Ans.

2d3P

3

80. If the proposed polynomial contain several terms affected with the same power of the principal letter, we must arrange the polynomial in the manner explained in division (Art. 20.); and in applying the above process we shall be obliged to perform several partial extractions of the square roots of the coeffi cients of the different powers of the principal letter, before we can arrive at the root required. Such examples however very rarely occur.

Before quitting this subject we may make the following remarks :—

I. No binomial can be a perfect square; for the square of a monomial is a monomial, and the square of the most simple polynomial, that is, a binomial, consists of three distinct terms, which do not admit of being reduced with each other. Thus, such an expression as a + b2 is not a square; it wants the term + 2 a b to render it the square of (a + b).

II. In order that a trinomial, when arranged according to the powers of some one letter, may be a perfect square, the two extreme terms must be perfect squares, and the middle term must be equal to twice the product of the square roots of the extreme terms. When these conditions are fulfilled, we may obtain the square root of a trinomial immediately, by the following

RULE.

Extract the square roots of the extreme terms, and connect the two terms thus found by the sign+, when the second term of the trinomial is positive, and by the sign —, when the second term of the trinomial is negative. Thus the expression

is a perfect square; for the two extreme terms are perfect squares, and the middle term is twice the product of the square roots of the extreme terms, hence the square root of the trinomial is

Or,

√9a64ab b1

2

Заз 8ab2

2

An expression such as 4 a2 + 12 a 6-9 b2 cannot be a perfect square, although 4 a 2 and 9 b2, considered independently of their signs, are perfect squares, and 12 a b = 2 a X 6 b, for 9b2 is not a square, since no quantity, when multiplied by itself, can have the sign

III. In performing the operations required by the general rule, if we find that the first term of one of the remainders is not exactly divisible by twice the first term of the root, we may immediately conclude that the polynomial is not a perfect square.

IV. We may apply to the square roots of polynomials which are not perfect squares, the simplifications already employed in the case of monomials (art. 51). Thus, in the expression,

The quantity under the radical is not a perfect square, but it may be put under the form

Vab (a 2 +ab + 4 b2)

The factor within brackets is manifestly the square of a + 2 b, hence

2

√ a3 b + 4 a 3 b2 + 4 a b 3 = √ a b ( a 2 + 4 a b + 4 b 2)

= √ a b (a + 2 b ) *

= (a + 2 b) vão

81. Let us next proceed to form the Cube of x + a.

(x +α)3 = (x + a) × (x + a) × (x + a)

3 = x2 + 3 x2 a + 3 x a2 + a3 by rules of multiplication.

Let it be required to form the cube of a trinomial (x +a+b); represent the two last terms a + b by the single letter

then

= x2+3x2 (a + b) + 3 x (a + b)2 + (a + b) 3

3

2

=x3 + 3 x a + 3x2b+ 3 x a2 + 6x a b + 3 x b2 + a 3

This expression is composed of the sum of the cubes of all the terms, together with three times the sum of the squares of each term, multiplied by the simple Dower of each of the others in succession, together with six times the product of the simple power of all the terms.

By following a process of reasoning analogous to that employed in (Art. 78), we can prove that the above law of formation will hold good for any polynomial of whatever number of terms. We shall thus find

(a+b+c+d)3 =a3+b3+c3+d3+3a2 b+3a2 c+3a2 d+3b2 a+3b2 c+362 d +3c a+3c b+3c d+3d a+3d' b+3d c+6abcd (2a3 —4ab+3b 2)3 = 8 a o — 64 a 3 b 3 + 27 b 6 — 48 a 3 b +36 a1 b2 + 96 a 1 b 2 + 144 a2 b1 + 54 a2 b1 108 a b3

1443 b3

=8a
· 48 a3 b+132 a1 b2 —208 a3 b 3+198 a2 b1 ~~108 a b
+27 66

5

in a similar manner, we can obtain the 4th, 5th, &c. powers of any polynomial.

82. We shall now explain the process by which we can extract the cube root of any polynomial, a method analogous to that employed for the square root, and which may easily be generalized, so as to be applicable to the extraction of roots of any degree.

