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Here a=10, and b=108; whence ((a−b)=»(100—108)

-2, and n-3 (db)n =20, or non=20, where it readily appears, from inspection, that n=2. Whence (10+x108) =itin 1444X-2)=l+N 12=l+»3, and V10108)

N14-4X-2)=1-İN 12=1—^3. 2. Fnd the cube root of 2 + 11N -1, or 2£N-121.

Here a=2, and b=121; whence (a+b)=(4+121) =5, no_3{♡(a+b)}n=2a, or n—15=4, where it appears, from

—, inspection, that n=4. Whence (2+1-121)=fin (16— 4x5)=2+in–432+x-1, and V (2 - N - 121) =

(16-4X5=2-41-42-N-1. 3. Required the cube roots of 45+29X2, 9+485, 9 £180, 20+68N-7, 35+691–6, and 81-EN-2700. Answers : 3+12 and 3 - 72; + $5 and } 3+

İN5; 5+-7 and 5–1–7; 5+^-6 and 5-7-6; -3 + 2 N45 and —3—283.

For Trinomial, Quadrinomial Surds, fc. RULE. Divide half the product of any two radicals by a third, gives the square of one radical part of the root; this repeated with different quantities, will give the squares of all the parts of the root, to be connected by + and But if any quantity occur oftener than once, it must be taken but once. For if x+y+z be any trinomial surd, its square will be zi+y+z2+2xy+2x2+2yz; then if half the product of any two rectangles, as 2xy + 2xz, (or

2x*yz 2x*yz) be divided by some third 2yz, the quotient =x', must

2yz needs be the square of one of the parts; and the like for the rest. Extract the root of 10+ (24)+ (40)+ (60).

„(24XV (40) Here

-3, and 21 (60)

2 (40) „(40) XV (60)

„(255. And the root is „2+ 3+»5. 2 (24) Find the square root of 14+ (32)-> (48)+^ (80)-> (24)

✓(32X48) V (24) tw(40)—(60). Here

this produces 2N (80) ✓(32 X 48)

) nothing. Again

EN 2N

27 (24) (32X40)

(48X24) „(25)=5; and

=N4=2; and 2/ (60)

21 (32) (32X80) = 3; and

=N(16)=4, &c., therefore the parts of

2 (40) the root are 4, 75, 73, 72, 74, &c., and the roots of 2+ ^2-^3+»5; for, being squared, it produces the surd.

-2, and V (24) X(60)

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20 (24)=N(16)=4. And V(40X60)

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2c

n r

4+1

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Case XIII. To extract any root (c) of a binomial surd A+B.

Rule. Let A2-B=D, take Q such, that QD=n', the last in. teger power. Let „{(A+B) X VQ? = r, the nearest integer number. Reduce ANQ to the simplest form PN's; let the nearest nti

IN SEN (t's— n) integer =t; then the root =

if it 2NS

Q can be extracted. Note. + is for the binomial A + B, and for the residual A B.

1. What is the cube root of 968 25.

Here D=343=7X7X7, and Qx7n, Q=1, n=7, then ^{(A+B)XVQ=56+=r=4, and AVQ=1968 = A}=

=:

rt 222=PNs, and „s=2

== 2, and tN =

2s 22 =212; „(t's—n)=(

87)=1; Q=1, and the root 2N2+1

=2^2+1, which I find to succeed. 1 2. Find the cube root of 68-4374.

Here D=250=5X5X5X2, and 58X2=4D=QD=n; and =4, n=2X5=10, and {(A+B)XNQ}=»(134X2)=6

r+, 7323 =r; ANQ=136/l=PNs, and ✓s=l;

21s 2 6

INS~~ (t's— ~) (16-10) 4t, and ts=4, and the root

^2

♡4 4-16

; for its cube is 68—2776. 2 3. Find the 5th root of 29/6+41/3.

Here D=3, n=3, Q=81, r=5, NS=6,t=1, INS=6, ^{(t's -n)} N3, Q=V81 ♡9, and the root to be tried V6+3

Scholium. If the quantity be a fraction, or has a 9 common divisor, extract the root of the denominator, or of that common divisor, separately The demonstration may be seen in Sir Isaac Newton's Universal Arithmetic, Gravesend's, or MacLaurin's, or Dr. Waring's Med. Algebra. CASE XIV. To find such a multiplier, or multipliers, as will make

any binomial surd rational. RULE 1. When one, or both of the terms, are any even roots ; multiply the given binomial, or residual, by the same expression, with the sign of one of its terms changed, and repeat the operation in the same way, as long as there are surds; when the last result will be rational.

In like manner, a trinomial surd may also be rendered rational, by changing the sign of one of its terms for the multiplier ; and a

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3

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quadrinomial surd by changing the signs of two of its terms, &c.

2. When the terms of the binomial surds are odd roots, the rule becomes more complicated ; but for the sum, or difference, of two cube roots, which is one of the most useful cases, the multiplier will be a trinomial surd consisting of the squares of the two given terms, and their product, with its sign changed.* Given surd 5+^3

N5+N3 5+3 given surd. Multiplier 5-13 ✓5-13 5-13 Ist mult.

25+5/3 5+15 75-73 1st prod.

