Note. When n is an even number, employ equation (2), and when it is an odd number, equation (3), in order to rationalize Va+Vb. EXAMPLES. (1) Find a multiplier to rationalize 11-37. Employing equation (1), we have a=11, b=7, and n=3; hence required 112+/11.7+7=121+77+49. = multiplier = 7 =4, a rational product. 5+4. Here we have a=5, b=4, n=3, an odd number; hence by equation (3) we have multiplier required, 25-20+16; and, by multiplication, (5+V4)(25—20+16)=5+4=9= a rational number. (3) What multiplier will render the denominator of the fraction a rational quantity? 5 1 V7 - V2 (4) Change into a fraction that shall have a rational denominator. V4-V2 104. To extract the square root of a binomial surd. Before commencing the investigation of the formula for the extraction of the square root of a binomial surd, it will be necessary to premise two or three lemmas. Lemma 1. The square root of a quantity can not be partly rational and partly irrational. For, if va=b+c, then, by squaring, we have a=b2+c+2b √c; therefore, √c= a-b2-c that is, an irrational equal to a rational quantity, which is absurd. Lemma 2. If a±√b=x± √y be an equation consisting of rational and irrational quantities, then a=x, and √b=√y; i. e., the rational and irrational parts of the two members of an equation must be separately equal. For, if a be not equal to x, let a-x=d; then we have ±√y F√b=a-r; but a-x=d; therefore .. a=x, and, taking away these equals, √b=√y. Lemma 3. If √a+√b=x+y, then √a-√b=x-y; where x and y are supposed to be one or both irrational quantities. For, since a+ √b=x2+y2+2xy; and since x and y are both rational, 2ry must be irrational. By Lemma 2, we have Let it now be required to extract the square root of a+ √b. Assume √a+√b=x+y; then √a−√b=x—y ...a+ √b=x2+ y2+2xy a= √b=x2+ y2 — Qxy =2(x2+y2), or a=x2+y2. Again, Va+vbx √a- √b=x2-y, or √a2-b=xa—y3. .. By addition, 2a where c√a-b; and, therefore, a2-b must be a perfect square; and this is the test by which we discover the possibility of the operation proposed.* * When the quantity a2-b is not a square, the values of a and b are no longer rational; but it is clear that the formulas (1) and (2) will still give true results. As, however, these are more complicated than the original expressions themselves, they are rarely employed; yet, when bis imaginary, the result merits attention. In order to examine this case, change b into —b2; a+√b becomes a+b√—1. The remarkable circumstance just alluded to is this, that the square root of a+b√-1 has the same form as this quantity itself. This is shown from the formula (1), for since c=√a2+b2, when b is changed into —b2, the second member becomes +√a2+b2+, 2 The quantity under the first radical is positive, and that under the second negative, since Va+b2 is greater than EXAMPLES. (1) What is the square root of 11+ √72, or 11+6 √2 ? Here a=11; b=72; c=√ a2—b=√121-72=7 √a+c+ √ a = c=3+ √2. ..√11+6√2=1 2 (2) What is the square root of 23—8 √7? Here a=23; b=82x7=448; c=√ a2 —b= √ 529 —448-9 √a+c_ 2 2 (3) What is the square root of 14±6√5? (6) To what is √np+2m2−2m √np+m3 equal? Ans. 3+ √5. Ans. √√T. Ans. 7+3√5. Ans. √np+m2—m. (7) Simplify the expression16+30-1+√16-30-1. Ans. 10. (8) To what is √28+10√3 equal? (9) Vbc+2b √bc-b2+√bc-2b √bc-b2±26 √ab+4c (10) Vab+4c2-d2+2 √4abc2 —abd2= √ ab+ √√ 4c2—d2. (11) What is the square root of (12) What is the square root of 3-4 √ −1? Ans. 5+ √3. Ans. 1-√-1. Ans. 2-√-1. 