EXAMPLES.

1. Let a+x be involved, or raised to the 5th power. Here the terms, without the coefficients, are

a5, a1x, a3x2, a2x3, axa, x5.

And the coefficients, according to the rule, will be 5X4 10X3 10X2 5×1

1,5,

;

10,

or 1, 5, 10,

Whence the entire 5th power of a+x is

a5+5a4x+10a3x2 +10a3x3+5ax1+x3.

3. Let a-x be involved, or raised, to the 6th power. Here the terms, without their coefficients, are

а®, a3x, a1x2, a3x3, a2x2, αx5, x®.

And the coefficients, found as before, are

6X5 15X4 20X3 15X2 6X1

1,6,

a°-6a5x+15a4x320a3x3+15a2x4-6αx5+x6.

3. Required the 4th power of a+x, and the 5th power Ans. a4+4a3x+6a2x2+4ax3 +xa, and a5

of a-x.

-

5a4x+10a3x2 - 10a2x35ax1 —x3. 4. Required the 6th power of a+b, and the 7th power Ans. as+6a5b+15a4b2+20a3b3+15a2b++ 6ab5+bo, and a77a6y+21a5y2-35 a4y3+35a3y-21a2y5+7ay—y7.

of a-y.

a-bx+c.

5. Required the 5th power of 2+x, and the cube of Ans. 324-80x+80x+40x3+10x4+x5, and a3+3a3 c-+Sac2+c3- 3a2 bx-6acbx3c2 bx+3ab2x2+3cb2x2 - b3x3.

EVOLUTION.

EVOLUTION, or the extraction of roots, is the reverse of involution, or the raising powers; being the method of finding the square root, cube root, &c, of any given quantity.

CASE 1.

To find any root of a simple quantity.

RULE.

Extract the root of the coefficient for the numeral part, and the root of the quantity subjoined to it for the literal part; then these, joined together, will be the root required.

And if the quantity proposed be a fraction, its root will be found by taking the root both of its numerator and denominator.

-.

Note. The square root, the fourth root, or any other even root, of an affirmative quantity, may be either + or Thus, a2+a ora, and b1=+b or −b, &c. But the cube root, or any other odd root, of a quantity, will have the same sign as the quantity itself. Thus, 3/a3=α; 3/-a3-a; and 5/-a5=-a, &c.*

It may here, also, be farther remarked, that any even root of a negative quantity, is unassignable.

Thus, a cannot be determined, as there is no quantity, either positive or negative, (+ or -), that when multiplied by itself, will produce-a2.

EXAMPLES.

1. Find the square root of 9x2; and the cube root of 8x3.

Here/9x2=√9× √x2=3×x=3x. Ans.

And /8x3/8X/x3=2Xx=2x. Ans.

The reason why a and -a are each the square root of a2 is obvious, since by the rule of multiplication, (+a) × (+a) and (-a) X (-a) are both equal to a2.

And for the cube root, fifth root, &c. of a negative quantity, it is plain, from the same rule, that

(-a)X(-a)X(-a);

And consequently

-a3; and (3) X(+a2)=-α5,
-a3a, and 1⁄2

2. It is required to find the square root of

4c2 √4c3 2c

3. It is required to find the square root of 4a2x6.

4. It is required to find the cube root of

5. It is required to find the 4th root of 256a1x3.

6. It is required to find the square root of

Ans. 4ax2.

8a3

7. It is required to find the cube root of

8. It is required to find the 5th root of

CASE II.

To extract the square root of a compound quantity.

RULE.

1. Range the terms, of which the quantity is composed, according to the dimensions of some letter in them, beginning with the highest, and set the root of the first term in the quotient.

2. Subtract the square of the root, thus found from the first term, and bring down the two next terms to the remainder for a dividend.

F

3. Divide the dividend, thus found, by double that part of the root already determined, and set the result both in the quotient and divisor.

4. Multiply the divisor, so increased, by the term of the root last placed in the quotient, and subtract the product from the dividend; and so on, as in common arithmetic.

EXAMPLES.

1. Extract the square root of x1 —4x3+6x2 -—4x+1. x4-4x+6x-4x+1(x2-2x+1

Ans. x2-2x+1, the root required.

2. Extract the square root of 4a2+12a3x+13a2x2+6 ax3+x1.

4a4+12a3x+13a2x2+6ax3+x1 (2α2+3αx+x2

Note. When the quantity. to be extracted has no exact root, the operation may be carried on as far as is thought necessary, or till the regularity of the terms shows the law by which the series would be continued.

EXAMPLE.

1. It is required to extract the square root of 1+x.

Here, if the numerators and denominators of the two last terms be each multiplied by 3, which will not alter their values, the root will become

X x2 2x3 3.5x4 3.5.7x5

+

2.4 2.4.6 2.4.6.8 2.4.6.8.10'

where the law of the series is manifest.

&c.

EXAMPLES FOR PRACTICE.

2. It is required to find the square root of a4+4a3x+ 6a2x2+4ax+x1. Ans. a2+2ax+x2.

3

3. It is required to find the square root of x4 - 2x3†

1

16'

Ans. x2-x+1.