Having found the first term of the root, we must consider the rest of the square, namely, 2ab+b2, to see how we can derive from it the second term of the root. Now this remainder, 2ab+b2, may be put under the form (2a+b)b; whence it appears that we shall find the second term of the root if we divide the remainder by 2a+b. The first part of this divisor, 2a, is double of the first term already determined; the second part, b, is yet unknown, and it is necessary at present to leave its place empty. Nevertheless, we may commence the division, employing only the term 2a; but as soon as the quotient is found, which in the present case is b, we must put it in the vacant place, and thus render the divisor complete. The whole process, therefore, may be represented as follows: a2+2ab+b2 a+b=the root, a2 2ab+b2 2a+b=the divisor. (156.) Hence, to extract the square root of a polynomial, we have the following RULE. 1. Arrange the terms in the order of the powers of some one letter; take the square root of the first term for the first term of the required root, and subtract its square from the given polynomial. 2. Divide the first term of the remainder by double the root already found, and annex the result both to the root and the divisor. QUEST.-Give the rule for extracting the square root of a polynomia. 3. Multiply the divisor thus increased by the last term of the root, and subtract the product from the last remainder. 4. Double the terms of the root already found for a new divisor, and divide the first term of the remainder by the first term of the divisor, and annex the result both to the root and the divisor. 5. Multiply the divisor thus increased by the last term of the root, and subtract the product from the last remainder. Proceed in the same manner to find the additional terms of the root. Ex. 1. Required the square root of the polynomial a'+2a'x+3a2x2+2ax2+x*. The operation may be represented as follows: a'+2a"x+3a2x2+2ax3+x1|a2+ax+x2=the root, a1 2a'x+3a'x' 2a+ax=the first divisor, 2a'x+ a2x2 2a2x2+2ax2+x1|2a2+2ax+x2= the second The terms of the polynomial being arranged in the order of the powers of the letter a, we extract the square root of a* and obtain a', which is the first term of the root. Having subtracted a from the given polynomial, we divide 2ax, the first term of the remainder, by 2a, and obtain +ax, which is the second term of the root, which we annex both to the root and also to the divisor. We then multiply the complete divisor, 2a+ax, by ax, and subtract the product from the last remainder. We now double the terms of the root a+ax, and obtain 2a+2ax for a new divisor. Dividing 2a'x' by 2a', we obtain x2, the third term of the root, which we annex both to the root and to the divisor. Multiplying the complete divisor by x', and subtracting from the last remainder, nothing remains. Therefore a+ax+x' is the required root. For verification, multiply the root by itself, and we shall obtain the original polynomial. Ex. 2. Required the square root of the polynomial a-2a3x+3a2x2 —2ax3+x*. Ans. a-ax+x2. Ex. 3. Required the square root of the polynomial a'-4a3x+6a2x2-4ax2+x'. Ans. a2-2ax+x2. Ex. 4. Required the square root of the polynomial a*+4a3x+4a2x2-12ax-6a2+9. Ans. a2+2ax-3. Ex. 5. Required the square root of the polynomial a*-4a'b+8ab'+4b". Ans, a'-2ab-2b'. Ex. 6. Required the square root of the polynomial 4x+8ax3+4a2x2+16b2x2+16ab2x+16b*. Ans, 2x2+2ax+4b'. Ex. 7. Required the square root of the polynomial 9x-12x2+10x*—10x3+5x2−2x+1. Ans. 3x3-2x2+x-1. Ex. 8. Required the square root of the polynomial a'+2ab+2ac+b2+2bc+c. Ans. a+b+c. Ex. 9. Required the square root of the polynomial 4x-12x+13x2-6x+1. Ans. 2x-3x+1. QUEST.-How may the result obtained be verified? I Ex. 10. Required the square root of the polynomial a°+2a+3a*+4a3+3a2+2a+1. Ans. a+a+a+1. (157.) No binomial can be a perfect square. For the square of a monomial is a monomial; and the square of a binomial consists of three distinct terms, which do not admit of being reduced to a less number. Thus the expression a2+b2 is not a perfect square; it wants the term ±2ab to render it the square of a±b. This remark should be continually borne in mind, as beginners often put the square root of a2+b' equal to a+b. (158.) A trinomial is a perfect square when two of its terms are squares, and the third is the double product of the roots of these squares. Thus the square of a+b is a'+2ab+b2, and the square of a-b is a'-2ab+b2. Therefore the square root of a2±2ab+b2 is a±b. (159.) Hence, to obtain the square root of a trinomial when it is a perfect square, we have the following RULE. Extract the roots of the two terms which are complete squares, and connect them by the sign of the other term. Ex. 11. Required the square root of a2+4ab+4b3. The two terms a' and 46' are complete squares, and QUEST.-Why can not a binomial be a perfect square? When is a trinomial a perfect square? Give the rule for the square root of a trinomial. the third term, 4ab, is twice the product of the roots a and 26; hence a+2b is the root required. Ex. 12. Required the square root of 9a2-24ab+166* Ans. 3a-4b. Ans. Ex. 13. Required the square root of 9a-30a3b +25a2b2. Ex. 14. Required the square root of 4a2+20ab+361. Ans. Ans. Ex. 15. Required the square root of 16a-26a*b* +49a*b*. Ex. 16. Required the square root of 64a-496*. Ans. There are other roots which may be obtained by successive extractions of the square root. Thus, The fourth root is equal to the square root of the square root. The eighth root is equal to the square root of the fourth root, &c. Ex. 17. Required the fourth root of 81a-216a'b +216a'b'-96ab'+166*. Ans. 3a-2b. Ex. 18. Required the fourth root of 16a+b*+x^ +24a2b2+24a2x2 + 6b2x2 +32a b-32a3x+8ab3-8ax3 -46'x-4bx3-48a'bx-24ab'x+24àbx2. Ans. 2a+b-x. Ex. 19. Required the fourth root of 16y+128y'x +384y'x'+512yx*+256x*. Ans. 2y+4x. Ex. 20. Required the fourth root of 16a+160ab +600a b'+1000ab*+6256". Ans. 2a+56' |