Equivalent Exponents.' (213.) Two Exponents which are numerically equivalent, are also equivalent exponentially; and may therefore be substituted, the one for the other. Thus a-a; the fourth root of a2 is equivalent to the square root of a. produces a2. 3 In like manner x6=x2; =y3; a‡=a2; &c. To give an example in numbers, the fourth root of the square of 9, that is, of 81, is 3, and the square root of 9 is also 3. On this principle a fractional exponent may always be taken in its lowest terms. The Sign to be Prefixed to a Root. (214.) Every odd Root of a quantity has the same sign as the quantity itself. For the quantity itself is an odd power with reference to such root, (206), and an odd power has the same sign as the quantity of which it is a power, (202). (215.) Every even Root of a positive quantity is ambiguous, that is, the quantity itself does not determine whether the root is positive or negative. This follows from the principle that every even power of a negative, as well as of a positive quantity, is positive, (202). The square of -a, as well as of +a, is +a2; hence the square root of a2 is a or else -a; and when it is not known whether a2 Iwas derived from a or -a, it is uncertain which of these two is the square root. This uncertainty as to the sign of the root, is expressed by the ambiguous sign plus or minus; thus the square root of a2 is ±a. For a like reason the fourth root of a1 is The square root of 9 is ±3; the fourth root of 16 is ±2; &c. (216.) An even Root of a negative quantity is impossible, since there is no quantity which can be raised to an even negative power, (202). Thus no quantity multiplied into itself will produce a2; this quantity therefore has no square root. For a like reason -a has no square root or fourth root. In like manner -9 has no square root; -16 has no square root or fourth root; -25a2 has no square root, &c. EXERCISES On the Roots of Monomials. 1. Find the square root of 4a2x4. 3 6. Find the cube root of a3xy2. 7. Find the square root of 25a3: 8. Find the cube root of -64y2. 9. Find the square root of 36x6. 10. Find the cube root of 8ay*. 11 Find the square root of 1a4y2. 12 Find the cube root of -8a3x3n. 13. Find the square root of xy1. 14. Find the cube root of -ax3. 15. Find the fourth root of 16a1y2. 16. Find the cube root of Ans. 2ax2. Ans. ±3a2x2. Ans. ±4x+. Ans. ±5a2. Ans. -4y3 Ans. 6x3. Ans. 2a3y3. Ans. ±fx1y2. Ans. ±2ay2. Ans. -fx. 2. ROOTS OF POLYNOMIALS. (217.) The method of extracting any required Root of a Polynomial, may be discovered from the manner in which the corresponding power of a polynomial is formed. This subject will be elucidated under the appropriate Rules. RULE XVIII. (218.) To Extract the Square Root of a Polynomial. 1. Arrange the Polynomial according to the powers of one of its letters, and take the square root of the left hand term, for the first term of the root. 2. Subtract the square of the root thus found from the given Polynomial; divide the remainder by twice the root already found, and annex the quotient to both the root and the divisor. 3. Multiply the divisor thus formed by the last term in the root; subtract the product from the dividend; divide the remainder by twice the root now found; and so on, as before. EXAMPLE. To extract the Square Root of a2+2ab+b2. The required Root, we already know, is a+b, since the square of this binomial is the given trinomial :—our object is to show that the root a+b would be found by the foregoing Rule. The first term a of the root, is the square root of a2, the left hand term of the given polynomial. Subtracting the square of a, we have the remainder 2ab+b2. We have now to discover a divisor of this remainder, which will give, for a quotient, the next term b of the root. 2a, that is, twice the root already found, divided into 2ab, gives b; and b annexed to 2a makes the divisor 2a+b. This divisor multiplied by b, produces 2ab+b2,-which completes the operation. The Rule is framed in accordance with the process thus discovered. EXERCISES On the Square Root of Polynomials. 1. Find the square root of the polynomial a4-4a3+6a2-4a+1. 2. Find the square root of the polynomial 3. Find the square root of the polynomial 1—6y+13y2-12y3+4y*. 4. Find the square root of the polynomial 4x6-4x+12x3+x2-6x+9. 5. Find the square root of the polynomial Ans. a2-2a+1. Ans. a2+2ax+x2. Ans. 1-3y+2y2. Ans. 2x3-x+3. 4a4+12a3x+13a2x2+6ax3+x1. Ans. 2a2+3ax+x2. 9a4-12a3y+28a2y2-16ay3+16y*. Ans. 3a2-2ay+4y2. 9. Find the square root of the polynomial 25+34x2+12x3 +20x+9x4. Ans. 5+2x+3x2. 10. Find the square root of the polynomial a4+4a3y-12ay+4a2y2+9-6a2. Ans. a2+2ay-3. 11. Find the square root of the polynomial Ans. 2-4x+2x2. 4+24x2-16x+4x4-16x3. 12. Find the square root of the polynomial 1—2y+7y2 —2y3+5y*+12y5+4y6. Ans. 1−y+3y2+2y3. SQUARE ROOT OF NUMBERS. The preceding Rule is substantially the same as the Arithmetical Rule for extracting the square root of Numbers. To show its application, however, to the latter purpose, we must premise the following principles.. (219.) If a Number be separated into periods of two figures each,from right to left, these periods will correspond, respectively, to the units, tens, hundreds, &c., in the Square Root of the number. For since the square of ten is 100, the square of the tens figure in the root will leave two vacant places in the right of the given number; these two places must therefore correspond to the units in the root. And since the square of a hundred is 10000, the square of the hundreds in the root leaves four vacant places in the right of the number; and the first two corresponding to the units, the next two must correspond to the tens in the root. In like manner it is shown that the third period of two figures corresponds to the hundreds in the square root of the given number; and so on. (220.) If a Number be divided into any two parts, the Square of the number will be equal to the square of the first part + twice the first the second + the square of the second part. For, a representing the first part of the number, and b the second, a+b will be equal to the number; and (a+b)2=a2+2ab+b2. With these principles established we may proceed to the following EXAMPLES. 1. To extract the Square Root of 529 5'29 ( 23 43) 129 129 The first period 29 corresponds to the units figure, and the 5 to the tens figure in the root, (219). The greatest integral square root of the 5 is 2 tens, the square of which is 4 hundreds, and this subtracted leaves 129. |