Let P be the given polynomial, R its cube root. Let these two polynomials be arranged according to the powers of some one letter, a for example. It fol lows, from the law of formation of the cube of a polynomial, that the cube of R contains two terms, which are not susceptible of reduction with any others; these are, the cube of the first term, and three times the square of the first term multiplied by the second term; for it is manifest that these two terms will involve a affected with an exponent higher than any that is to be found in the succeed. ing terms. Consequently these two terms must form the two first terms of P Hence, if we extract the cube root of the first term of P, we shall obtain the first term of R, and then, dividing the second term of P by three times the square of the first term of R thus found, the quotient will be the second term of R. Having thus determined the two first terms of R, cube this binomial, and subtract it from P. The remainder P' will contain three times the product of the square of the first term of R by the third, together with a series of terms involving a, affected with a less exponent than that with which it is affected in this product, which consequently forms the first term of P'. Dividing the first term of P' by three times the square of the first term of R, the quotient will be the third term of R. Forming the cube of the trinomial root thus determined, and subtracting this cube from the original polynomial P, we obtain a new polynomial P", which we may treat in the same manner as P', and continue the op ration till the whole root is determined.

EXTRACTION OF THE SQUARE ROOT OF NUMBERS.

83. We have already given rules in our Arithmetic, by which we are enabled to extract the Square and Cube Roots of any proposed number; we shall now proceed to explain the principles upon which these rules are founded.

The numbers

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100, 1000,

when squared become

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 10000, 1000000,

and reciprocally, the numbers in the first line are the square roots of the num bers in the second.

Upon inspecting these two lines we perceive, that among numbers expressed by one or two figures, there are only nine which are the squares of other whole numbers; consequently, the square root of all the rest must be a whole number, plus a fraction.

Thus the square root of 53, which lies between 49 and 64, is 7 plus a frac tion. So also, the square root of 91 is 9 plus a fraction.

84. It is however very remarkable, that the square root of a whole number, which is not a perfect square, cannot be expressed by an exact fraction, and i therefore incommensurable with unity.

To prove this, let

a

a fraction in its lowest terms, be, if possible, the

a

square root of some whole number N, then the square of or b

must be

f29 equal to N. But since a and b are, by supposition, prime to each other, a2 and a2

b2 are also prime to each other; therefore, is an irreducible fraction, b and cannot be equal to a whole number.

85. The difference between the squares of two consecutive whole numbers is greater in proportion as the numbers themselves are greater; the expression for this difference can easily be found.

Let a and a +1 be two consecutive whole numbers;
Then

chat is to say, the difference of the squares of two consecutive whole numbers, is equal to twice the less of the two numbers, plus unity.

Thus the difference between the squares of 348 and 347 is equal to 2 X 347 +1, or 695.

Let us now proceed to investigate a process for the extraction of the square root of any number, beginning with whole numbers.

Extraction of the square root of whole numbers.

86. If the number proposed consist of one or two figures only, its root may be found immediately, by inspecting the squares of the nine first numbers in (A.t. 83.). Thus, the square root of 25 is 5, the square root of 42 is 6 plus a fraction, or 6 is the approximate square root of 42, and is within one unit of the true value; for 42 lies between 36, which is the square of 6, and 49, which is the square of 7.

Let us consider, then, a number composed of more than two figures, 6984 for example.

Since this number is comprised between 100, which is the square of 10, and 10000, which is the square of 100, its root must necessarily consist of two figures, that is to say, of tens and units. Designating the tens in the root sought by a, and the units by b, we have

60'84

78

49

148 118'4

118/4
0.

which shows, that the square of a number consisting of tens and units, is composed of the square of the tens, plus twice the product of the tens by the units, plus the square of the units.

This being premised, since the square of a certain number of tens can contain nothing lower than hundreds, it follows that the squares of the tens contained in the root must be found in the part 60 (or 60 hundreds), to the left of the two last figures of 6084 (which written at full length is 6000+80+ 4); we therefore separate the two last figures from the others by a point. The part 60 is comprised between the two squares 49 and 64, the roots of which are 7 and 8, hence 7 is the figure which expresses the number of tens in the root