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8. Extract the square root of a'++*.

22

26

5x8 2a 8a%

£, &c. answer.

12847 a?

1+1)1+1-82, &c.

1

2+1)+1

4a

+1+4
2 24 24

2+1-3-5 8a3 4a?

-Fito x2 26

#et, &c.

&c
Fa +
4a? 8a4
26

28
43 +16a 6406
26 Xco

2012 + +

1649

F

6428 256ato
x x2

2C6
5.28 528

2012

&c. and 4a? sas 128a?

64a 6498 256a109 80 on to an infinite number of terms.

a

toka

士、

2a+

a

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10

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16259

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tatta?

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2a +

zona

&c. root. 2a

2bx+22 45°x? +4b2° ** 2a to.)

ai_262-la2a'

2a

8a3
*
x2

a?
26x+x.

*
20

-2bx x
x2 za
2a

462x2 +4622+24 8a3 4a

- 26x 2 +

42%
22 2013 x2
+

46°2? +46x2+x
4a?
8a4 6406 Rem.

4a
2013 24
Rem. +

8a4 6406
a +40+6a’+42+1(a +2a+1 root.

at
20°+4a)4a+60?

a-2a+2a-attiamati 4af4d

a*
2a?+40+1)+20°+42+1 2a--a-2a3+20?
2a? T4a+1

-2a ta?
2a*_2a+1)a_a+1

a?
b 62 18
ai_abla-

&c. the root.
2 8a 162
a?
7

1–2

142*(1-32-424-16-, &c. 20- -ab

1

root. 62 -abti

2-12) 4

- +1x* 72 2ab

2-22-428)—***
8a'

12+3x+420
78 74
-482 tota

Remainder, 720—6420
8a 64a?
72 73 73 74
20b

4a

16a? 8a 64a" 16. Extract the square root of 4x4—16x8+24x?_-16x+4.

Ans. 2.x*_-42+2. 17. Extract the square root of 16x* +24x+89x® +60x+100.

Ans. 4x2+3x+10. 18. Extract the square root of 1+y.

y

ya 5y* 8 +16 128

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y

Ans.1+

516128+, &c.

19. Extract the square root of 9x9—122+10x*-28x®+172 -8x+16.

Ans. 328–22 +1-4.

To extract the Cube Root of a Compound Quantity.

56. Rule. Arrange the terms according to the dimensions of some letter, as in division, and extract the root of the first term, which must be a cube. Place this root in the quotient, subtract its cube from the first term, and there

a +3a+b+3ab2+63(a+b will be no remainder. Bring

a> down the three next terms for a dividend, and put three

3a?+3ab+b)3a%b+3ab? +88

3a+b+3ab? +88 times the square just found in the divisor's place, and see how often it is contained in the first term of the dividend, and the quotient is the next term of the root. Add three times the product of the two terms of the root, plus the square of the last term, to the terms already in the divisor's place, and the divisor will be complete. Multiply the divisor by the last term of the root, subtract the product from the dividend, and bring down the next three terms for a dividend, and proceed as before. flexa)373(axb)+(6x0)}=3a*+3ab+8=the first divisor.

la+b)2x3+3(a+b)c+exc=3a? +6a)+382+3ca +306+c, the divisor for the third letter in the root. 3(a++c)+3d(a++c)+(dxd=divisor for the fourth term of the root, &c. &c. 1. Extract the cube root of 2—628715x*—202*+-15x2–6c+1.

200—628+1524—20x8+152–6x+1(22_2x+1.

26
*32*—62? +42|46x+15x9_2033

4x?

-6x 112.x*_ 8r 3x4—12x3+1522—6x+13x4—12x+15x*—6x+1

|3x*— 12x8 +15x2—6x+1 2. Extract the cube root of 28 +628–40x8+962–64.

