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1. Required the cube or third power, of 2a?.
2. Required the biquadrate, or 4th power, of 2aax.

, of


За2 x 4. Required the biquadrate, or 4th power

of 5. Required the 4th power of a+x; and the 5th pow.


er of a-y.


A binomial or residual quantity may also be readily raised to any power whatever, as follows :

1. Find the terms without the coefficients, by obserying that the index of the first, or leading quantity, begios with that of the given power, and decreases continually by 1, in every term to the last ; and that in the following quantity, the indices of the terms are 1, 2, 3, 4, &c.

2. To find the coefficients, observe that those of the first and last terms are always 1 ; and that the coefficient of the second term is the index of the power of the first : and for the rest, if the coefficient of any term be multiplied by the index of the leading quantity in it, and the prodoct be divided by the number of terms to that place, it will give the coefficient of the term next following.

Note. The whole number of terms will be one more than the index of the given power ; and when both terms of the root are t, all the terms of the power will be t; but if the second term be --, all the odd terms will be t, and the even terms ; or, which is the same thing, the terms will be + and - alternately (n).

(n) The rule here given, which is the same in the case of integral


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

a5, a4x, a3x, a2x3, ax4, 35.
And the coefficients, according to the rule, will be

5X4 10X3 10+% 5X1
2 3

4 1

; or 1, 5, 10,

5, 1. Whence the entire 5th power of a +x is

as +5a4x+1023x2 + 10a2x3 +50x4 + x5

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

ab, a5x, a*x, a 3x3, a2x4, ax5, 26. And the coefficients, found as before, are

6 X5 15 X4 20 X3 15 X2 6X1
& 1, 6,
2 3


5 6 or 1, 6, 15, 20. 15, 6, 1. Whence the entire 6th power of a-x is

a® – 625x+15a4x• - 20a3x3 +15a2 x4 – 6ax5 +6


powers as the binominal theorem of Newton, may be expressed in general terms, as follows:

m-1 m2 fato)maamfomam.167m.

am-363, &c 2

2 3 1

1 m-2 (a-bm=qm-mam-16+m.

-am-2b2 - m.,

am-363. &c. 2

2 3

Fam-2b2 +m.


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which formula will, also, equally hold when m is a fraction, as will be more flly explained hereafter.

It may, also, be farther observed, that the sum of the coefficients in every power, is equal to the number 2 raised to that power, Thus i+1=2, for the first power; 1+2+i=4-22, for the square ; 1+3+u+1-8-23, for the cube, or third


power ;

so on,

3. Required the 4th power of a+x, and the 5th power of a-X.

4. Required the 6th power of a+x, and the 7th power

of a-y.

5. Required the 5th power of 2+x, and the cube of a-bx+c.


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


To find any root of a simple quantity.


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, va

- + a or-a, and i64 + bor-6, &c. But the cube root, or any other odd root, of a quantity, will have the same sign as the quantity itself. Thus,

Vaiza; V-a3=-a; and v-ar-a, &c. (0)

(6) The reason why + a and - a are each the square root of a is obvious, since, by the rule of multiplication, (ta)x(te) and (-a)X(-a) are both equal to a?.

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

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


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


Here 9x2=V9XV x2 =3Xx=3x2 Ans.
And V8x3=3/8X/X3=2 Xx=2x. Ans.

a? x2
2. It is required to find the square root of and

4c? the cube root of


2703 a2x2 va x2

8a8 x 3


2703 3c 3. It is required to find the square root of 4a2x6. 4. It is required to find the cube root of - 125a3.6. 5. It is required to find the 4th root of 256a4x8. 6. It is required to find the square root of




Here ✓

and V

9.x2 ya

8a3 7. It is required to find the cube root of


32a5 x16 8. It is required to find the 5th root of


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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)=-a3; and (-93)*(+a)=-45,

And consequently -235-, and -25=-a


To extract the square root of a compound quantity.


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.

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 arith. metic.


1. Extract the square root of x4 - 4x3 +6x2 - 4x+1,

x4--4x3 +6x2 - 4x+1(x2 – 2x+1

2x2 – 2x) - 423 +6x3

- 4.3 +4x2

2x2 - 4x+1)2:0° - 4x+1

2x2 - 4x+1

Ang x2 - 2.+-1, the root required. 2. Extract the square root of 404 +12a3x+1322 ya 60x3 +**.

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