The 12 will satisfy the question in its algebraic sense, and considered without reference to its sign, will be the answer to the following arithmetical question:-A person purchased a number of yards of cloth for 240 cents: if he had paid the same sum for 3 yards more, it would have cost him 4 cents less per yard. How many yards did he purchase?

REMARK. In the solution of a problem, both roots of the equation will satisfy the enunciation, understood in its algebraic sense. If the enunciation, considered arithmetically, admits of a double interpretation, when translated into the language of Algebra, the solution of the equation will make known the fact: and hence, while one root resolves the question in its arithmetical sense, the other resolves another similar question also in its arithmetical sense; and both questions will be stated by equations of the same general form, having equal numerical roots with contrary signs.

3. A man bought a horse, which he sold for 24 dollars. At the sale, he lost as much per cent. on the price of his purchase, as the horse cost him. What did he pay for the horse? Let x denote the number of dollars that he paid for the horse: 24 will express the loss he sustained.

then, x

per cent. by the sale, he must have lost

But as he lost x

100

upon a dollars he loses a sum denoted by 100

we have then

Both of these values satisfy the question.

For, in the first place, suppose the man gave 60 dollars for the horse and sold him for 24, he then loses 36 dollars. But, from the enunciation he should lose 60 per cent. of 60, that is, 60 × 60

60

of 60 =

100

ciation.

100

= 36; therefore 60 satisfies the enun

If he pays 40 dollars for the horse, he loses 16 by the sale;

7

1080

124

CHAP.

EQUATIONS

EQUATIONS OF THE SECOND DEGREE.

40

143

for, he should lose 40 per cent. of 40, or 40 × = 16; there

fore 40 verifies the enunciation.

100

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?

many 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 is 3; and if 18 be added to it, the Ans. 24.

digits will be inverted? (U-X)X=2/

7. 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.

8. Two persons, A and B, departed from different places at the same time, and travelled toward each other. On meeting, it appeared that A had travelled 18 miles more than B; and that A could have gone B's journey in 15 days, but B would have been 28 days in performing A's journey. How far did each travel? A 72 miles.

Ans.

{

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 had 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?

x + x-6= X + bujns. 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 stock; 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 travelled 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.

141. Thus far, we have only resolved particular problems involving equations of the second degree, and in which the known quantities were expressed by particular numbers.

We propose now, to explain the general properties of these equations, and to examine the results which flow from all the suppositions that may be made on the values and signs of the known quantities which enter into them.

142. It has been shown (Art. 140), that every complete equation of the second degree can be reduced to the form

x2 + 2px = 9 (1),

in which p and q are numerical or algebraic quantities, whole numbers, or fractions, and their signs plus or minus.

If we make the first member a perfect square, by adding p2 to both members, we have

x2 + 2px + p2 = q + p2,

which may be put under the form

(x+p)2 = 9+ p2.

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

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

x+p-m=0, whence, x = − p +m,

or, by making

x + p + m = 0, whence, x =

-p-m;

and by substituting for m its value, we have

x = − p + √ q + p2, and x = -p-√q + p2.

Now, either of these values being substituted for x in its corresponding factor of equation (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 every equation of the second degree may be decomposed into two binomial factors of the first degree with respect to x, having x for a first term, and the two roots, taken with their signs changed, for the second terms.

being resolved gives x 4 and x = 7; either of which values will satisfy the equation. We also have

(x − 4)(x+7)= x2+3x-28.

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

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 roots

143. If we designate the two roots of any equation by x' and x', we shall have

x = − p + √1\+ p2, and a' = − p −√√9+p2;

by adding the roots, we obtain,

x+x=2p;

and by multiplying them together,

1st. The algebraic sum of the two roots is equal to the co-efficient of the second term of the equation, taken with a contrary sign.

2d. The product of the two roots is equal to the absolute term. taken also with a contrary sign.

144. 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, the to become negative in succession, and after that, both to become negative together, we shall have all the combinations of signs which can arise; and the complete equation of the second degree will, therefore, always be expressed under one of the four following forms :

In order that the value of x, in these equations, may be found either exactly or approximatively, it is necessary that the quantity under the radical sign be positive (Art. 126).

Now, p2 being necessarily positive, whatever may be the sign of p, it follows, that in the first and second forms all the values