For, in the first

the horse and sold

place, suppose the man gave 60 dollars for him for 24, he then loses 36 dollars. But,

from the enunciation, he should lose 60 per cent. of 60, that is,

therefore, 60 satisfies the problem.

If he pays 40 dollars for the horse, he loses 16 by the sale; for, he should lose 40 per cent. of 40, or

therefore, 40 satisfies the conditions of the problem.

4. 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 on those he sold: how many did he buy? Ans. 75.

5. A merchant bought cloth for which he paid £33 15s., which he sold again at £2 8s. per piece, and gained by the bargain as much as one piece cost him: how many pieces did he buy?

Ans. 15.

6. What number is that, which, being divided by the product of its digits, the quotient will be 3; and if 18 be added to it, the order of its digits will be reversed? Ans. 24.

7. Find a number such that if you subtract it from 10, and multiply the remainder by the number itself, the product will be 21. Ans. 7 or 3.

8. Two persons, A and B, departed from different places at the same time, and traveled towards each other. On meeting, it appeared that A had traveled 18 miles more than B; and that A could have performed B's journey in 153 days, but B would have been 28 days in performing A's journey. How far did each travel? A 72 miles.

Ans.

B 54 miles.

9. 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 nad 10s. apiece more to pay than before: how many were there in the company?

Ans. 7.

10. What two numbers are those whose difference is 15, and of which the cube of the lesser is equal to half their product? Ans. 3 and 18.

11. Two partners, A and B, gained $140 in trade: A's money was 3 months in trade, and his gain was $60 less than his tock: B's money was $50 more than A's, and was in trade 5 months: what was A's stock?

Ans. $100.

12. Two persons, A and B, start from two different points, and travel toward each other. When they meet, it appears that A has traveled 30 miles more than B. It also appears that it will take A 4 days to travel the road that B had come, and B 9 days to travel the road that A had come. What was their distance apart when they set out?

Ans. 150 miles.

Discussion of Equations of the Second Degree involving but one unknown quantity.

115. It has been shown that every complete equation of the second degree can be reduced to the form (Art. 113)

in which p and q are numerical or algebraic, entire or fractional, and their signs plus or minus.

If we make the first member a perfect square, by completing the square (Art. 112*), we have

x2 + 2px + p2 = q + p2,

which may be put under the form

(x + p)2 = q + p2.

Now, whatever may be the value of q+p2, its square root may be represented by m, and the equation put under the form

( x + p)2 == m2, and consequently (x+p)2 - m2 = 0.

But as the first member of the last equation is the difference between two squares, it may be put under the form

(x+p−m) (x + p + m) = 0 - - .

- (2),

in which the first member is the product of two factors, and the second 0. Now, we can make this product eq al to 0, and consequently satisfy equation (2) only in two different ways. viz., by making

x+p―m = 0, whence, x = − p + m,

or, by making

x + p + m = 0, whence, x=- -p· m.

Now, either of these values being substituted for x in equa tion (2), will satisfy that equation, and consequently, will satisfy equation (1), from which it was derived. Hence, we conclude, 1st. That every equation of the second degree has two roots, and only two.

2d. That the first member of every equation of the second degree, whose second member is 0, can be resolved into two binomial fac tors of the first degree with respect to the unknown quantity, having the unknown quantity for a first term and the two roots, with their signs changed, for second terms.

For example, the equation

x2+3x-28 = 0

being solved, gives

x = 4 and x=

;

We also have

either of which values will satisfy the equation.

(x − 4) (x + 7) = x2 + 3x − 28 = 0.

If the roots of an equation are known, we can readily form the binomial factors and deduce the equation.

EXAMPLES.

1. What are the factors, and what is the equation, of which the roots are 8 and - 9?

2. What are the factors, and what is the equation, of which the roots are 1 and + 1?

3. What are the factors, and what is the equation, whose

116. If we

article, by x'

designate the two roots, found in the preceding and x', we shall have,

or substituting for m its value √ + p2,

x' = · p + √√ l + p2,

Adding these equations, member to member, we get

x' +x"=2p;

and multiplying them, member by member, and reducing, we find

Hence, after an equation has been reduced to the form of

1st. The algebraic sum of its two roots is equal to the co-efficient of the first power of the unknown quantity, with its sign changed.

2d. The product of the two roots is equal to the second member, with its sign changed.

If the sum of two quantities is given or known, their product will be the greatest possible when they are equal.

Let 2p be the sum of two quantities, and denote their difference by 2d; then,

pd will denote the greater, and p-d the less quantity. If we represent their product by q, we shall have

p2 — d2 = q.

Now, it is plain that q will increase as d diminishes, and that it will be the greatest possible, when d = 0; that is, when the two quantities are equal to each other, in which case the product becomes equal to p2. Hence,

3d. The greatest possible value of the product of the two roots, is equal to the square of half the co-efficient of the first power of the unknown quantity.

Of the Four Forms.

117. Thus far, we have regarded p and q as algebraic quantities, without considering the essential sign of either, nor have we at all regarded their relative values.

If we first suppose p and q to be both essentially positive, then to become negative in succession, and after that, both to become negative together, we shall have all the combinations of signs which can arise. The complete equation of the second degree will, therefore, always be expressed under one of the four following forms: