b If ✔ is irrational, (that is, not a square), the addition or subtraction can be only made by connecting the surds by the signs + or -1, as they are. STURMIUS, in his Mathesis Enucleata, has also given a method similar to the above. Ex. 4. To transform √2+√3 to a general surd. Ex. 5. To transforma-2/x to a universal surd. Ans. (a+4x-4/ax). Ex. 6. To transform 3/1+/72 to a universal surd. Ans. 3/9. § V. METHOD OF EXTRACTING THE SQUARE ROOT of BINOMIAL SURDS. 363. The square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let √n=a+m; then by squaring both sides, n=a3+ 2a/m+m, and 2a/mn-a-m; therefore, ✔m= n—a2. m a rational quantity, which is contrary to 2a the supposition. A quantity of the form a is called a quadratic surd. 364. In any equation x+y=a+b, consisting of rational quantities and quadratic surds, the rational parts on each side are equal, and also the irrational parts. If x be not equal to a, let x=a+m; then a+m+ √y=a+√b, or m+√y=√b; that is, b is partly rational, and partly a quadratic surd, which is impossible, (Art. 363); .. xa, and y=b. 365. If two quadratic surds x and y, cannot be reduced to others which have the same irrational part, their product is irrational. If possible, let✔xy=rx, where r is a whole number or a fraction, Then xy=r2x2, and y=rx; y =r; that is, y and x may be so reduced as to have the same irrational part, which is contrary to the supposition. 366. One quadratic surd, x, cannot be made up of two others, m and ✔n, which have not the same irrational part. If possible, let√x=√m+√n; then by squaring both sides, x=m+2mn+n, and x-m-n=2/mn, a rational quantity equal to an irrational, (Art. 365), which is absurd. 1 y, one 367. Let (a+b)=x+y, where c is an even number, a a rational quantity, b a quadratic surd, x and or both of them, quadratic surds, then (a+b)=x-y, By involution, a+b=x°+cx2-1y+c. x2--3 y2 + &c., and since c is even, the odd terms of the series are rational, and the even terms irrational; .. C- 1 C- -1 2 C 1 C 2 3 +c. -x-2y+&c., and bcxy+c. 2 x2--Sy3+ &c. (Art. 364); hence, a—b⇒xˆ—cx¤1y+c. 2 -x-2y2- &c.; and consequently, by evolution, 1 (a—b) c=x—y. 368. If c be an odd number, a and b, one or both quadratic surds, and x and y involve the same surds that a and b do respectively, and also (a+b)% = x+y; then (a—b)=x—y. 1 By involution, a+b=x°+cxc ̈1y+c. C -1 2 &c., where the odd terms involve the same surd that ~ does, because c is an odd number, and the even terms, the same surd that y does; and since no part of a can consist of y and its parts, (Art. 366), a=x +c. -x-3y2+&c., and b-cxy+c.. 2 C- -1 c. 2 3 -2 c-1 y3+ &c.; hence, a-b-x-cx-1y+c. 20-2 yz -&c.;.. by evolution, (a—b)=x—y. 369. The square root of a binomial, one of whose terms is a quadratic surd, and the other rational, may sometimes be expressed by a binomial, one or both of whose terms are quadratic surds. Let a+b be the given binomial, and suppose' √(a+b)=x+y; where x and y are one or both quadratic surds; then √ a-√b)=x—y, (Art. 367); .. by multiplication, √✓ (a2—b)=x2 —y3, also, by squaring both sides of the first equation, a+√ b= x2+2xy+y3, and a=x2+y3 (Art. 364); ..by addition, a+√(a2—b)=2x2, and by subtraction, a-(a2-b)=2y2; and the root x+y=√[}α + {√(a2 —b)]+√[}a¬}√(a2 —b)]. From this conclusion it appears, that the square root of a+b can only be expressed by a binomial of the form x+y, one or both of which are quadratic surds, when a2 -b is a perfect square. By a similar process it might be shown that the square root of a√b is √[la+√(a2 —b)] — √ [{a -1√(a2 —b)], subject to the same limitation. Ex. 1. Required the square root of 3+2√2. Let (3+22)=x+y; then (3-2/2)=x y; by multiplication, √(9-8)=x2-y; that is, x2-y2=1. Also, by squaring both sides of the first equation, 3+22=x2+2xy+y2, and x2 +y2=3, (Art. 264); .. by addition, 2x2=4, and x=√2. Again, by subtraction, 2y=2;.'