275. When two imaginary expressions differ only in the sign of the imaginary part, they are said to be conjugate. Thus a-b√-1 is conjugate to a +b√ −1.

α

Similarly √2+3√−1 is conjugate to √2 −3√—1.

276. The sum and the product of two conjugate imaginary expressions are both real.

For

a+b√1+a-b√-1=2 a.

Again (a+b√− 1)(a — b√− 1) = a2 − (− b2) = a2 + b2.

277. If the denominator of a fraction is of the form a+b√-1, it may be rationalized by multiplying the nu merator and the denominator by the conjugate expression a-bv-1. For instance,

=

c+d√1_(c + d√ − 1) (a - b√ — 1)
a+b√-1 (a+b√− 1) (a - b√ — 1)

Thus, by reference to Art. 271, we see that the sum, difference, product, and quotient of two imaginary expressions is in each case an imaginary expression of the same form.

278. Fundamental Algebraic Operations upon Imaginary Quantities.

Ex. 1. Find value of √ aa + 5 √ − 9 a1 − 2 √ — 4 aa.

a2√-1 15 a2√ 1

5√-9 a1 = 5√9 a1 (− 1) =
-2√-4a=-2√4 a1(−1) = — 4a2√−1

(2√3√ − 1) (3√2√− 1) = 6√6(√— 1)2= — 6√6.

2+3√1_(2 + 3 √ − 1) (2 − √− 1) _7 +4√−1_7+4√=1 − −

=

(2 + √− 1) (2 −√− 1)

=

279. The method of Art. 262 may be used in finding the square root of a+b√— 1.

Ex. Find the square root of — 7 — 24 √— 1.

We have now to find two quantities whose sum is - 7 and whose product is 144; these are 9 and - 16;

hence

-7-24 √19+(-16) - 2 √9 x (16)

5. 2√— a2x2 + 7 √ − 4 a2x2 + 12 √ — 36 a2x2.

6. √−−√={+√=}}+√=}. 9. (2√−2+ √−3)(√−3-√−5).

CHAPTER XXV.

PROBLEMS.

280. In previous chapters we have given collections of problems which lead to simple equations. We add here a few examples of somewhat greater difficulty.

Ex. 1. A grocer buys 15 lbs. of figs and 28 lbs. of currants for $2.60; by selling the figs at a loss of 10 per cent, and the currants at a gain of 30 per cent, he clears 30 cents on his outlay: how much per pound did he pay for each ?

Let x, y denote the number of cents in the price of a pound of figs and currants respectively; then the outlay is

15x+28 y cents.

.. 15 x + 28 y = 260.

1

(1).

The loss upon the figs is x 15 x cents, and the gain upon the

10

From (1) and (2) we find that x = 8, and y = 5; that is, the figs cost 8 cents a pound, and the currants cost 5 cents a pound.

Ex. 2. At what time between 4 and 5 o'clock will the minute-hand of a watch be 13 minutes in advance of the hour-hand?

Let x denote the required number of minutes after 4 o'clock; then, as the minute-hand travels twelve times as fast as the hour-hand, the hour-hand will move over minute divisions in x minutes. At 4

х 12

o'clock the minute-hand is 20 divisions behind the hour-hand, and finally is 13 divisions in advance; therefore the minute-hand moves over 20 + 13, or 33 divisions more than the hour-hand.

If the question be asked as follows: "At what times between 4 and 5 o'clock will there be 13 minutes between the two hands?" we must also take into consideration the case when the minute-hand is 13 divisions behind the hour-hand. In this case the minute-hand gains 2013, or 7 divisions.

Ex. 3. Two persons A and B start simultaneously from two places, c miles apart, and walk in the same direction. A travels at the rate of p miles an hour, and B at the rate of q miles; how far will A have walked before he overtakes B?

c miles.

Suppose A has walked x miles, then B has walked x
A, walking at the rate of p miles an hour, will travel x miles in

х

p

с

hours: these two times

9

Ex. 4. A train travelled a certain distance at a uniform rate. Had the speed been 6 miles an hour more, the journey would have occupied 4 hours less; and had the speed been 6 miles an hour less, the journey would have occupied 6 hours more. Find the distance.

Let the speed of the train be x miles per hour, and let the time occupied be y hours; then the distance traversed will be represented by xy miles.

On the first supposition the speed per hour is x + 6 miles, and the time taken is y - 4 hours. In this case the distance traversed will be

represented by (x + 6) (y − 4) miles.

On the second supposition the distance traversed will be represented by (x-6) (y + 6) miles.

All these expressions for the distance must be equal;

.. xy=(x+6)(y — 4) = (x − 6) (y+6).

From these equations we have

Ex. 5. A person invests $ 3770, partly in 3 per cent Bonds at $102, and partly in Railway Stock at $84 which pays a dividend of 4 per cent; if his income from these investments is $136.25 per annum, what sum does he invest in each ?

Let x denote the number of dollars invested in Bonds, y the number of dollars invested in Railway Stock; then

Therefore he invests $ 2720 in Bonds and $1050 in Railway Stock.

EXAMPLES XXV.

1. A sum of $100 is divided among a number of persons; if the number had been increased by one-fourth each would have received a half-dollar less: find the number of persons.