values are negative if p is positive in the equation; that is, if x2 + px ==q, which gives

and both positive if p is negative in the equation, that is, a3. px=

which gives

When both values are negative, neither of them answers directly to the conditions of the question; but if -x be put into the original equation instead of x, the new equation will show what alteration is to be made in the enunciation of the question; and the same values will be found for x as before, with the exception of the signs.

If in this equation q is greater than 2, the quantity

4

becomes negative, and the extraction of the root

cannot be performed. The values are then said to be imaginary.

1. It is required to find two numbers whose sum is p, and whose product is q.

This example presents the case above mentioned, in which p and q are both negative.

x=

= 15 ± (225 — 54 )2 = 15 ± (225 216.) }

4

4

The values are 9 and 6, both positive, and both answer the conditions of the question. And these are the two numbers required, for 9+6= 15, 9 x 654. This ought to be × so, for x in the equation represents either of the numbers indifferently. Indeed whichsoever x be put for, p-x will represent the other; and p x-x2 will be their product.

Here (— 8) is an imaginary quantity, therefore both valucs are imaginary.

In order to discover why we obtain this imaginary result, let us first find into what two parts a number must be divided, that the product of the two parts may be the greatest possible quantity. In the above example, p represents the sum of the two numbers or parts, let d represent their difference, then

dis greater than zero; but when d= 0, the expression becomes

p which is the second power of P. Therefore the greatest

4

2

possible product is when the two parts are equal.

In the above example ? = 8, and P

2

4

= 64. This is the

greatest possible product that can be formed of two numbers whose sum is 16. It was therefore absurd to require the product to be 72; and the imaginary values of x arise from that absurdity.

2. It is required to find a number such, that if to its second power, 9 times itself be added, the sum will be equal to three times the number less 5.

x2+9x=3x-5.

x2+6x=5.

This equation is in the form of x2 + p x = −q, which gives

The values are-1 and -5, both negative. Consequently neither value will answer the conditions of the question. This shows also that those conditions cannot be answered.

But if we change the sign of x in the equation, that is, put in - instead of r, it becomes

This shows that the question should be expressed thus:

It is required to find a number, such, that if from 9 times itself, its second power be subtracted, the remainder will be equal to 3 times the number plus 5.

The values will both be positive in this, and both answer the conditions.

The values are 5 and 1 as before, but now both are positive, and both answer the conditions of the question.

3. There are two numbers whose sum is a, and the sum of whose second powers is b. It is required to find the numbers.

Examine the various cases which arise from giving different values to ɑ and b. Also how the negative value is to be interpreted. Do the same with the following examples.

4. There are two numbers whose difference is a, and the sum of whose second powers is b. Required the numbers.

5. There are two numbers whose difference is a, and the difference of whose third powers is b. Required the numbers.

6. A man bought a number of sheep for a number a of dollars; and on counting them he found that if there had been a number b more of them, the price of each would have been less by a sum c. How many did he buy?

7. A grazier bought as many sheep as cost him a sum a, out of which he reserved a number b, and sold the remainder for a` sum c, gaining a sum d per head by them. How many sheep did he buy, and what was the price of each?

8. A merchant sold a quantity of brandy for a sum a, and gained as much per cent. as the brandy cost him. What was the price of the brandy?

XXXVI. Of Powers and Roots in General.

Some explanation of powers both of numeral and interal quantities was given Art. X. The method of finding the roots of the second and third powers, that is, of finding the second and third roots of numeral quantities, has also been explained; and their application to the solution of equations. But it is

frequently necessary to find the roots of other powers, as well as of the second and third, and of literal, as well as of numeral quantities. Preparatory to this, it is necessary to attend a little more particularly to the formation of powers.

The second power of a is a × a = a2.

The fifth power of a is a xa xa xa x a = a3.

If a quantity as a is multiplied into itself until it enters m times as a factor, it is said to be raised to the mth power, and is expressed a". This is done by m This is done by m-1 multiplications; for one multiplication as a X a produces a2 the second power, two multiplications produce the third power, &c.

We have seen above Art. X. that when the quantities to be multiplied are alike, the multiplication is performed by adding the exponents. By this principle it is easy to find any power of a quantity which is already a power. Thus

The second power of a' is a X a2 = a313 = ao.

—

The third power of a2 is a2 × a2 × a2 = a2+2+3 = a®.

a2m.

The second power of am is am × am = am+m = a2

The third power of am is am × am x am = am+m+m = a3m . The mth power of a3 is a3 × a3 × a2 × a2 x =q2+2+2+3+......, until a is taken m times as a factor, that is, until the exponent 2 has been taken m times. Hence it is expressed a1m.

....

The nth of am is am X am X am power = am+m+m+.. until m is taken n times, and the power is expressed amn.

N. B. The dots..... in the two last examples are used to express the continuation of the multiplication or addition, because it cannot come to an end until m in the first case, and n in the second, receive a determinate value.

In looking over the above examples we observe;

1st. That the second power of a3 is the same as the third power of a2, and so of all others.