After having arranged the polynomial with reference to a, extract the square root of 25a, this gives 5a2, which is placed to the right of the polynomial; then divide the second term, -30a3b, by the double of 5a2, or 10a2; the quotient is -3ab, and is placed to the right of 5a2. Hence, the first two terms of the root are 5a2 — 3ab. Squaring this binomial, it becomes 25a1-30ab+9a2b2, which, subtracted from the proposed polynomial, gives a remainder, of which the first term is 40a2b2. Dividing this first term by 10a2, (the double of 5a2), the quotient is +462; this is the third term of the root, and is written on the right of the first two terms. Forming the double product of 5a2-3ab by 462, and the square of 462, we find the polynomial 40a2b2-24ab3+16b4, which, subtracted from the first remainder, gives 0. Therefore 5a2—3ab+4b2 is the required root.

25a1b2 — 40a3b3c+76a2b2c2—48ab2c3+36b2ca—30a2bc+24a3bc2

129. We will conclude this subject with the following remarks.

1st. A binomial can never be a perfect square, since we know that the square of the most simple polynomial, viz. a binomial, con

tains three distinct parts, which cannot experience any reduction amongst themselves. Thus, the expression a2-+b2 is not a perfect square; it wants the term +2ab in order that it should be the square

of a±b.

2d. In order that a trinomial, when arranged, may be a perfect square, its two extreme terms must be squares, and the middle term must be the double product of the square roots of the two others, Therefore, to obtain the square root of a trinomial when it is a perfect square; Extract the roots of the two extreme terms, and give these roots the same or contrary signs, according as the middle term is positive or negative. To verify it, see if the double product of the two roots gives the middle term of the trinomial. Thus,

since

9a-48ab2+64ab is a perfect square,

√9a3a3, and √64a2b1-8ab2,

and also 2X3a3x-8ab2-48ab2, the middle term.

But 4a2+14ab+962 is not a perfect square: for although 4a3 and +962 are the squares of 2a and 3b, yet 2×2a×3b is not equal to 14ab.

3d. In the series of operations required in a general problem, when the first term of one of the remainders is not exactly divisible by twice the first term of the root, we may conclude that the proposed polynomial is not a perfect square. This is an evident consequence of the course of reasoning, by which we have arrived at the general rule for extracting the square root.

4th. When the polynomial is not a perfect square, it may be simplified (See Art. 125.).

Take, for example, the expression

√a3b+4a2b2+4ab3.

The quantity under the radical is not a perfect square; but it can be put under the form ab(a2+4ab+462). Now, the factor between the parenthesis is evidently the square of a+2b, whence we may conclude that,

√a3b+4a2b2+4ab3=(a+2b) √ab.

Of the Calculus of Radicals of the Second Degree.

130. A radical quantity is the indicated root of an imperfect power.

The extraction of the square root gives rise to such expressions as √a, 3 √b, 7 √ 2, which are called irrational quantities, or radicals of the second degree. We will now establish rules for performing the four fundamental operations on these expressions.

131. Two radicals of the second degree are similar, when the quantities under the radical sign are the same in both. Thus, 3√ and 5c √ are similar radicals; and so also are 9 2 and 7√2.

Addition and Subtraction.

132. In order to add or subtract similar radicals, add or subtract their co-efficients, then prefix the sum or difference to the common radical.

Thus, And

In like manner,

And

3a √b +5c √b =(3a+5c) √b.

3a vb-5c √b=(3a-5c) √b.

7 √2a+3 √2a=(7+3) √2a=10 √2a.

7 √2a-3 √2a=(7—3) √2a= 4√2a.

Two radicals, which do not appear to be similar at first sight, may become so by simplification (Art. 125).

For example,

√48ab2+b√75a=4b √3a+5b √3a=9b √3a,

and 245-3 √5-6 √5-3 √5-31/5.

When the radicals are not similar, the addition or subtraction can only be indicated. Thus, in order to add 3 √b to 5 √a, we write 5 √a+3 √b.

Multiplication.

133. To multiply one radical by another, multiply the two quan tities under the radical sign together, and place the common radical over the product.

Thus, Vax √b Vab; this is the principle of Art. 125, taken in an inverse order.

When there are co-efficients, we first multiply them together, and write the product before the radical. Thus,

2a Va2+b2 X-3a √ a2+b2 =—6a2(a2+b2).

Division.

134. To divide one radical by another, divide one of the quantities under the radical sign by the other and place the common radical over the quotient.

a

sions are equal to the same quantity ; hence the expressions

b

themselves must be equal. When there are co-efficients, write their quotient as a co-efficient of the radical.

135. There are two transformations of frequent use in finding the

numerical values of radicals.

The first consists in passing the co-efficient of a radical under the sign. Take, for example, the expression 3a √5b; it is equivalent to 9a X √5b, or V9a2.5b = √ 45a2b, by applying the rule for the multiplication of two radicals; therefore, to pass the co-efficient of a radical under the sign, it is only necessary to square it.

The principal use of this transformation, is to find a number which shall differ from the proposed radical, by a quantity less than unty. Take, for example, the expression 6 √13; as 13 is not a perfect square, we can only obtain an approximate value for its root. This root is equal to 3, plus a certain fraction; this being multiplied by 6, gives 18, plus the product of the fraction by 6; and the entire part of this result, obtained in this way, cannot be greater than 18. The only method of obtaining the entire part exactly, is to put 6 √13 under the form....... √62×13 = √36×13= √ 468. Now 468 has 21 for the entire part of its square root; hence, 6 √13 is equal to 21, plus a fraction.

In the same way, we find that 12 √7=31, plus a fraction.

136. The object of the second transformation is to convert the

quantities, a and p being any numbers whatever, and q not a perfect square. Expressions of this kind are often met with in the resolution of equations of the second degree.

Now this object is accomplished by multiplying the two terms of the fraction by p-√q, when the denominator is p+ √q, and by p+q, when the denominator is p-q. For multiplying in this manner, and recollecting that the sum of two quantities, multiplied by their difference, is equal to the difference of their squares, we have