Ex. 2. Find a multiplier which shall render V3- Væ rational, and determine the product. Ex. 3. Find multipliers which shall render V3- Vāc rational, and determine the product. 238. 3d. When the surd is a trinomial. When a trinomial surd contains only radicals of the second degree, we may reduce it to a binomial surd by multiplying it by the same expression, with the sign of one of the terms changed. Thus, Va+v6+Vc multiplied by Va+V6-Vc gives for a product a+b-c+2Vab, which may be put under the form of m+2 Vab. Ex. 1. Find multipliers that shall make V5+V3–V2 rational, and determine the product. Ex. 2. Find multipliers that shall make 1+12+V3 rational, and determine the product. a To transform a Fraction whose Denominator is a Surd in such a Manner that the Denominator shall be Rational. 239. If we have a radical expression of the form Võ+v% or V-va it may be transformed into an equivalent expression in which the denominator is rational by multiplying both terms of the fraction by Vo=Vc. Hence the a RULE. Multiply both numerator and denominator by a factor which will render the denominator rational. EXAMPLES. Reduce the following fractions to equivalent fractions having a rational denominator: 2 2V3 1. Ans. 3 V5+V2 2. Ans. 3 240. The utility of the preceding transformations will be seen if we attempt to compute the numerical value of a fractional surd. Ex. 1. Let it be required to find the square root of }; that is, to find the value of the fraction V7 V21 Making the denominator rational, we have and the value of the fraction is found to be 0.6546. 715 Ex. 2. Compute the value of the fraction V11+13 Ans. 3.1003. V6 V7+ V3 2V8+375-772 Ans. 0.7025. 9+2 V10 Ex. 5. Compute the value of the fraction 9-2V10 Ans. 5.7278. Square Root of a Binomial Surd. 241. A binomial surd is a binomial, one or both of whose terms are surds, as 2+13 and 15-V2. A quadratic surd is the square root of an imperfect square. If we square the binomial surd 2+V3, we shall obtain 7+4V3. Hence the square root of 7+4V3 is 2+V3; that is, a binomial surd of the form avo may sometimes be a perfect square. 242. The method of extracting the square root of an expression of the form a Vī is founded upon the following principles: 1st. The sum or difference of two quadratic surds can not be equal to a rational quantity. Let va and V denote two surd quantities, and, if possible, suppose VaVo=c, where c denotes a rational quantity. By transposing Vī and squaring both members, we obtain #Vo_b+c2-a 2c The second member of the equation contains only rational quantities, while Vo was supposed to be irrational; that is, we have an irrational quantity equal to a rational one, which is impossible. Hence the sum or difference of two quadratic surds can not be equal to a rational quantity. 243. 2d. In any equation which involves both rational quantities and quadratic surds, the rational parts in the two members are equal, and also the irrational parts. Suppose we have x+Vy=a+V6. m = Then, if æ be not equal to a, suppose it to be equal to a+m; then a+m+Vy=a+VO, 60 that =Vo-Vy; that is, a rational quantity is equal to the difference of two quadratic surds, which, by the last article, is impossible. Therefore x=a, and consequently Vy=V. 244. To find an expression for the square root of a= V. (3.) (4.) Subtracting (4) from (3), we have q– Võ=3–2Vzy+y. (5.) By evolution, va-Vo=V-Vy. (6.) Multiplying (1) by (6), we have va_b=x-Y. (7.) Adding (3) and (7), a+va-b=2x. (8.) Hence, (9.) 2 Subtracting (7) from (3), y= (10.) 2 It is obvious from these equations that x and y will be rational when a--b is a perfect square. If a -6 be not a perfect square, the values of Vic and Vy will be complex surds. Hence, to obtain the square root of a binomial surd, we proceed as follows: Let a represent the rational part, and Võ the radical part, and find the values of x and y in equations (9) and (10). Then, if the binomial is of the form at V6, its square root will be Væ+ Vy. If the binomial is of the form a-V, its square root will be vã– Vý. EXAMPLES. = 1. Required the square root of 4+2V3. Here a=4 and Vā=2V3; or b=12. 4+v16-12 Hence -3, -=1. 2 Hence Vāc+Vy=V3+1, Ans. Verification. The square of V3+1 is 3+2V3+1=4+2V3. 2. Required the square root of 11+672. Here a=11 and Võ=672; or b=72. x=9 and y=2. Va+Vy=3+V2, Ans. 3. Required the square root of 11–2V30. Ans. VÕ-Vā. 4. Required the square root of 2+13. Ans. VI+VI 5. Required the square root of 7+2V10. Ans. V5+V2. 6. Required the square root of 18+8V5. Ans. V10+2V2. 245. This method is applicable even when the binomial contains imaginary quantities. 7. Required the square root of 1+4V 3. Here a=1 and Võ=4V — 3. Hence b=-48 and a - b=49. Therefore x=4 and y=-3. The required square root is therefore 2+ V-3, Ans. 8. Required the square root of — }+V — 3. Ans. }+V-3. 9. Required the square root of 2v1 or 0+2V – 1. Ans. 1+V1. |