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Prob. 10. The difference of two numbers is 3, and the difference of their cubes is 117; what are those numbers?

Ans. 2 and 5.

Prob. 11. A company at a tavern had £8 15s. to pay for their reckoning; but, before the bill was settled, two of them left the room, and then those who remained had 10s. a-piece more to pay than before: how many were there in company? Ans. 7.

Prob. 12. A grazier bought as many sheep as cost him £60, and, after reserving 15 out of the number, he sold the remainder for £54, and gained 2s. a head by them; how many sheep did he buy?

Ans. 75.

Prob. 13. There are two numbers, whose difference is 15, and half their product is equal to the cube of the lesser number; what are those numbers?

Ans. 3 and 18.

Prob. 14. A person bought cloth for £33 15s. which he sold again at £2. 8s. per piece, and gained by the bargain as much as one piece cost him; required the number of pieces?

Ans. 15.

Prob. 15. What number is that, which, when divided by the product of its two digits, the quotient is 3; and if 18 be added to it, the digits will be inverted? Ans. 24.

Prob. 16. What two numbers are those, whose sum multiplied by the greater is equal to 77; and whose difference multiplied by the lesser is equal to 12?

Ans. 4 and 7.

Prob. 17. To find a number such, that if you subtract it from 10, and multiply the remainder by the number itself, the product shall be 21.

Ans. 7 or 3.

Prob. 18. To divide 100 into two such parts, that the sum of their square roots may be 14. Ans. 64 and 36.

Prob. 19. It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their difference. Ans. 10 and 14.

Prob. 20. The sum of two numbers is 8, and the sum of their cubes is 152; what are the numbers? Ans. 3 and 5.

Prob. 21. The sum of two numbers is 7, and the sum of their 4th powers is 641; what are the numbers?

Ans. 2 and 5.

Prob. 22. The sum of two numbers is 6, and the sum of their 5th powers is 1056; what are the numbers ? Ans. 2 and 4.

Prob. 23. Two partners, A and B, gained £140 by trade; A's money was 3 months in trade, and his gain was £60 less than his stock; and B's money, which was £50 more than a's, was in trade 5 months; what was A's stock?

Ans. £100.

Prob. 24. To find two numbers such that the difference of their squares may

be equal to a given number, q2; and when the two numbers are multiplied by the numbers a and b respectively, the difference of the products may be equal to a given number, so

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Prob. 25. 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.

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Prop. 26. 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: q.

Ans.

(a+b)p+(ab)2p2+4abpq 2(p-g)

Prob. 27. A wine merchant sold 7 dozen of sherry and and 12 dozen of claret for 501.; he sold 3 dozen more of sherry for 107. than he sold of claret for 61. Required the price of each.

Ans. Claret, 31.; and sherry, 27. per dozen.

Prob. 28. 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.

Prob. 29. A regiment of foot receives orders to send 216 men on garrison

duty, each

company sending the same number of men; but before the detach

How many companies

ment 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. 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.

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ON THE NATURE OF EQUATIONS.

171. The valuable improvements recently made in the process for the determination of the roots of equations of all degrees, render it indispensably necessary to present to the notice of the student a concise view of the present state of this interesting department of analytical investigation. The researches of Messrs. Atkinson and Horner on the method of continuous approximation to the roots of equations, and the beautiful theorem of M. Sturm for the complete separation of the real and imaginary roots, have given a fresh impulse to this branch of scientific research, and entirely changed the state of the subject of numerical equations. Indeed, the elegant process of Sturm for discovering the number of real roots, and their initial figures in any numerical equation, combined with the admirable method of continuous approximation as improved by Horner, fully complete the theory and numerical solution of equations of all degrees.

We do not intend to enter at great length into the theory of equations; but it is hoped that the portion of it which we have introduced into the present treatise will be discussed in a simple and perspicuous manner, and be found amply sufficient for most practical purposes.

DEFINITIONS.

1. An equation is an algebraical expression of equality between two quantities.

2. A root of an equation is that number, or quantity, which, when substituted for the unknown quantity in the equation, verifies that equation.

3. A function of a quantity is any expression involving that quantity; thus, a x2+b a x2+b, ax3+cx+d, a are all functions of x; and also ax2-by',

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cx+d'

·y2+y.x+x2+a2+b+2, are all functions of x and y.

