To avoid radicals, let 2x2 represent the number of bees at first;

Whence, x=6, or -11; but the latter value, being fractional, though satisfying the equation, is excluded by the nature of the question; the number of bees is 2×62-72.

4. Divide a into two parts, whose product shall be b2.

Let x= one part; then, a-x= the other.
Therefore, x(α-x), or ax-x2-b2.
Whence, x=(a±√a2-462); that is,

x=}(a±√a2—4b2), and a—x={(a‡√a2—462), are the parts required, and the two parts are the same, whether the upper or lower sign of the radical quantity be used. Thus, if the number a is 20, and b 8, the parts are 16 and 4, or 4 and 16.

The forms of these results enable us to determine the limits under which the problem is possible; for it is evident that if 462 be greater than a2, √a2—462 becomes imaginary.

The extreme possible case will be, when

a2-462=0, in which

case x=¿α, and a-x=ja; also, b2={a2, and b=ja.

In the following examples, that value of the unknown quantity only is given, which satisfies the conditions of the question in an arithmetical sense:

5. What two numbers are those whose sum is 20 and product 36? Ans. 2 and 18.

6. Divide 15 into two such parts that their product shall be to the sum of their squares as 2 to 5. Ans. 5 and 10. 7. 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. Divide 24 into two such parts that their product shall be equal to 35 times their difference. Ans. 10 and 14.

9. Divide the number 346 into two such parts that the sum of their square roots shall be 26. Ans. 112 and 152.

SUGGESTION.-Let x= the square root of one of the parts, and 26-x, of the other.

10. What number added to its square root gives 132? Ans. 121.

11. What number exceeds its square root by 483?

Ans. 561.

12. What two numbers are those, whose sum is 41, and the sum of whose squares is 901? Ans. 15 and 26.

13. What two numbers are those, whose difference is 8, and the sum of whose squares is 544? Ans. 12 and 20.

14. A merchant sold a piece of cloth for $24, and gained as much per cent. as the cloth cost him. Required Ans. $20.

the first cost.

- 15. Two persons, A and B, had a distance of 39 miles to travel, and they started at the same time; but A, by traveling of a mile an hour more than B, arrived one hour before him; find their rates of traveling.

Ans. A 31, B 3 mi. per hr.

16. A and B distribute $1200 each among a number of persons; A gives to 40 persons more than B, and B gives $5 apiece to each person more than A; find the number of persons. Ans. 120 and 80.

17. From two towns, distant from each other 320 miles, two persons, A and B, set out at the same instant to meet each other; A traveled 8 miles a day more than B, and the number of days before they met was equal to half the number of miles B went in a day; how many miles did each travel per day? Ans. A 24, B 16 mi. 18. A set out from C toward D, and traveled 7 miles a day. After he had gone 32 miles, B set out from D toward C, and went every day of the whole journey; and after he had traveled as many days as he went miles in one day, he met A. Required the distance from C to D. Ans. 76, or 152 miles.

19. A grazier bought a certain number of oxen for $240, and after losing 3 sold the remainder for $8 a head more

than they cost him, thus gaining $59 by his bargain. What number did he buy?

Ans. 16.

20. Divide the number 100 into two such parts that their product may be equal to the difference of their Ans. 38.197, and 61.803 nearly. squares.

21. Two persons, A and B, jointly invested $500 in business; A let his money remain 5 months, and B only 2, and each received back $450, capital and profit. How much did each advance? Ans. A $200, B $300.

22. It is required to divide each of the numbers 11 and 17 into two parts, so that the product of the first parts of each may be 45, and of the second parts 48. Ans. 5, 6, and 9, 8.

45

45

Represent the four parts by x, 11—x, and 17- and put the

product of the second and fourth equal to 48.

x

23. Divide each of the numbers 21 and 30 into two parts, so that the first part of 21 may be three times as great as the first part of 30; and that the sum of the squares of the remaining parts may be 585.

Ans. 18, 3, and 6, 24.

24. Divide each of the numbers 19 and 29 into two parts, so that the difference of the squares of the first parts of each may be 72, and the difference of the squares of the remaining parts 180. Ans. 7, 12, and 11, 18.

DISCUSSION OF THE GENERAL EQUATION IN QUADRATICS.

234. The discussion of the general equation in quadratics consists in investigating its general properties, and in interpreting the results derived from making particular suppositions on the different quantities which it contains.

Taking the general form, (Art. 231,) x2+2px=q, and completing the square, we have x2+2px+p2=q+p2.

Now, x2+2px+p2=(x+p)2. For the sake of simplicity,

But, since the left member is the difference of two squares, it may be resolved into two factors (Art. 93); this gives

(x+p+m)(x+p-m)=0.

Now, this equation can be satisfied in two ways, and in only two; that is, by making either of the factors equal to 0. If we make the second factor equal to zero, we have

x+p―m=0;

Or, by transposing, x=-pm-P+√9+p2.

If we make the first factor equal to zero, we have x+p+m=0

Or, by transposing, x——p—m▬▬p−√9+p2. Hence, we have

Property 1st. Every quadratic equation has two roots, (or values of the unknown quantity,) and only two.

From the equation (x+p+m)(x+p—m)=0, we derive

Property 2d. Every affected equation, reduced to the form x2+2px=q, may be decomposed into two binomial factors, of which the first term in each is x, and the second, the two roots with the signs changed.

Thus, the two roots of the equation, x2-7x+10=0, are x=2 and 5. Hence, x2-7x+10=(x-2)(x-5).

It is now evident that the direct method of resolving a quadratic trinomial into its factors, is to place it equal to zero, and then find the roots of the resulting equation.

In this manner let the trinomials, Art. 94, be solved.

By reversing the operation, we can readily form an equation whose roots shall have any given values. Thus,

Let it be required to form an equation whose roots shall

1. Find an equation whose roots are 4 and 5.

Ans. x2-9x=-20.

2. Whose roots are - and +}. Ans. x2+x=.

3. Find an equation, without fractional coëfficients, whose roots are and . Ans. 15x2-22x——8.

4. Find an equation whose roots are m+n and m―n. Ans. x2-2mx—n2—m2.

Resuming the equation x2+2px-q, and denoting the two roots by x and x", we have

x'

-P+√q+p2, x"=-p−√q+p2.

Adding, x+x"——2p.

But, 2p is the coëfficient

of x, taken with a contrary sign. Hence, we have

Property 3d. The sum of the two roots of a quadratic equation, reduced to the form x2+2px=q, is equal to the coefficient of the first power of x, taken with a contrary sign. If we take the product of the roots, we have

But

q is the known term of the equation, taken with a contrary sign. Hence, we have

Property 4th. The product of the two roots of a quadratic equation, reduced to the form x2+2px=q, is equal to the known term taken with a contrary sign.