Imágenes de páginas
PDF
EPUB

But if the quantities have no common surd, they can only be subtracted by means of the sign

EXAMPLES.

1. To find the difference between ✔320 and 80. First,/320 (64×5)=85;and/80/(16×5)=45. Then, 8/5 4/5 4/5 the difference sought.

=

2. To find the difference between 3/128 and 2/54.

First,/128/(64×2)=4/2; and/54/27×2)=33/2. Then, 4/23/2/2, the difference required.

48.

Ans. 3.

/256 and

32.

Ans 22/4.

and

and ✓.

Ans. √3. Ans.√6.

3. Required the difference of 75 and
4. Required the difference of
5. Required the difference of
6. Find the difference of √

7. Required the difference of 3 and /. Ans./75. 8. Find the difference of 24a2b2 and √✓/54"*.

PROBLEM VI.

Ans. (36-2ab)/6.

To multiply Surd Quantities together.

REDUCE the surds to the same index, if necessary; next multiply the rational quantities together, and the surds together; then annex the one product to the other for the whole product required; which may be reduced to more simple terms if necessary.

EXAMPLES.

1. Required to find the product of 4/12 and 3./2. Here, 4×3×

√12×√/2=12/(12×2)=12/24=12/(4x6)

= 12×2× √6=24/6, the product required.

2. Required to multiply by √}.

Here

[ocr errors]

××//=//=//==/ 18 18, the product required.

[ocr errors]

3. Required the product of 3/2 and 2/8.

Ans. 24.

4. Required the product of 1/4 and 33/12. Ans. 1/6. 5. To find the product of and. Ans. 15. 6. Required the product of 23/14 and 33/4. Ans. 122/7. 7. Required the product of 2a3 and a3. Ans. 2a2. 8. Required the product of (a+b)3 and (a+b)*".

9. Required the product of 2x+b and 2x - √b.

10: Required the product of (a+2/6)1, and (a−2/b)*.

I

11. Required the product of 2x and 3xй.

12. Required the product of 4 and 2y.

PROBLEM VII.

To divide one Surd Quantity by another.

REDUCE the surds to the same index, if necessary; then take the quotient of the rational quantities, and annex it to the quotient of the surds, and it will give the whole quotient required; which may be reduced to more simple terms if requisite.

EXAMPLES.

1. Required to divide 6√96 by 3/8.

Here 63. (96 ÷ 8) = 2/12 = 2 √ (4×3) = 2 × 2√3 4/3, the quotient required.

=

2. Required to divide 123/280 by 33/5.

Here 123 = 4, and 280 ÷ 5 = 56 = = 8 X 723 . 7 ; Therefore 4 × 2 ×3/7 = 83/7, is the quotient required:

3. Let 4/50 be divided by 25.

4. Let 63/100 be divided 33/5.

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small]

Ans. 2/10.

Ans. 23/20.

Ans. 5. Ans./30.

Ans. §ao.

PROBLEM VIII.

To involve or raise Surd Quantities to any Power.

RAISE both the rational part and the surd part. Or multitiply the index of the quantity by the index of the power to which it is to be raised, and to the result annex the power of the rational parts, which will give the power required.

[blocks in formation]

EXAMPLES.

1. Required to find the square of fat.

First, (?)2 = 2, × 3 =, and (a1)2 = a1× 2 = a+ = a.

Therefore, (a)

=

a, is the square required.

2. Required to find the square of a3.

First, }=}, and (a3)2 = at
X i,

= a/a;

Therefore (a)2= a/a is the square required. 3. Required to find the cube of

6 or × 6a.

First, ()'= × × 3 = 4, and (64)3 = 6a

Theref. (6) = × 6√6 = 4

4. Required the square of 22/2.

= 6 √6;

6, the cube required.

5. Required the cube of 3, or 3.
6. Required the 3d power of v3.
7. Required to find the 4th power of √2.

I

8. Required to find the mth power of a".
9. Required to find the square of 2+ 3.

[blocks in formation]

PROBLEM IX.