These functions are usually written ƒ(x), and ƒ(x, y).

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when divided by x-a, will leave a remainder, which is the same function of a that the given polynomial is of x.

Let f(x)=x"+px"1+q x"2+

.; and, dividing f(x) by x—a, let

Q denote the quotient thus obtained, and R the remainder which does not involve x; hence, by the nature of division, we have

f(x)= Q(x—a) + R.

Now this equation must be true for every value of x; hence, if a=a, we have

f(a)=0+ R;

for R is altogether independent of x, and therefore the remainder R is the same function of a that the proposed polynomial is of x.

EXAMPLES.

(1.) What is the remainder of x2-6x+7 divided by x-2, without actually performing the operation?

(2.) What is the remainder of x3-6x2+8x-19, divided by x+3?
(3.) What is the remainder of x1+6x3+7x2+5x-4, divided by x-5?
(4.) What is the remainder of x3+p x2+q x+r, divided by x-a?

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the first member of the equation is divisible by x—a.

Instead of deducing the remainder as in the last proposition, we shall actually perform the division, either by the usual way, or the preferable method of synthetic division, as it keeps the work within the breadth of the page.

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Now this remainder is the same function of a that the first member of the proposed equation is of x; and, therefore, since a is a root of the equation, the remainder vanishes, and the polynomial, or first member of the equation, is divisible by x-a.

Conversely, if the first member of an equation, f(x)=0, be divisible by x-a, then a is a root of the equation.

For, by the foregoing demonstration, the final remainder is ƒ(a); but since f(x), or the first member of the equation, is divisible by x-a, the remainder must vanish; hence ƒ(a)=0, and therefore, a being substituted for x in the equation f(x)=0, verifies the equation, and consequently a is a root of the equation.

PROPOSITION III.

Every equation containing but one unknown quantity has as many roots as there are units in the highest power of the unknown quantity.

Let f (x)=0 be an equation of the nth degree; then if a, be a root of this equation, we have, by last proposition,

(x—a1) ƒf, (x)=ƒ(x)=0,

where f(x) represents the quotient arising from the division of f(x) by -a. Now if a, is also a root of the equation f(x)=0; it is obvious that

f(x) must be divisible by x-a2, for x-a, is not divisible by x-a1; hence, if f(x) represent the quotient of f (x) divided by x-a,, we have

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and, consequently, there are as many roots as factors, that is, as units in the highest power of x, the unknown quantity; for the last equation will be verified by any one of the n conditions,

x=a, x=ɑ2, x=α3, x=α49 .... x=α„:

and since the equation contains n factors, there are n roots.

Cor. When one root of an equation is known, the depressed equation containing the remaining roots is readily found by synthetic division; and if two or more roots are known, the equation containing the remaining roots is found by two or more corresponding divisions.

EXAMPLES.

(1.) One root of the equation 21—25x2+60x—36=0 is 3; find the equation containing the remaining roots.

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is the equation containing the remaining roots.

(2.) Two roots of the equation x1—12x3+48x2—68x+15=0, are 3 and 5; find the quadratic containing the remaining roots.

1-12 +48-68+15 (3
327+63-15

9 +21-5(5
5-20

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is the equation containing the two remaining roots.

(3.) One root of the cubic equation ¿a3—6x2+11x—6=0 is 1; find the quadratic containing the other roots. Ans. x2-5x+6=0.

(4.) Two roots of the biquadratic equation 4x1—14x3—5x2+31x+6=0 are 2 and 3; find the reduced equation. Ans. 4x2-6x+1=0.

(5.) One root of the cubic equation x3+3x2—16x+12=0 is 1; find the remaining roots. Ans. 2 and -6. (6.) Two roots of the biquadratic equation x-6x2+24x-16=0, are 2 and -2; find the other two roots. Ans. 3+√5.

PROPOSITION IV.

To form the equation whose roots are a1, α2, aз, ɑşi

The polynomial, f(x), which constitutes the first member of the equation required, being equal to the continued product of x-ɑ1, x—Ɑ2, X—A3,

a-a, by the last proposition, we have

(x—a,) (x—a2) (x—a,)

(x—a1)=0;

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