EXAMPLES XXXII. e. Find the value of 1. (9/2-7)(9√2+7). 2. (3+5/7)(3-5/7). 3. (5/8-2/7)(5√8+2√7). 4. (2/11+5/2) (2/11-5/2). 5. (√a+2√b)(Ja−2↓b). 6. (3c-2x) (3c+2,√x) 7. (√a+x-√a)(√a+x+↓a). 8. (√2p+3q-2√q)(√2p+3q+2√q). 9. (√a+x+√a ̄x) (√a+x−√a−x). 10. (5√x2-3y2+7a) (5√√x1 −3y2−7a). 11. 29 (11+3/7). 13. (3/2-1)÷(3√2+1). 15. (2x-√xy)÷(2√xy− y). 12. 17÷(3/7+2√3). 16. (3+√5)(√/5 − 2)÷(5−√√5). 274. It is possible to find a factor which will rationalize any binomial surd. CASE I. Suppose the given surd is ab. Let a=x, b=y, and let n be the L.C.M. of p and 9; then x" and y" are both rational. Now x-y" is divisible by x−y for all values of n, and Let x, y, n have the same meanings as before; then (1) If n is even, x-y" is divisible by x+y, and xn−yn=(x+y) (xn−1— xn−2y+. Thus the rationalizing factor is +xyn—2—yn−1). and the rational product is x2 — y”. (2) If n is odd, x2+y" is divisible by x+y, and x+y=(x+y)(xn−1 — xn−2y+...... - xyn-2+yn-1). Thus the rationalizing factor is xn-1-xn-2y+. and the rational product is x2+yn. —xyn−2+yn−1; Example 1. Find the factor which will rationalize √3+ 3/5. Let x=3, y=5; then 26 and y6 are both rational, and x − yε=(x+y) (x − x1y+x3y2 — x2y3 + xy1 — y5); thus, substituting for x and y, the required factor is or 3-3.5+3.5a—3a. 5a +34, 55, and the rational product is 3-5—38—52—2. Example 2. Express (5*+9)÷(5a—9*) as an equivalent fraction with a rational denominator. To rationalize the denominator, which is equal to 5—3a, put 5x, 3y; then since x-y= (x − y) (x2+x2y+xy2+y3) the required factor is 5+5.3+5.37+37; PROPERTIES OF QUADRATIC SURDS. 275. The square root of a rational quantity cannot be partly rational and partly a quadratic surd. that is, a surd is equal to a rational quantity; which is impossible. 276. If x+ √y=a+√b, then will x=a and y=b. For if x is not equal to a, let x=a+m; then that is, х which is impossible. Therefore and consequently, If therefore we must also have 277. It appears from the preceding article that in any equation of the form 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. 278. If √a+√b=√x+√y, then will √a−√b=√x−√y• For by squaring, we obtain 280. From the values just found for x and y, it appears that each of them is itself a compound surd unless a2-b is a perfect square. Hence the method of Art. 279 for finding the square root of a+√b is of no practical utility except when a2-b is a perfect square. Example. Find the square root of 16+2√√55. Assume Then √16+2√/55=√x+ √¥. .. x+y=16.. 2√xy=2√/55 .(1), .(2). x=11, or 5, and y=5, or 11. That is, the required square root is √11+√5. In the same way we may shew that √16-2/55/11 – √ɔ̃, NOTE. Since every quantity has two square roots, equal in magnitude but opposite in sign, strictly speaking we should have = However it is usually sufficient to take the positive value of the square root, so that in assuming a b√xy it is understood that x 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-√147. Since 175-147=√7 (√/25 −√21)=√7 (5 − √21). ... 175-1477.√5-√21. And, proceeding as in the last article, √5-√21 = √ √ ··√√175-√147= 2/7 (√1⁄2-√i) 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√6x5 .. √/11+2√30= √6+√5. 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; |