105. It is manifest, from what has been said above, that algebraic polynomials may be raised to any power merely by applying the rules of multiplication. We can, however, in all cases obtain the desired result without having recourse to this operation, which would frequently prove exceedingly tedious. When a binomial quantity of the form x+a is raised to any power, the successive terms are found in all cases to bear a certain relation to each other. This law, when expressed generally in algebraic language, constitutes what is called the "Binomial Theorem." It was discovered by Sir Isaac Newton, who seems to have arrived at the general principle by examining the results of actual multiplication in a variety of particular cases, a method which we shall here pursue, and give a rigorous demonstration of the proposition in a subsequent article of this treatise. a; representing the quantity under the first radical by a2, and that under the second by-82, the expression takes the form a+ẞV-1; hence Let us form the successive powers of x+a by actual multiplication. x + a x + a x2+ xa + xa+a2 x2+2x a+a2 x + a x3+2x2a+ xa2 +a+ 2x a2+a3 x3+3x2a+ 3x a2+a3. x + a x+3x3a+ 3x2a2 + x a3 + r3a+ 3x2a2+ 3x a3+a1 x+4x3a+6x2a2+ 4x a3+a1. x + a x+4x+a+6x3a2+ 4x3a3+ xa1 x + a x+5xa+10x a2+10x3a3+ 5x2a*+ x a3 x + a x2+6x*a+15x3a2+20x1a3+15x3a*+ 6x2a3+ xas +a+ 6xa2+15x1a3+20x3a*+15x2a3+6xa+a x2+7x®a+21xa2+35x*a3+35x3a*+21x2a3+7xa®+a”. ・ ・ In order that these results may be more clearly exhibited to the eye, we shall arrange them in a table. TABLE OF The powers of x+a. (x+a)x+a (x+a)2x2+2xa +a2 (ra) r'+3x2a+ 3ra2+a3 (x+a)*x+4x3a+ 6x2a2+ 4xa3 +a1 (x+a)3.x5+5x1a+10x3a2+10x2a3+ 5xa* +a (x+a) x+6x3a+15x^a2+20x3a3+15x2a1+ 6xa3 +as (x+a) x2+7x6a+21x3a2+35x^a3+35x3a*+21x2a3+ 7xa® +a (x+a) +8x7a+28xoa2+56x3a3+70x1a*+56x3a3+28x2a®+8xa2+a®. In the above table, the quantities in the left-hand column are called the expressions for a binomial raised to the first, second, third, &c., power; the cor responding quantities in the right-hand column are called the expansions, or developments, of those in the left. 106. The developments of the successive powers of x-a are precisely the same with those of x+a, with this difference, that the signs of the terms are alternately + and -; thus, (x-a)=x-5x1a+10x3a2-10x a3+5xa*—a3, and so for all the others. 107. On considering the above table, we shall perceive that, I. In each case the first term of the expansion is the first term of the binomial raised to the given power, and the last term of the expansion is the second term of the binomial raised to the given power. Thus, in the expansion of (a) the first term is x4, and the last term is a1, and so for all the other expansions. II. The quantity a does not enter into the first term of the expansion, but appears in the second term with the exponent unity. The powers of r decrease by unity, and the powers of a increase by unity in each successive term. Thus, in the expansion of (x+a) we have xo, x3a, x1a3, x3a3, xaa*, τας, αν. III. The coefficient of the first term is unity, and the coefficient of the second term is, in every case, the exponent of the power to which the binomial is to be raised. Thus, the coefficient of the second term of (x+a) is 2, of (x+a) is 6, of (x+a) is 7. IV. The coefficient of any term after the second may be found by multiplying the coefficient of the preceding term by the index of x in that term, and dividing by the number of terms preceding the required term. Thus, in the expansion of (x+a) the coefficient of the second term is 4; this multiplied by 3, the index of x in that term, gives 12, which, when divided by 2, the number of terms preceding the third term, gives 6, the coefficient of the third term. Again, 6, the coefficient of the third term multiplied by 2, the exponent of x in that term, gives 12, which, when divided by 3, the number of terms preceding the fourth term, gives 4, the coefficient of the fourth term. So, also, 35, the coefficient of the fifth term in the expansion of (x+a)7, when multiplied by 3, the index of x in that term, gives 105, which, when divided by 5, the number of terms preceding the sixth, gives 21, the coefficient of that term. By attending to the above observations we can always raise a binomial of the form (x+a) to any required power, without the process of actual multiplication. |