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EXAMPLES FOR PRACTICE.

2

1. Required the cube or third power, of 2a?.
2. Required the biquadrate, or 4th power, of 2ax.
3. Required the cube, or third power, of 3x2y3

raya
4. Required the biquadrate, or 4th power of

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

;

За2 x

er of a-y.

RULE II.

:

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

1. Find the terms without the coefficients, by observing that the index of the first, or leading quantity, begins 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 product 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

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

a5, a4x, a3ra, az x3, ax4, х5. And the coefficients, according to the rule, will be

5 X4 10 X3 10+2 5X1 1,5,

2 3 4 1 or 1, 5, 10, 10, 5, *1. Whence the entire 5th power of a +x is

a5 +5a4x+10a3x2 +10a2x3 +50x4 + 305

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

al, a5x, a4x2, a 3x3, a2x4, axb, x6. And the coefficients, found as before, are

6 X5 15 X4 20 X3 15 X2 6X1
(
2 3

;
4

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

a® — 6a5x+15a4x• 20a3x3 +15a2 x 4 - 6ax5 +26

1, 6,

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m-1

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

m-1 m-2 (a+b)mzamt mam-167m.

am-363, &c 2

2 3

-1 -2 (a-bm=am-mam-16+m.- Fam-2b2-m

am-363. &c. 2

2 3

-am-262 +m.

m-1

mon . &c.

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a

which formulæ will, also, equally hold when m is a fraction, as will be more f lly 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 11–2, for the first power; 1+2+1=4=22, for the square ; 1+3+0+1-8=23, for the cube, or third power ; and 3. Required the 4th power of atx, and the 5th power of a-x.

so on,

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.

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.

CASE I.

To find any root of a simple quantity.

RULE.

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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? = +aor-a, and 364 + 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=-Q; and v-a5 = -a, &c.(o)

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

It may here, also, be farther remarked, that any ever 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 a.

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EXAMPLES.

1. Find the square root of 9x2 ; and the cube root of 3.23.

Here 9x2=V9Xvx?=3Xx=3x2 Ans.
And V8x3=38 Xvx3=2Xx=2x. Ans.

a2x2 2. It is required to find the square root of and

402 8a3x3 the cube root of

2703 a2 x2

8a8 x 3 Qax

; and V 4c2

26

2703 3c 3. It is required to find the square root of 422 x6. 4. It is required to find the cube root of – 125a3x6. 5. It is required to find the 4th root of 256a4x8. 6. It is required to find the square root of

ax

Here v

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424

9.xo ya

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

125x6

32a5x10 8. It is required to find the 5th root of

243

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)=-43; and (-93)x(+a)=-45,

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

CASE II.

To extract the square root of a compound quantity.

RULE.

a

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 arithmetic.

EXAMPLES.

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

X4 --4x3 +6x2 - 4x +1(x2 - 2x+1
34

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Ans. 22 – 2x+1, the root required. 2 Extract the square root of 494 +12a3x+13a8+ 60x3 +*4.

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