. y=1, and x+y =2+1 the root required. = Or, the root may be found by substituting 3 for a, 2/2=√8 for ✓b, or 8 for b, in the above formula; thus, I x+y=√[}+{√(9—8)]+√[3—±√(9—8)=√(}+ 2 Ex. 2. Required the square root of 19+8√3. Ans. 4+√3. * Ex. 3. What is the square root of 12-140? Ans. 7-√/5. Ex. 4. Find the square root of 7+4/3. Ans. 2+√3. Ex. 5. Find the square root of 7-2/10. Ans. 5-2. Ex. 6. Find the square root of 31+12-5. Ans. 6+/-5. Ex. 7. Find the square root of 18-10-7. Ans. 5--7. Ex. 8. Find the square root of -1+4/-5. Ans. 2+/-5. 370. The cth root of a binomial, one or both of whose terms are possible quadratic surds, may sometimes be expressed by a binomial of that description. Let A+B be the given binomial surd, in which both terms are possible; the quantities under the radical signs whole numbers; and A is greater than B. Let /[(A+B)×√Q]=x+y; then/[(A-B) xQ]=x-y, (Art. 367); ... by multiplication, √[(A2-B2)×Q]=x2—y2 ; now let Q be so assumed, that (A2 -B2) ×Q may be a perfect cth power n°, then x2-y2=n. Again, by squaring both sides of the two first equations, we have [(A+B)3×Q]=x2+2xy+ya ✓[(AB)2 ×Q]= x2 -2xy + y2 C [(A+B)2 XQ]+°V[(A−B)2 ×Q] =2x2+2y2; which is always a whole number when the root is a binomial surd; take therefore s and t, the nearest integer values of √[(A+B)2 ×Q] and [(A—B)2 XQ], one of which is greater and the other less than the true value of the corresponding quantity; then since the sum of these surds is an integer, the fractional parts must destroy each other, and 2x+2y3= s+t, exactly, when the root of the proposed quantity can be obtained. We have therefore these two equations, x-y3n, and x2+y=s+t; .. by addition, 2x2=n+s+t, and x=√(2n+s+t); and by subtraction, 2y=1s+t-n, and y=√(s+-t—2n). Consequently, if the root of the binomial [(A+ B)XQ] be of the form x+y, it is √(2n+s+1) + V(s+1—2n); and the cth root of A+B is V (2n+s+1)+ v(s+1-2n) 22/Q Ex. 1. Required the cube root of 10+108. In this case, 108 is >10; ..A=108, B=10, 12-B108-100=8, and 8Q=n3. Now, since & is a cube number, Q may be taken equal to 1; then 3Q=8=3; ... n=2. Also, [(A+B)']=7+f; V[(A—B)']=1-f, where f is some fraction less than anity; ... s=7, t=1 ; and x+y= √12+2 2 = √3+1. If therefore the cube of 10+108 can be expressed in the proposed form, it is 13+1; which on trial is found to succeed. Ex. 2. Find the cube root of 26+15 √ 3. Ans. 2+ 3. Ex. 3. Find the cube root of 9 √3-11 1 2. Ans. 13-√2.: Ex. 4. Find the cube root of 45+8. ... 371.. In the operation, it is required to find a number Q, such, that (A2-B) XQ may be a perfect cth power; this will be the case, if Q taken equal to (A-B2); but to find a less number which will answer this condition, let A-B be divisible by a a,... (m); b, b, (n) ; d, d, . . . (r); &c. in succession, that is, let A-B3amb"dr &c.; also, let Q abd &c. Then (A2-B2).Q=um+xxbn+y× dit, &c., which is a perfect cth power, if x, y, z, &c., be so assumed that m+x,n+y, r+z, &c. are respectively equal to c, or some multiple of c. Thus, to find a number which multiplied by 2250 will produce a perfect cube, divide 2250 as often as possible by the prime numbers 2, 3, 5, &c. and it appears that 2 × 8 × 3 × 5 × 5 × 5=2 × 3a × 53=2250; if, therefore |