2. The even roots of a positive quantity are either positive or negative.

For the even powers of either a positive or a negative quantity are positive.

Thus,

a; that is, a* = ±a.

Va = a or Note. The sign, called the double sign, is when we wish to indicate that it is either + or

prefixed to a quantity

3. The even roots of a negative quantity are impossible.

For no quantity when raised to an even power can produce a negative result. Such roots are called imaginary quantities.

202. From Arts. 200 and 201 we derive the following rule :

Extract the required root of the numerical coefficient, and divide the exponent of each letter by the index of the root.

Give to every even root of a positive quantity the sign ±, and to every odd root of any quantity the sign of the quantity itself.

Note. Any root of a fraction may be found by taking the required root of each of its terms.

EXAMPLES.

1. Find the square root of 9a4b2c6.

By the rule, √9abc = ±3a2bc, Ans.

2. Find the fifth root of 32 a 10x5m.

5 32 a10x5m = 2a2xm, Ans.

Find the values of the following:

SQUARE ROOT OF POLYNOMIALS.

203. Since (a+b)2 = a2+2ab+b2, we know that the square root of a2+2ab+b2 is a + b.

It is required to find a process by which, when the quantity a2+2ab+b2 is given, its square root, a + b, may be determined.

a2+2ab+b2a+b

2a+b2ab + b2

2ab+b2

The square root of the first term is a, which is the first term of the root. Subtracting its square from the given expression, the remainder is 2ab+b2, or (2a+b) b. Dividing the first term of this remainder by 2a, or twice the first

term of the root, we obtain b, the second term. This being added to 2a, gives the complete divisor 2a + b; which, when multiplied by b, and the product, 2ab + b2, subtracted from the remainder, completes the operation.

From the above process we derive the following rule: Arrange the terms according to the powers of some letter. Find the square root of the first term, write it as the first term of the root, and subtract its square from the given expression.

Divide the first term of the remainder by twice the first term of the root, and add the quotient to the root and also to the divisor.

Multiply the complete divisor by the term of the root last obtained, and subtract the product from the remainder.

If other terms remain, proceed as before, doubling the part of the root already found for the next trial-divisor.

EXAMPLES.

204. 1. Find the square root of 9x - 30 a3x2 + 25 αo.

9 x2 - 30 a3x2 + 25 ao | 3x2 - 5a23, Ans.
9x4

The square root of the first term is 3x2, which is the first term of the root. Subtracting 9x4 from the given expression, we have - 30 a3x2 as the first term of the remainder. Dividing this by twice the first term of the root, 6x2, we obtain the second term of the root, - 5 a3, which, added to 6x2, completes the divisor 6x2 - 5a3. Multiplying this divisor by - 5a3, and subtracting the product from the remainder, there is no remainder. Hence, 3x2 - 5 a3 is the required square root.

2. Find the square root of

12x5-14x2 + 1

8x+9x+4x.

Arranging according to the descending powers of x,

9x + 12x - 8x4 – 14x2 + 4x+1|3x2+2x2-2x-1, 926

Ans.

It will be observed that each trial-divisor is equal to the preceding complete divisor, with its last term doubled.

Note. Since every square root has the double sign (Art. 201), the result may be written in a different form by changing the sign of each term. Thus, in Example 2, another form of the answer is

-3x2-2x2 + 2x + 1.

Find the square roots of the following:

3. a2-4a3+6a2-4a+1.

9.4a2+6464 – 20 a3b – 80 ab3 + 57 a2b2.

12. x2+4y2 + 9 z2 - 4 xy + 6 xz – 12 yz.

13. 9x + 30x + 25 x2 - 42x2 - 70x2 +49.

14. 16c-40c2 - 24 c32 + 25 c2+30c+9.

15.9+a+ 30 a − 4a + 13a2 + 14a – 14a3.

16. 4x – 4y - 3 xy2 - 6 xy2 + 5 x2y + 4 xy2 + 4y6.

17. 25 x2-44x2 - 40x + 4x + 25 +46 x2 - 12x.

19. 9x-12xy+10x+y2-16 x2y3 +9 x2y2-4 ху+4у.

Find to four terms the approximate square roots of the

following:

20.1+x.

21.1-2α.

22. a2 - 4 ab + b2.

23. 4x2+2y.

SQUARE ROOT OF NUMBERS.

205. The method of Art. 204 may be used to extract the • square roots of arithmetical numbers.

The square root of 100 is 10; of 10,000 is 100; etc. Hence, the square root of a number less than 100 is less than 10; the square root of a number between 10,000 and 100 is between 100 and 10; and so on.

That is, the integral part of the square root of a number of one or two figures, contains one figure; of a number of three or four figures, contains two figures; and so on. Hence,

If a point be placed over every second figure in any integral number, beginning with the units' place, the number of points shows the number of figures in the integral part of its square

root.

206. Let it be required to find the square root of 4624.

4624

a2 = 3600

60+8 = a+b

120+8 1024

Pointing the number according to the rule of Art. 205, we see that there are two figures in the integral part of the square root.

= 2a + b | 1024

Let a denote the value of the number in the tens' place in the root, and b the number in the units' place. Then a must be the greatest multiple of 10 whose square is less than 4624; this we find to be 60. Subtracting a2, that is, the square of 60 or 3600, from the given number, the remainder is 1024. Dividing the remainder by 2a or 120, we have 8 as the value of b. Adding this to 120, multiplying the result by 8, and subtracting the product, 1024, there is no remainder. Hence, 60+8 or 68 is the required square root.

The ciphers being omitted for the sake of brevity, the work will stand as follows:

4624 | 68

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