To evolve or extract the Roots of Surd Quantities*.

EXTRACT both the rational part and the surd part. Or divide the index of the given quantity by the index of the

* The square root of a binomial or residual surd, a + b, or a — b, may be found thus: Take va2 — b2 = c ;

then va+b vato + vete

=

[blocks in formation]

Thus, the square root of 4 + 2√3 = 1 + √3;

and the square root of 6-2√5 = √5 — 1.

But for the cube, or any higher root, no general rule is known.

For more on the subject of Surds, see Bon' ycastle's Algebra, the 8vo. edition, and the Elementary Treatise of Abgebra, by Mr. J. R. Young.

root to be extracted; then to the result annex the root of the rational part, which will give the root required.

EXAMPLES.

1. Required to find the square root of 16/6.

First, ✓16=4, and (61)1 = 61 ÷2 = 61;

theref. (166) = 4.6 = = 44/6, is the sq. root required.

2. Required to find the cube root of ✯ √3.

First, √/'+ = }, and (√3)'

V

[merged small][ocr errors][ocr errors][ocr errors]

= 3' ÷ 3 = 3a ;

1/3, is the cube root required.

3. Required the square root of 63. 4. Required the cube root of a3b. 5. Required the 4th root of 16a2.

Ans. 6/6.

Ans. 1a/b.

Ans. 24/a.

[merged small][ocr errors][merged small][merged small][merged small]

ARITHMETICAL PROPORTION is the relation which two quantities, of the same kind, bear to each other, in respect to their difference.

Four quantities are said to be in Arithmetical Proportion, when the difference between the first and second is equal to the difference between the third and fourth.

Thus, 3, 7, 12, 16, and a, a + b, c, c + b, are arithmetically proportional.

Arithmetical Progression is when a series of quantities either increase or decrease by the same common difference. Thus, 1, 3, 5, 7, 9, 11, &c. and a, a + b, a + 2b, a +36, a+4b, a +5b, &c. are series in arithmetical progression, whose common differences are 2 and b.

pro.

The most useful part of arithmetical proportion and gression has been exhibited in the Arithmetic. The same may be given algebraically, thus:

Let a denote the least term,

z the greatest term,

d the common difference,
n the number of the terms,

and s the sum of the series;

then the principal properties are expressed by these equations, viz.

[blocks in formation]

Moreover, when the first term a is 0 or nothing, the theorems become zd (n - 1)

and s = zn.

EXAMPLES FOR PRACTICE.

1. The first term of an increasing arithmetical series is 1, the common difference 2, and the number of terms 21; required the sum of the series?

First, 1+2 × 20 = 1 + 40 = 41, is the last term.

Then

1 + 41

2

× 20 = 21 × 20 = 420, the sum required.

2. The first term of a decreasing arithmetical series is 199, the common difference 3, and the number of terms 67; required the sum of the series?

First, 199 3.66

Then

quired.

199 +1
2

[blocks in formation]

× 67 = 100 × 67 6700, the sum re

3. To find the sum of 100 terms of the natural numbers 1, 2, 3, 4, 5, 6, &c.

And. 5050.

4. * Required the sum of 99 terms of the odd numbers 1, 3, 5, 7, 9, &c.

Ans. 9801.

*The sum of any number (n) of terms of the arithmetical series of odd numbers 1, 3, 5, 7, 9, &c. is equal to the square (n2) of that number. That is,

If 1, 3, 5, 7, 9, &c. be the numbers, then will

12, 22, 3, 42, 52,

=

be the sums of 1, 2 3, &c. terms,

1 or 12, the sum of 1 term,

Thus, 0+1 =
1+3 4 or 24, the sum of 2 terms,
4+5 = 9 or 32, the sum of 3 terms,
9716 or 43, the sum of 4 terms, &c.

« AnteriorContinuar »