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EXAMPLES XXXII. e.

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Find the value of 1. (9,12 7)(9/2+7). 2. (3+5/7)(3-5/7). 3. (578-2-7)(5/3+2/7). 4. (2/11+5.J2) (2/11--52). 5. (Ja+210) (Ja-2/6). 6. (3c-21x)(3c+21x) 7. (Na+ x -- Na)(Va+x+wa). 8. (2p+39–219) (W2p+39+229). 9. (Na+x+Na-x)(Va+x-da-x). 10. (5V

x2 – 3y2 +7a)(51x: – 3y2 7a). 11. 29-(11+3/7).

12. 17+(3/7+2./3). 13. (3/2-1):(31/2+1). 14. (2/3+7/2)--(5/3-4/2). 15. (2r – Vay)+(2-ay-ay). 16. (3+5)(15-2)=(5-5). Ja Ja+13

2/15+8 8./3-6.75 17.

18. Ja-na Nx

5+215 5/3-315 Rationalize the denominator of 25/3–41/2 10/6 - 2017

7 +12 19. 20.

21. 73-512

3./6+227. 9+ 2114 2/3+3./2

x2 22. 23.

24. 5+2/6

2+1x2 - y2 +-W1 - x2

2Na+6+31a 25,

26. Vi+22+11 - 22

2Na+b-da-b 19 +32-3

3+6 27.

28. 19+22+3

573-2/12-32+ 350 29. V10+15+13

30. (V3+V5)(V5+v2) V3+ 10-15

V2+ V3+15

Ja +a+a

Given V2=1.41421, V3=1.73205, V5=2.23607: find to four places of decimals the value of

1 31.

32.
3+ 5

34.

15-2 V3 15-2

4+15 9-45 35. 75+15 15-2 Х

36. (2-3)(7-4/3)=(33-5). V5-1

3+ 15

33. V5+ V3

9; then

274. It is possible to find a factor which will rationa'ize any binomial surd.

CASE I. Suppose the given surd is pa- b.
Let pa=x, yb=y, and let n be the l.c.m. of

р

and X” and yñ are both rational. Now an y" is divisible by x - y for all values of n,

and 2" - * =(x-y)(xn-1+xn-2y+x=8y2 + ...... +yn-1). Thus the rationalizing factor is

xn-1+xn-2y+3+=842 + +yn-?; and the rational product is x" ym.

CASE II. Suppose the given surd is pa+ b.
Let x, y, n have the same meanings as before; then
(1) If n is even, xn — yn is divisible by x+y, and

2" yn =(x+y)(n-1-2-2y + ...... +xyn-2-yn-1). . Thus the rationalizing factor is

29-1-2"-2y+ ·+xyn-2 yn-1; and the rational product is x" — - y". (2) If n is odd, xn + yn is divisible by x+y, and

20* + y* = (x+y)(xn-1-2-y+ ...... - xyn-2+yn-?). Thus the rationalizing factor is

xn-1-an-2y+ - Xyn–2+yn-; and the rational product is an + ym.

Example 1. Find the factor which will rationalize V3+ 35. Let x=34, y=5}; then 26 and yo are both rational, and

26-y=(x+y)(35 -- 4y + xy2 - 4y + xyt-y); thus, substituting for x and y, the required factor is

3% -37.53+33.53–33.53+34.54–55,

31-9.51+31.51-15+34.51-5; and the rational product is 3% – 5%=38 – 52=2.

Example 2. Express (57+98)=(51–96) as an equivalent fraction with a rational denominator.

To rationalize the denominator, which is equal to 54-37, put 51=X, 3=y; then since ** - yt =(x-4) (3 + x2y+ya+ya) the required factor is 51+58.31+57.31+3+;

or

and the rational denominator is 54-3=52—3=22. .. the expression =

(54+3})(51 +51.3:+52.31+3?)

22
5+2.5.31+2.57.37+2.51.37+3+

22
14+57.3:+5.31+51.37

11

PROPERTIES OF QUADRATIC SURDS. 275. The square root of a rational quantity cannot be partly rational and partly a quadratic surd.

If possible let vn=a + vm; then by squaring, n=a2 + m +2am;

n-a?- m .. vm=

2a that is, a surd is equal to a rational quantity; which is impossible.

276. If x+ vy=a + vb, then will x=a and y=b. For if x is not equal to a, let x=a+m; then

a+m+ vy=a + vb; that is,

vb=m+ vy; which is impossible.

[Art. 275.] Therefore

x=d, and consequently,

y=b. If therefore

X+ vy=a+ vb, we must also have X vy=a - Vb.

277. It appears from the preceding article that in any equation of the form X+VY=A+VB..

(1), we may equate the rational parts on each side, and also the irrational parts ; so that the equation (1) is really equivalent to two independent equations, X = A, and Y=B.

a

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278. If Va+b= Vx+ Vy, then will va-vb=vx-vy. For by squaring, we obtain

at vb=x+2Vxy+y;

..a=x+y, vb=2Vry. [Art. 277.] Hence

a-v=x-2xy +y, and

Va-v=vx-vy. 279. To find the square root of a + vb.

Suppose Vat vb=vx+ Vy; then as in the last article, x+y=a.

..(1), 2 vxy=0

(2). :: (x,y)2 = (x+y)2 - 4xy

=a? 6 from (1) and (2).

.. x-y= Va? 6. Combining this with (1) we find a+ Va2-b

Va2_6 and

y=
2

2
a + v(a? b)
.: Va+b=1

-v(a2-b)

+ 2

2 280. From the values just found for x and y, it appears that each of them is itself a compound surd unless a2-6 is a perfect square. Hence the method of Art. 279 for finding the square root of a+ vb is of no practical utility except when aż– b is a perfect square.

Example. Find the square root of 16+255.
Assume V16+2 55= vä+vy.
Then

16+2V55=x+2 Vxy+y;
.. x+y=16..

(1), 2vry=255

(2). .. (x-Y)2=(x+y)2 – 4xy

=162 – 4 x 55 ...by (1) and (2).

=4x9. .: X-Y=+6.

.(3). From (1) and (3) we obtain

x=11, or 5, and y=5, or 11. That is, the required square root is v11+ v5.

a

In the same way we may shew that

116 – 2/55= 11 – 15. NOTE. Since every quantity has two square roots, equal in magnitude but opposite in sign, strictly speaking we should have the square root of 16+21/55= +(11+25),

16-2/55= +(11-15). However it is usually sufficient to take the positive value of the square root, so that in assuming va-Vb=Nx - Ny it is understood that wc is greater than y. With this proviso it will be unnecessary in any numerical example to use the double sign at the stage of work corresponding to equation (3) of the last example.

281. When the binomial whose square root we are seeking consists of two quadratic surds, we proceed as explained in the following example.

Example. Find the square root of /175 1147.
Since 175-/147= /7 (1/25 -21)=7 (5 - 21).

:: 17175 - V147= 17.15-721. And, proceeding as in the last article,

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282. The square root of a binomial surd may often be found by inspection.

Example 1. Find the square root of 11 +2/30.

We have only to find two quantities whose sum is 11, and whose product is 30; thus

11+2/30=6+5+2/6 x 5

=(6+15) . 711 +2/30= 16+15. Example 2. Find the square root of 53 - 12/10.

First write the binomial so that the surd part has a coefficient 2 ; thus

53 - 12/10=53 – 2-/360. We have now to find two quantities whose sum is 53 and whose product is 360; these are 45 and 8;

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