15. Given x+y3=2xy(x+y), and xy-16, to find the Ans. x=2/5+2, y=2/5-2. values of x and y. 16. Given 3+y3=a, and x2y+xy2=a, to find the relative values of x and y. Ans. x=y. 17. Given x+y:x:: 5 : 3, and xy=6, to find x and y. Ans. x=±3, y=±2. 18. Given x+y: x 7:5, and xy+y=126, to find a and y. Ans. x=15, y=±6. 19. Given 2+y=a, and xy=b, to find the values of x and y. Ans. x=a+2b+1 Ja—26. y=±√a+2b-Ja-26. (Art. A)* Equations in the form of x-2ax2+a2-b, require for their complete solution, the square root of an expression in the form of a±√; for by extracting the square root of the equation, we have The right hand member of this equation is an expression well known among mathematicians as A BINOMIAL SURD. Expressions in this form may or may not be complete powers; and it is very advantageous to extract the root of such as are complete, for the roots will be smaller, and more simple quantities, in the form of a'±√b', or of Ja'±√b'. Let us now investigate a method of extracting these roots; and, for the sake of simplicity, let us square 3+7. By the rule of squaring a binomial, we have 9+6√7+7, Conversely, then, the square root of 16+6/7, is 3+√7. * That the same Articles may number the same in both the School and College Edition, we shall designate all additional Articles, in this volume, by A, B, &c. But when a root consists of two parts, its square consists of the sum of the squares of the two parts, and twice the product of the two parts. Now we readily perceive that 16 is the sum of the squares of the two parts expressing the root; and 6/7, the part containing the radical, is twice the product of the two parts. To find what this root must be, let a represent one part of the root, and y the other: Add equations (1) and (2), and extract square root, and we have x+y=/ 16+6/7 (3) Subtract equation (2) from (1), and extract square root, and we have x-y=16-6/7. Multiply (3) and (4), and we have (4) x2—y2—/256—252=√4=2; (5) Add (1) and (5), and we have 22-18, or x=3; Sub. (5) from (1), and 2y=14, or y=√7. Whence y, or the square root of 16+6/7 is 3+√7. We shall now be more general. In numerals, and, in short, in all cases, the sum of the squares of the two parts of the root, as (a+b), in the first square, and (a+b), in the second, contain no radical sign; and the sum of these rational parts may be represented by c and c', and the squares represented in the form of Hence, generally, if we represent the parts of the roots by x and y, we shall have x2+y2= the sum of the rational parts, and 2xy= the term containing the radical. The signs to x and y must correspond to the sign between the terms in the power. If that sign is minus, one of the signs of the root will be minus; it is indifferent which one. EXAMPLES. 1. What is the square root of 11+6/2? 2. What is the square root of 7+4√3? 3. What is the square root of 7—2/10? Ans. 3+√2. Ans. 2+√3. Ans. √5-√2 or √2—√5. 4. What is the square root of 94+42/5? 5. What is the square root of 28+10/3? 6. What is the square root of np+2m2-2m/np+m2? In this example put a=np+m2, and x and y to represent the two parts of the root, Then and x2+y2=m2+a, Ans. ±(√np+m2—m Ans. ±(b+√bc-b3). 7. What is the square root of bc+2b/bc-b2? 8. What is the sum of √/16+30√−1+√√/16—30,√—1? Ans. 10. 9. What is the sum of 11+6√2 and √√√7—2√10? Ans. 3+√5. 10. What is the sum of 31+12-5 and, -1+4/-5? Ans. 8. In a similar manner we may extract the cube root of a binomial surd, when the expression is a cube; but the general solution involves the solution of a cubic equation, and, of course, must be omitted at this place; and, being of little practical utility, we may omit it altogether. (Art. 92.) Fractional exponents are at first very troublesome to young algebraists; but such exponents can always be banished from pure equations by substitution. For the exponents of all such equations must be multiples of each other; otherwise they would not be pure, but complex equations. To make the proper substitution, put the lowest exponent of any letter, as x, equal to a simple letter, say P; and the lowest exponent of any other letter, as y, equal to another simple letter, say Q. And let this be a general rule. 2 By the above direction, put x3-P, and y3—Q. = By squaring equation (1), P2+2PQ+Q2=36. Subtracting equation (2), we have 2PQ=16. By extracting square root P-Q=±2 But by equation (1), P+Q= 6 Therefore, P=4 or 2, and Q=2 or 4, 2. Given xy+y=21, and x2y1+y=333, to find the values of x and y. By comparing exponents in the two equations, we perceive that if we put xy2=P, and y=Q, the equations become P+Q = 21 P2+Q=333 Solved as the preceding, gives P=18, Q=3. From which we obtain x=2, or y=3 or 18. 1089 to find the values of x and y Assume x=P, and y=Q. By squaring and cubing these assumed auxiliary equations, we have 4 2 y3 = Q2, Seek the common measure (if there be one) between 208 and Substitute this value of in equation (1), |