Let it now be required to extract the square root of a+√b. .. By addition 2a =2(x+y), or a = x2+y2. x2_y2 = √ a2 —b=c, suppose. Therefore, by addition and subtraction we have where c= √/a2-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 EXAMPLES. (1.) What is the square root of 11+√72, or 11+6√2? (2.) What is the square root of 23—8√√/7? Here a=23; b=82x7=448; c=√/a2-b= √√/529-448=9 ..√/23-8√7 = √atc √ a = c = = 4-√7. (8.) What is √28+10√3 equal to? (9.) √bc+2bbc-b2+√bc-2b√bc—b2=+2b. Ans. 3+√5. Ans. √7+√11 Ans. 7+3/5. Ans. √np+m2-m. (7.) Simplify the expression 16+30-1+16-30/1. Ans. 10. Ans. 5+√3. (10.) √ab+4c-d2+2√/4abc—abd2=√ab+√4c2—d3. (11.) What is the square root of -2√=1? (12.) What is the square root of 3-41/=1? Ans. 1-1. Ans. 2-√-T. (13.) What is the square root of 112+20✅✅12. ? Ans. (1+√2). (5+√3) BINOMIAL THEOREM. 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. Let us form the successive powers of x + a by actual multiplication. 6 4 x+5x3u+10 x1 a2+10x3 a3 + 5 x2 a + + xa 3 3 3 3 3 4 6 +a+, 5 x1 a2+10x3 a3+10x2a +5 x a3 + a® x+6x3a + 15 x1 a2 + 20 x x + a .........2d power 3d power. ...4th power. .5th power. a3 + 15 x 2 a1 + 6 x a3 + a ® .........6th power. x2+6x6a+ 15 x3 a2+20 x* a3 + 15 x3aa + 6 x2Û3 + xa° + x6a+ 6 x3 a2 + 15 x1 a3 + 20 x 3 a* + 15 x2 a3 +6 xa® + a' x2+7x®a+21 x3 a2 +35 x 1 a3 +35 x3 a * +21 x 2 a3 +7 xao +a2 7th power. 4 4 In order that these results may be more clearly exhibited to the eye, we snail arrange them in a table. In the above table, the quantities in the let hand suums are sulest in pressions for a binomial raised to the first, second, surd BE DIET. IN SUPPL sponding quantities in the right hand couana are cunt de EJUSDENTS JUM velopements of the others. 106. The developements of the successive power of 2 — 1 20 per same with those of z + a, with this dierence aux the agus sé le verus me alternately and -; thus, (z — a ) 3 = z 3 — 5 z + e + 10 z 1 e 2 — i z1 and so for all the others. 107. On considering the above talle we stal perene aut 1. In each case the first term of the expatmne a de fans vra of the stomak mased to the given power, and the last tern sé de er passa a te verant vers of the binomial raised to the given power. Thun an de ezpatan of 2 the first term is z and the last term is a * aut in fur all wie erste IL The quantity a does not enter ano de fra vra pears in the second term with the export m.cy. by unity, and the powers of a arzeme by using a cara muren ve beya In the expansion of (z + a) * we have, z', z ' in ay III. The quefficient of the first term is unty and an coille aus, ut za mene d term is in every case the exponent of the proto via de kanal do as to beg raised. Thus the coefficient of the second term of (x + a)2 is 2, of (x + a) ' is 6, of (x + a)' is 7. IV. If the coefficient in any term be multiplied by the index of z in that term and divided by the number of terms up to the given place, the resulting quotient will be the coefficient of the succeeding term. Thus in the expansion of (x+a)* the coefficient of the second term is 4; this multiplied by 3, the index of z in that term, gives 12, which when divided by 2 the number of terms up to the given place 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 up to the given place, gives 4, the coefficient of the 4th term. So also 35, the coefficient of the 5th term in the expansion of (x+a), when multiplied by 3, the index of x in that term, gives 105, which, when divided by 5, the number of terms up to the given place, gives 21, the coefficient of the succeeding term. By attending to the above observations, we can always raise a binomial of the form (a) to any required power, without the process of actual multiplica tion, 9 6 (x + a) ' = x ' + 9 x 3 a + 36 x 1 a2 + 84 x ® a 3 +126 x 3 a* + 126 x'a'+ (x — a) 10 - 252 x3 a3 + 210 x 1 ao — 120 x 2 a 2 + 45 x 2 a 3 — 10 x a 2 + « 1o. 108. The labour of determining the coefficients may be much abridged by attending to the following additional considerations · 4 V. The number of terms in the expanded binomial is always greater by unity than the index of the binomial. Thus the number of terms in (x + a) 1 is 4+ 1, or 5, in (z + a)1o is 10 + 1, or 11. VI. Hence, when the exponent is an even number, the number of terms in the expansion will be odd, and it will be observed, on examining the examples already given, that after we pass the middle term the coefficients are repeated in a reverse order; thus, VII. When the exponent is an odd number, the number of terms in the expansion will be even, and there will be two middle terms, or two contiguous terms, each of which is equally distant from the corresponding extremities of the series; in this case the coefficient of the two middle terms is the same, and then the coefficients of the preceding terms are reproduced in a reverse order; thus, 109. If the terms of the given binomial be affected with coefficients or exponents, they must be raised to the required powers, according to the principles already established for the involution of monomials; thus: (a2+}a! j' = (a2)' + a3) × (3ab)+36 a3) × (3ab) +84 (a3) × (3ab)• +126(a) X (3ab ) + 126(a')' × (3ab) +84(a)3 ×(3ah)® +36 (a3 ja × (3ab)' + 9a3 × (3ab) + (3ab)3 = a+27ab+324a3ba + 2268 aab3 +10206a1b* + 30618ab +61236 a13b + 78732 a11b2 + 59049 a11 b3 + 19683 a3b3 110. We shall now proceed to exhibit the binomial theorem in a general form let a be required to raise any binomial (r+a) to the power represented by |