we shall have, from the first,

x=

By the first solution, the second part of the proposed num

a

and

, are, as was required in the enunciation of the

question, both less than the number proposed. By the second solution, we have

=a+

1

; and the two parts

1-√m'

are

Their signs being contrary, the number a is not, properly speaking, their sum, but their difference.

400. When we make m=n, that is, if we suppose that the squares of the two required parts are equal, we have ✔m' =1; the first solution gives two equal parts, and a, a result which is evident, whilst the second solution gives two in

finite results (Art. 166), namely, or, and or-.

0

a 1-1

These are proper results, according to the above enunciation, since that the quantities required must be infinitely great, with respect to their difference a, if we can suppose the ratio of their squares equal to unity.

Now, if a=18, m≈25, and n=16; then substituting these values in the formula

and 8 equal to the two parts required, the same as in Ex. 2., which is a particular case of this general problem.

Prob. 10. What two numbers are those, whose sum is to the greater as 10 to 7; and whose sum, multiplied by the less, produces 270 ? Ans. 21 and +9. Prob. 11. What two numbers are those, whose difference is to the greater as 2 to 9, and the difference of whose squares is 128 ? Ans. 18 and ±14.

Prob. 12. A mercer bought a piece of silk for 16l. 4s. ; and the number of shillings which he paid for a yard was to the number of yards as 4: 9. How many yards did he buy, and what was the price of a yard?

Ans. 27 yards, at 12s. per yard.

Prob. 13. Find three numbers in the proportion of 1, 3, and ; the sum of whose squares is 724.

Ans. 12, 16, and +18. Prob. 14. It is required to divide the number 14 into two such parts, that the quotient of the greater part, divided by the less, may be to the quotient of the less divided by the greater as 16: 9. Ans. The parts are 8 and 6. Prob. 15. What two numbers are those whose difference is to the less, as 4 to 3; and their product, multiplied by the less, is equal to 504 ? Ans. 14 and 6. Prob. 16. Find two numbers, which are in the proportion of 8 to 5, and whose product is equal to 360.

Ans. +24, and 15. Prob. 17. A person bought two pieces of linen, which, together, measured 36 yards. Each of them cost as many shillings per yard, as there were yards in the piece; and their whole prices were in the proportion of 4 to 1. What were the lengths of the pieces? Ans. 24 and 12 yards.

Prob. 18. There is a number consisting of two digits, which being multiplied by the digit on the left hand, the pro-. duct is 46; but if the sum of the digits be multiplied by the same digit, the product is only 10. Required the number.

Ans. 23.

Prob. 19. From two towns, C and D, which were at the distance of 396 miles, two persons, A and B, set out at the same time, and met each other, after travelling as many days as are equal to the difference of the number of miles they travelled per day; when it appears that A has travelled 216 miles. How many miles did each travel per day?

Ans. A went 36, and B 30, Prob. 20. There are two numbers, whose sum is to the greater as 40 is to the less, and whose sum is to the less as 90 is to the greater. What are the numbers ?

Ans. 36, and 24. Prob. 21. There are two numbers, whose sum is to the less as 5 to 2; and whose difference, multiplied by the difference of their squares, is 135. Required the numbers. Ans. 9, and 6.

pro

Prob. 22. There are two numbers, which are in the portion of 3 to 2; the difference of whose fourth powers is to the sum of their cubes as 26 to 7. Required the numbers. Ans. 6, and 4.

Prob. 23. A number of boys set out to rob an orchard, each carrying as many bags as there were boys in all, and each bag capable of containing 4 times as many apples as

there were boys. They filled their bags, and found the number of apples was 2916. How many boys were there? Ans. 9 boys.

Prob. 24. It is required to find two numbers such, that the product of the greater, and square of the less, may be equal to 36; and the product of the less, and square of the greater, may be 48. Ans. 4, and 3. Prob. 25. There are two numbers, which are in the proportion of 3 to 2; the difference of whose fourth powers is to the difference of their squares as 52 to 1. Required the numbers. Ans. 6, and 4.

Prob. 26. Some gentlemen made an excursion, and every one took the same sum. Each gentleman had as many servants attending him as there were gentlemen; and the number of dollars which each had was double the number of all the servants; and the whole sum of money taken out was $3456. How many gentlemen were there?

Ans. 12. Prob. 27. A detachment of soldiers from a regiment, being ordered to march on a particular service, each company furnished four times as many men as there were companies in the regiment; but those becoming insufficient, each company furnished 3 more men; when their number was found to be increased in the ratio of 17 to 16. How many companies were there in the regiment ? Ans. 12. Prob. 28. A charitable person distributed a certain sum among some poor men and women, the numbers of whom were in the proportion of 4 to 5. Each man received onethird of as many shillings as there were persons relieved; and each woman received twice as many shillings as there were women more than men. Now the men received all to

gether 18s. more than the women. each?

How many were there of Ans. 12 men, and 15 women. Prob. 29. Bought two square carpets for 621. 1s. ; for each of which I paid as many shillings per yard as there were yards in its side. Now had each of them cost as many shillings per yard as there were yards in a side of the other, I should have paid 17s. less. What was the size of each ?

Ans. One contained 81, and the other 64 square yards. Prob. 30. A and B carried 100 eggs between them to market, and each received the same sum. If A had carried as many as B, he would have received 18 pence for them; and if B had only taken as many as A, he would have received 8 pence. How many had each ? Ans. A 40, and B 60. Prob. 31. The sum of two numbers is 5 (s), and their product 6 (p): What is the sum of their 5th powers? Ans. 275 (s-5ps3+5p3s).

CHAPTER X.

ON

QUADRATIC EQUATIONS.

401. Quadratic equations, as has been already observed, (Art. 388), are divided into pure and adfected. All pure equations of the second degree are comprehended in the formula x2=n, where n may be any number whatever, positive or negative, integral or fractional. And the value of x is obtained by extracting the square root of the number n; this value is double, for we have, (Art. 295), x=±√n, and in fact, (±√n)2=n. This may be otherwise explained, by observing, (Art. 106), that x2-n=(x+√n).(x−√n)=0, and that any product consisting of two factors becomes nought, when there is no restriction in the equality to zero of that product, by making each of its factors equal to zero.

We have, therefore, x=-√√n, x=+√√n, or x=±√n. 402. Now, since the square root is taken on both sides of the equation, x2=n, in order to arrive at x=±√n; it is very natural to suppose that, x being the square root of x2, we should also affect x with the double sign; and, therefore, in resolving the equation xn, we would write +x=± n; but by arranging these signs in every possible manner, namely:

+x=+√n, +x=-√n,

-x=—√n, —x=+√n,

we would still have no more than the two first equations, that is, +x=±√n; for if we change the signs of the equations -x=-√n and −x=+√n, they become +x+√n and +x=n, or n=±√n.

403. If, in the formula x2=n, n be negative, or which is the same thing, if we have x2=-n, where n is positive;. then, x=-n= ± √nx √−1, and in fact (±√n)3×(v -1)2=nX-1=-n; therefore, the two roots of a pure equation are either both real or both imaginary.

404. All adfected quadratic equations, after being properly reduced according to the rules pointed out in the reduction of simple equations, may be exhibited under the following

general forins; namely +nx=o, and x2+nx=n'; where n and may be any numbers whatever, positive, or negative, integral or fractional.

405. The solution of adfected quadratic equations of the form x2+nxo, is attended with no difficulty; for the equation x2+-nxo, being divided by x, becomes a+n=o, from which we find x=-n, though we find only one value of x, according to this mode of solution, still there may be two values of x, which will satisfy the proposed equation.

In the equation, x2=3x, for example, in which it is required to assign such a value of x, that x2 may become equal to 3x, this is done by supposing x=3, a value which is found by dividing the equation by x; but besides this value, there is also another which is equally satisfactory; namely, x=0; for

then x2=0, and 3x=0.

406. An adfected quadratic equation is said to be complete, when it is of the form x2+nx=n'; that is, when three terms are found in it; namely, that which contains the square of the unknown quantity, as x; that in which the unknown quantity is found only in the first power, as nx; and lastly, the term which is composed only of known quantities; and, as there is no difficulty attending the reduction of adfected quadratic equations to the above form by the known rules: the whole is at present reduced to determining the true value of x from the equation x2+nx=n'.

We shall begin with remarking, that if x2+nx were a real square, the resolution would be attended with no difficulty, because that it would be only required to extract the square root on both sides, in order to reduce it to a simple equation.

407. But it is evident that x2+nx cannot be a square; since we have already seen (Art. 288), that if a root consists of two terms, for example, x+a, its square always contains three terms, namely, twice the product of the two parts, besides the square of each part, that is to say, the square of x+a is x2+2ax+a2.

408. Now, we have already on one side x2+nx; we may, therefore, consider 2 as the square of the first part of the root, and in this case nx must represent twice the product of x, the first part of the root, by the second part: consequently, this second part must be in, and in fact the square of x+ in is found to be r2+nx+in2.

409. Now x2+nx+n2 being a real square, which has for its root+n, if we resume our equation x2+nx=n', we have only to add in2 to both sides, which gives us x2+nx+¦n2=n' +in2, the first side being actually a square, and the other