22. A and B with C pulling against them would raise a weight in 5 hours; A and C with B pulling against them in 7; and B and C with A pulling against them in 8. Required the time which it would occupy each alone to raise it.

23. A and B run a race of 2 miles. The length of the course is 3520 feet, so that each has to go over it three times. A gets first to the further end, and in returning meets B a minute before he reaches that end. Having returned to the starting point, he has advanced 1 minute on the journey back when he again meets B. Required the speed of each. 24. A steamer, voyage in 8 hours.

with the wind favourable, performs her

On starting for the return voyage the wind has sunk to half its former force; and this continues for 4 hours, when the wind dies away altogether for 2 hours. A favourable wind then arises of double the force which it had on the outward voyage, and the voyage, including a detour of half a mile, is completed in 24 hours more. In calm weather the double voyage is performed in 17 hours. Required the distance, and the respective rates due to steam and to the wind on the outward voyage.

25. Two persons, A and B, are travelling on roads which intersect at right angles. B is 540 yards short of the point of crossing when A passes it, and in 2 minutes they are equally distant from that point; also, in 8 minutes more they are again equally distant from it. Required the speed of each.

26. A and B entered into partnership and gained £200. Now, 6 times A's accumulated stock (capital and profit), made up 5 times B's original stock; and 6 times B's profit exceeded A's original stock by £200. Required the original stock of each.

27. A, B and C entered into partnership and gained £1500. Now A's stock fell short of B's by £700; and A's gain fell short of C's by £200. Also, the difference between B's stock and C's gain exceeded by a third of A's stock the difference between A's stock and his gain. Required their respective

stocks and gains.

PURE QUADRATIC EQUATIONS.

100. Quadratics are either pure or adfected. square of the unknown quantity appears, the and is solved at once by extracting the root. explain our meaning.

When only the quadratic is pure, An example will

We need not stop to point out the existence of two solutions in this case; the reason of it is obvious, from inspection of the question.

2. Given √(x2 + 5) + √√(x2 + 6) = 1, to find x.

√√(x2 + 5) = 1 − √(x2 + 6)

x2+5=1-2√√(x2 + 6) + x2 + 6

2√(x2+6)=2

√(x2+6)=1

x2+6=1

x2=-5

x= √ −5.

In this case the solution is imaginary, or impossible, as it is sometimes termed; i. e., no numerical quantity will satisfy it. The solution is nevertheless a proper solution of the equation. In this particular case, we have no difficulty in making this evident, for, by considering 2 (without reference to its meaning the square of x) as a simple quantity, and solving the equation

for this quantity, our result thus far is intelligible enough. At the same time, since the equation contains this quantity under the form of a square, we are quite sure that no numerical quantity can satisfy the conditions. In this sense the solution is impossible; but solutions of this class may nevertheless admit of interpretation.

101. The general form of an adfected quadratic equation, after all reductions, is x2 + px=q ; where Ρ and q are any given quantities whatever. The mode of proceeding is to add such a quantity to the first side of this equation, that it may become a complete square. Now the square of x+a is x2+2ax+a2. By comparing this with the given equation, we

find that Ρ stands in the place of 2a, or a

p
2

if, therefore, we

add the square of 2(a) to each side of the equation x2 + px − q,

the first side will be a perfect square, and, by extracting its root, 2 will disappear. The result is,

Hence the RULE. Having reduced the equation to the form x2 + px = q, by bringing all the x's to one side, and dividing by the coefficient of x2; add to both sides the square of half the coefficient of the second term (or of x), and extract the square root of both sides.

Ex. 1. Given x2 - 4x 5, to find x.

=

We must add the square of half the coefficient of x to both sides, and we get

therefore 5 and 1 are the solutions; and, by substitution, it is manifest that each of them satisfies the given equation.

2. Given x2+2x=80, to find x.

Adding 1 to each side,

Multiply by 2, and (~ − 1)2 + 1 (∞ − 1) = 18.

(x 2

Now, instead of squaring x1, we may at once complete the square for x 1, considering it as a simple quantity, by

adding to each side the square of

1

4