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277. The process of clearing a quantity of radical signs by multiplication, is called Rationalization, and we will now consider how we may find the factor which will rationalize a surd quantity, in some of the more important cases.

278. To find a factor which will rationalize any monomial.

It is evident that the factor in this case will be the monomial itself, with an index equal to the difference between unity and the given index, For, we have in general terms,

exac

a.

a а" Xa 1. Rationalize ya.

The factor is a; for, VaXy'a = a, a rational quantity.

2. Rationalize zt.
The factor is zó; for, zbXxx = x* = x.

279. To find a factor which will rationalize a binomial in the form of a="/b, or m/a/b, m and n being each some power of 2.

The product of the süm and difference of two quantities is equal to the difference of their squares. Hence, if we multiply the given binomial in this case, by the same terms connected by the opposite sign, the indices of the product must be respectively the halves of the given indices; and a repetition of the process will ultimately rationalize the quantity, provided m and n are any powers of 2.

1. Rationalize aty c.
The factor is a-vc; for we have

(a+vc) (a–Vc)=a-c. 2. Rationalize a-Vx.

(va-Vx) (Va+vx)= a-vx;

(a-x) (aty x)= a-X, a rational result; and the complete multiplier is

(Va+vx) (a+y).

A trinomial may be treated in a similar manner, when it contains only radicals of the 2d degree. 3. Rationalize y5+12–13.

V5+12-13
75+12 +13
5+y10-15
V10+ 2-16

V15+1 643
4+27/10
4-2710

16-40=-24 a rational result; and the complete multiplier is

(15+12+13) (4—210). Thus we perceive that it is necessary simply to change the sign of one of the terms of the trinomial, and multiply by the result, repeating the process with the product.

280. To find a factor which will rationalize any binomial surd whatever. Let the binomial be represented by the general form,

a to.

1

Put x = a'and y=b'; and let n be the least common multiple of r and s. Then 2* + y" will be rational. But x+y will exactly divide x*+y" when n is odd, and x*—y" when n is even ; and xwill exactly divide an-y" whether n is odd or even; (89). These quotients will therefore be the factors that will rationalize the respective divisors. Hence, let a represent the required factor; then (1)

- for a+++, when n is oda; x+y

x" —y” (2) 9= y for a

to', when n is even ; x+y

9=

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1

1

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for a"-6', n being odd or even.

9 =

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1. Rationalize the binomial ał +.
Since n =

6, an even number, we have from (2),
a'-6

ak-abitabdi-ab+a113_08,
3
x+y
the factor required; and

(aštok) x (as_a'titalo-abtali_0%) = a*—0, the rational result.

The foregoing methods may be applied in the solution of the following

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V 10-V 15 2. Reduce

to a fraction whose denominator shall be rational.

2v15_3V 10 Ans.

6

V6

3. Reduce to a fraction whose denominator shall be rational.

Ba

Ans. Va

12 4. Reduce to a fraction whose denominator shall be rational.

09

V72

Ans.

3 5 5. Reduce

to a fraction whose denominator shall be

V7+13 rational.

Ans. 5(17—V3)

4 6. Reduce

va

to a fraction whose denominator shall be

Va-vc rational.

Ans. a+Vac

7. Reduce /11+V5 to a rational denominator.
V11-15

Ans. 8+1 55

3
8
8. Reduce

11+13

to its simplest form. Ans. Vīl-//3. V10+V6 9. Reduce V10_16

to its simplest form. Ans. 4+1 15.

Ans. 1-1_15.

15_V3 10. Reduce

to its simplest form. 75+1-3

4

11. Simplify
(3+13) (3+15) (V5—2)

Ans. $V15. (5—15) (V3+1)

1+a+11a 12. Simplify

Ans.

1+V1-a? 1+a-V1-a 13. Find the factor which will rationalize 5-V2.

Ans. 575+V 10+5V2+V8.

a

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281. A Radical Equation is one in which the unknown quan. tity is affected by the radical sign.

282. In order to solve a radical equation, it is necessary in the first place to rationalize the terms containing the unknown quantity. In case of fractional terms, this may be effected in part by methods already explained. But the process is commonly one of involution.

The following are examples of simple equations containing radical quantities.

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Væ+11+1+1=5; by transposition,

V +11=5-12-4; squaring and reducing, +11=x+21–10V 247; transposing and reducing, V 2-4 =1; squaring and reducing,

= 5. V x-V-5

42-35 2. Given

to find x. Vx+1-5 5

X

OPERATION.

V/x-Vă=

4.x-35

5 Vx+1

VxMultiplying both terms of

2x–5–21

2–5x 4.x-35 the fraction by the nume

; 5

5 rator, (279), clearing of fractions, etc.,

2V x*_5x=23-30; dividing by 2,

V Z_5x = 3-15; squaring and reducing,

25x = 225; whence,

X = 9.

mya 3. Given +

to find x.

V x-ya The least common multiple of the denominators is x—a= (Vx+ya) (1/x-Va); and the solution will be as in the fol. lowing

OPERATION.

m

Vatr

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