28 +625—40x8 +962–64(2+2x–4

3x4 +6x3+4x2|6x5_40.28

1626 +1224 +828
3x4 +12x3—24x+16-1224_482®+962–64

-12x4–482 +963–64

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3. Extract the cube root of 27x9_54.x8 +63x?—71.00 +57x5—3624 +2228_9% e o +3x~1.

Dividend. 272-54x+63.7–71.xo+57:25—36x*+222°—934+3.x-1 27.709

Root or quotient 3x3–228 +3+1 273-188 +4.2*|-54.c®+63x7–71.

-54.28 +36x?— 828 27x6—3625+2124—6x3+x27x?–63x®+57x*—36x4 +2223

27x736x6 +21.25 60+ 18 2724—3629+30x4-212+92-3x+1 -27x+362—30x4+21.28–9x?+3.c-1

-272 +36x5—30x4 +21x3_9x?+3x—1

+1.

7. Extract the cube root of 27x0—5426 +63x4—442+2122462 6. Extract the cube root of 8x®+-362*+54x+27. Ans. 2x+3. 3(a+b)?+3(a+b)c+)3(a+b)*c+3(a+b)ctes

3(a+b){c+31a+b)c Tin the same manner that b was derived from a.

(a+b)*+3(a+b)*c+3(a+b)c%+0(a+b+c by first finding a and b, as above, and then deriving c from (a+b) If the root consists of three terms, a, b, c, they may be obtained

Ans. 32–2x+1. .

4. Extract the cube root of a+3a2b+3ab? +33 +3a’c+6abc+36*c+3ac? + 36c++.c.

Ans. a+b+c.

EVOLUTION.

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[From p. 41.] Here the numerator being the least compounded, and b rising therein to a single dimension only, I divide the same into the parts 6a 4a%c", and 15a46 10a2bc", which, by inspection, appear to be equal 2a*(3a —2c%), and 5a+b(3a’ — 2c); .: 3a:—2c is a divisor to both parts, and likewise to the whole, expressed by (30—2c") X (2a + 5ab); so that one of these two fac

X tors, if the fraction given can be reduced to lower terms, must also measure the denominator; but the former will be found to succeed: thus 30-2co)9a6—27abc-6abc? +18bc (3ab9bc .-. 3—2c.

Here, the denominator being the least compounded, and d rising therein to a single dimension only, I divide the same into the parts 4dd4acd, and -2ac2 + 2co : which, by inspection, appear to be equal to 4ad(a—c), and — 2c'(ac). ..ac is a divisor to both the parts, and likewise to the whole, expressed by (4a%d — 2c) X (–c); so that one of these two factors, if the fraction given can be reduced to lower terms, must also measure the numerator ; but the latter will be found to succeed : thus, a—c)a’d—c*da'c+cada+cd-ac-; ..a-c is the greatest

; ad-cd-aid tot ad+cd-ac-63 act

the Answer. 4ad4acd2ac? f 2c 4ad2c2 Solution to the 8th. Multiplying the numeator by 5 and I have. 154* - 22 + 10x - x + 2) 452€ + 10+20x*— 5x +40(3x

?

45x5 6x* + 3028 3x2 + 63

X by.5, 6.24—202 +2322—11x +5 1524–2x+10x?—+2)30x*—100x9 +115x2–55x+25(+2 30x4_42° +202?_2

+4.

–962 +95x2–53x+21 Multiply the last divisor by 32, and we shall have -96x8+95x53x+21(480x4—64x®+320x2—32x+641–53

480x47528 +265x?–105x

mult by 32, 41128 +552?+73x+64 -96x9+75x2–53c+21)13152x+1760x*+2336x+2048(–137

13152x3 +130152+72612—2877 Divide this by 4925, it becomes 32 — 2+1; which by another operation exactly divides — 962*+9522—512 +21; and .. in

3x8+2+1 the greatest common measure and the reduced form, is

5x + 2+2

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SIMPLE EQUATIONS. 1. An equation is a proposition which declares the equality of 2 quantities expressed algebraically. This is done by connecting these quantities by the sign (=): thus, 2- -456- X is an equa

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