-5N3–3 -N1543 75+3 2d mult. Product 25—3=22 5—3=2 543=2 Ans. ratio. N5+73-72 given surd. 49+35+25 mult. Ņ5+^3+12 Ist mult. 7-5 given surd. 5+ 15-710

Ņ343 +245+175 +1543-46

^245—9175mw 125 tw10+^ 6–2

343 * 3257-5 6+2^15 1st product.

Given 7+W3 surd: 6+21/15 2d multiplier. Mult. 372–321+13 -36—12/15

7-147+63 +12x15+60

+147-163+3 60–36=24, Ans.

Ans. 1037 +

3 1. Let (

wat b) Xab=wa-wb, and again I have (Wabxwa+b=wa-, and 3d (wa - b) X (Na+b)=ab rational.

2. Let ato be given; then (a+b)X(a,b)=ab and (a*-)xla’twb=a—b, the 2d product.

3. Let (2) X(ata/2+4)=-2, product. 4. Let (wa+30) Wa-ab+1)=a+b, Answer. 5. Let V5+3 be given, or reduced (5+ 9)(1/125 – (52X9)+(5X92)— 799) =5—9—44, product.

Or (w9+5) X{1798 – (99 x 5) + (9X52)—5}=9 -5=4.

*If a multiplier be required, that shall render any binomial surd, whether it consist of even or odd roots, rational, it may be found by substituting the given numbers, or letters, of which it is composed, in the places of their equals, in the following general formula : Binomial

at Multiplier a"'Fa"-"btam? Fra*-*+&c. where the upper sign of the multiplier must be taken with the upper sign of the binomial, and the lower with the lower; and the series continued to n terms.

*

*

a

Va multiply

a

a

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; and

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XV5+12

CASE XV. To reduce a fraction, whose denominator is a surd,

to another that shall have a rational denominator. RULE 1. When the proposed fraction is a simple one, of the

8 form multiply each of its terms by va, and the resulting va'

bna fraction will be

Or when it is of the form
ba?

' the terms by Na', and the result will be

And for the gen6 eral form multiply by a"-, & the result will be

bami Na 2. If it be a compound surd, find such a multiplier, by one of the rules of the last case, as will make the denominator rational; and multiply both the numerator and denominator by it.

2 3

3 1. Reduce the fractions

and

to a frac13'15 75-72 tions whose denominators shall be rational. 2 2

3 Here

35
V3_203
X

X
V3 13 3

5^3 53 *X5 653 6

3
5
-59125. Here

7572 75+2 3x5+312 315+32_N5+12

=N5+12, Ans. 52

3

1

2 2. Reduce

to a fraction, whose denominator shall be

3 ✓2 rational. Here 72

2x(342) 312+2 2+3x22 3-72 73-72) X (3+x2) 9-2 H2, Ans. 2 3

3
12

4 3. Reduce ;

;
13' 175: 75-12; or

;
3

93 76

12
a-b

10
; or

; or
; or

; or N7+3 3 t. IX X a +

7-15 V3

4
2

8

; or 9+10

75+ 73 „3+12+1 1 and

each of the 14 preceding questions to oth75+7+wil' ers that shall have rational denominators. 76 N7–13_42 718_42 - 718

Ans. V7+13977-73 7-3

4 3-NA 3x-XNX

a-nbaAns.

3+wa

X atwia

Not?

V2

i or

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N

; or

34+15; or

x

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2

Xz-Ni

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;

Х

N5)95, Ans.

,

157 1/3 by 15—13, and it becomes

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(amb) '+b-2ab 10 372 + 35+35

X ab ab 377-75*77* + 735+45 =5X(49+\35+25), Ans. . V3 81–990+100_309-3310 + 9300

X
V9+ 910 92–90 + 102

19
Answer.
4 .W4-W5W4-W5

X 34+25 74-574-75

= 710 - 2012 + (2

- —
-
35—3 95+3

X
N5+ /3^/5-W3W543

15-^3 153 21W 125w75+^ 45—127)

= 125-75+ 45/27, 5—3=2 the answer required. Or thus; multiply the terms of the fractions, 2

,
2X5-2/3

again, +

75-73 multiply the terms of the last fractions by 5+^3, and it be-573: +3851—372

) becomes

Ans.
5—3—2
8

V3+x2–1_8N3+8N2_8
^3+2+1^73+x2–1 5+276-
4x3+4x244
2+76

; again, multiply by --2 + 76, and it becomes -883–812+8+4x18+4N12—416

=4+2/187212 6-4-2 -216-413–412=4+62+4N3—2864493—4N2 =4+ 2x2–2x6, Ans.

Cor. A binomial becomes rational after one operation, a trinomial after two, and a quadrinomial after three, &c.

Cor. The number of operations is equal to the power of 2 in the index.

A and B began to trade with equal .sums of money. In the first year A gained 40 dollars and B lost 40; but in the second A lost one third of what he then had, and B gained a sum less by 40 dollars than twice the sum that A had lost; when it appeared that B had twice as much money as A. What money

did each begin with ?

Let x=the number of dollars each had at first; then x+10= the sum A had after the first year, and 2—40= the sum B had ; also ý.(2+40) the sum A had after the second

year, 40+ (2+40)—40—the sum B had ; .: H.(x +40) =-40+ 3.(2+40) - 40, and 3.(+40)=1-80; 2. 2x+80=3c-240, and x =320.

X

and x

=

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