantities have then the same modulus.

If we make b=1, the expression atbv -1 reduces to a. Thus, the formula x=a+b V-1 may represent all quantities real or imaginary, a representing the algebraic sum of the real quantities, and b that of the coefficients

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