PROCESS. (Correct to two deci2x20=40 150 mal places) 975 NOTE. In the process above we have first written 550 in the form 550.0000. If, instead of this, we had written it with six ciphers, that is, 550.000000, and then carried the process forward until all these were used below, we should have had the root correct to three places of decimals instead of two. In general, half the number of ciphers added in this way at the beginning is the number of decimal places to which the root obtained is correct. The square roots of decimal numbers, like 334.796, are obtained like those for whole numbers except that in the beginning the separation of the number into periods of two figures each must be carried out both ways from the decimal point. Thus, 334.796 would first be written as 3'34.'79'60. Similarly, 3.67893 would be written 3. '67'89'30. The extraction of the root is carried out from this point on just as shown in the process above. EXERCISES Find (by inspection or by the process shown in § 135) the square root of each of the following numbers. 1. 49. 3. 64. 6. 625. 7. 3844. 2. 81. 4. 225. 6. 529. 8. 2209. Find the square root correct to two decimal places, for each of the following numbers : 16. 382. 20. 11. 23. 17. 671. (Hint. Write as 11.'00'00.] [Hint. Write as .75.) 18. 1211. 21. 7. 24. 11. 19. 12.96 22. 26. 25. .08 136. Extracting Square Roots in Algebra. In extracting square roots in algebra we distinguish between several cases, according to the number of terms in the given expression. Monomials. The square root of a monomial can usually be seen by inspection. Thus, V36 m4n2=6 man, because 6 mn X6 m?n=36 m^n. Similarly, V49 xøy4z8 = 7 x3y2z4. Trinomials. If a trinomial is a perfect square, its square root can be obtained by what has been said in $ $ 63 and 64. Thus, suppose we wish to find the square root of 9x2+12 xy+4 y2. This trinomial is a perfect square, because its terms 9 z2 and 4 ya are squares and positive, while its remaining term, 12 xy, is equal to 2x3 xX2 y (see § 63). Being a perfect square, it can be expressed as the product of the two equal binomial factors (3 x+2 y)(3 x+2 y), as explained in § 64. Whence, the desired square root of 9 x2+12 xy+4 y2 is 3 x+2 y. Ans. Similarly, V4 52–4 8+1=2 s-1, because 452–4 s+1=(2 s-1)(2 s–1). Polynomials. To find the square root of a polynomial (other than a trinomial) we use a process much like that followed in obtaining square roots in arithmetic. (See §. 135.) This process is illustrated below. EXAMPLE. Find the square root of 4 x4 +12 x3 – 3 x2 – 18 x+9. 4 x4 PROCESS. 4 x4 +12 x3 – 3x2 – 18x+9 | 2x2 + 3x – 3 Ans. Trial divisor, 4x2 12 x3 – 3 x2 Complete divisor, 4 x2 +3 x 12 x3 + 9x2 Trial divisor, 4x2+62 · 1-12 x2 – 18 x+9 Complete divisor, 4 x2+6 x–3 –12 x2 – 18 x+9 EXPLANATION. We first arrange the terms of the polynomial in the descending (or ascending) powers of some letter. In the example, the arrangement is in descending powers of x. Extract the square root of the first term, write the result as the first term of the root (giving the 2 x2 in the answer), and subtract its square from the given polynomial (giving the 12 x3–3 x? in the second line of the process). Divide the first term of the remainder by twice the root already found, used as a trial divisor, and the quotient, 3 x, is the next term in the root. Write this term in the root, and annex it to the trial divisor to form the complete divisor (giving the 4 x2+3 x). Multiply the complete divisor by this term of the root, and subtract the product from the first remainder (giving – 12 x— 18 x+9). Find the next term of the root by dividing the first term of the remainder by the first term of the new trial divisor. This gives the -3 of the answer. Form the complete divisor and continue in this manner until a remainder of zero is obtained. . EXERCISES Find (by inspection or by the process shown in § 136) the square root of each of the following expressions. Square your answer in each case, and see if it gives (as it should) the given expression. 1. 9 cog. 5. 225 x@y2. 9. 529 98912 6. 625 m4nøq?. 10. a2x2m. 18. 9 m2 – 6 mx+x2. 19. x2+xy+1 y. 15. 4 a2+12 ab+9 62. 20. 4 x4 – 52 x2+169. 16. 4 x2+4 xy+ya. 21. (a+b)2 – 4(a+b)+4. 17. C2 — 4 ac+4 a?. 22. x4+2 x3+3 x2+2 x+1. 23. 9 x4 – 12 x3 +10 x2 – 4 x+1. 24. x6 — 2 x5 +3 x4 – 4 x3+3 x2 – 2 x+1. 25. x4 – 6 x3y+13 x y2 — 12 xy:+4 y4. [Hint. This Example differs slightly from the one worked in § 136 because we here have two letters, x and y, in the polynomial. The process, however, remains as before.] 26. 28+2 a®x2 — a4x4 — 2 a2x6 +a8. (HINT. First arrange in descending powers of x. That is, 28-2 a2x6 – aʻx4 ...] 27.9 x2 – 6 xy + y2 +12 x2—4 yz+4 22. NOTE. We are here extracting the root of an expression, 1+x, which is not a perfect square. As with arithmetic numbers that are not perfect squares, the process leads to an answer which is correct up to the point where it is stopped and which affords an approximation that approaches more closely the exact result the farther the process is continued. Compare $ 135. 31. Find (as in Ex. 30) the first four terms of the square root of the following expressions : (a) 1-x. (b) x2 +1. (c) x2 – 1. For further exercises on this topic, see Appendix, p. 308. 137. The Double Sign of the Square Root. We know that 3 is the square root of 9 because 3X3=9. But we also have (-3) X(-3)=9. Therefore, – 3 can also be regarded as a square root of 9. In other words, 9 has two square roots, +3 and -3, which are opposite in sign, but otherwise the same. Similarly, 16 has the two square roots, +4 and – 4. In general, aż has the two square roots a and — a. The double sign † is sometimes used. Thus, we say that the square root of 9 is +3. This is merely a way of saying briefly that the two roots are +3 and -3. In order to avoid all confusion, it is to be understood that the radical sign ✓ when placed over a number means the positive square root of that number. If it is desired to |