Here (27+48)=27+2/1296+48-147; 27+ √48=√147/49 X3=73.

Ex. 3. To transform 3/320-3/40 to a general surd. Here (2/320-2/40)=320-33/4096000+3/512000

-40=40;./320-3/40=23/5.

361. This transformation is very useful, since, by means of it, we can always reduce the sum or difference of any two surd quantities, if they admit of the same irrational part, to a single surd. This may be proved, in general, thus ; if a and /b admit of the same irrational part, they must be of the forma" and "b"m; and (a""m+/b'")"=a'"m+n"/ n(n-1) L

!^/(d'n(n—»m2-2b1a¤‚2)+&c. . b'"m =a'"m+na'n-1×mb'" +&c. . . . . b'rm ::~/a+b="/(a"m +nman-16""+&c. b'"m)= the nth root of a rational quantity. Hence the product of a by b is rational if a and b admit of the same irrational part; also, aa× /h, or /ax/b2, is rational, if a and b admit of the same irrational part; and, in general, /a/b, or/a /b", is rational, if a and b admit of the same irrational part.

362. It is proper to observe, that, for the addition or subtraction of two quadratic surds, the following method is given in the BIJA GANITA, or the Algebra of the HINDOos, translated by STRACHEY. Thus, to find the sum or difference of two surds, ✔a and √b, for instance.

RULE.

Call a+b the greater surd; and, if a Xb is rational, (that is, a square), call 2/ab the less surd, the sum will be (a+b +2√ab), (=(√a+b)2), and the difference (a+b-2√ ab). If axb is irrational, the addition and subtraction are impossible; that is, they can only be indicated.

Example. Required the sum and difference of 2 and 8. Here 2+8=10=> surd; 2X8=16, ..√16=4, and 2、/16 =2×4=8=<surd. Then 10+8=18, and 10—8=2; .. √18= sum, and 2 difference.

ANOTHER RULE.

α

Divide a by b, and write in two places. In the first

b

place add 1, and in the second subtract 1; then we shall have

α

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.

Ans. √(5+2√6).. Ex. 5. To transforma-2x to a universal surd.

Ans. (a+4x-4√/ax).

Ex. 6. To transform 33/1+/72 to a universal surd.

Ans. 32/9.

SV. 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=a2+2am+m, and 2a/m= n—a2—m; therefore, ✔✅m= a rational quantity,

20

which is contrary to the supposition.

A quantity of the forma 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+vyb; 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=r2x; :.√y=r√x; that is, ✔y and 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

367. Let (a+b) °=x+y, where c is an even number, a a rational quantity, b a quadratic surd, x and y, one or both of them,

quadratic surds, then (a−b)o =x-y.

C 1
2

By involution, a+b=x°+cxc¬1y+c.- x-2y2+ &c., and since c is even, the odd terms of the series are rational, and the even terms irrational; ..a=x°+c.—x¤¬22+&c., and

2

x-3y3+ &c., (Art. 364); hence, a—b

202&c.; and consequently, by

evolution, (a-b) =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)o" =x—y.

1

c-1
2

By involution, a+b=x2+ cxc-1y+c. -x c~-2y2+ &c., where the odd terms involve the same surd that x 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 C-1 2

parts, (Art. 366), a=x°+c.—x-2y2+ &c., and b-cx2-1y+

-2

C- - 1

2o ̄μ3+&c.; hence, a−b= x2-cx2¬1y+c. —

--&c.; ..by evolution, (a—b)=x-y.

2

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 quadra

tic 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+y2, and a=x2+ y2 (Art. 364);

by addition, a+√(a3—b)=2x, and by subtraction, a✓

(a2-b)=2y; and the root x+y=√[+a+!√(u2—b)]+√ [a-1√(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 [a+}√(a2—b)]—√[{a—±√(a3—b)], subJect to the same limitation.

Ex. 1. Required the square root of 3+2/2.

Let √(3+2√2)=x+y; then √(3—2√√2)=x-y; by multiplication, (9—8)=x2—y2; that is,

y=1. Also, by squaring both sides of the first equation, 3+2√2 x2+2xy-y2, and x2+32=3, (Art. 264); ..by addition, 2x2 =4, and x=".

Again, by subtraction, 2y=2; ..y=1, and x+y=√?+! 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, x+y=√{}+{√(9−8)]+√[{− }√(9−8)=√(} + 1) + √

(3−1)=√2+1.

Ex. 2. Required the square root of 19+8/3.

Ans. 4+3.

Ex. 3. What is the square root of 12-140 ?

Ex. 4. Find the square root of 7+4/3.
Ex. 5. Find the square root of 7—2/10.

Ans. /7-5.

Ans. 2+3.

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,

(A-B3) ×Q]=x2—y2; now let

Q be so assumed, that (A2-B2) XQ may be a perfect cth

power =nc, then x2-y2=n.

Again, by squaring both sides of the first two equations, we have

°√[(A+B)3×Q]=x2+2xy+ y2

✅/[(A—B)3× Q]=x2—2xy+y2

:: V/[(A+B)3×Q]+°✓/[(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)3× Q] and [(A-B)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 2x2+2y2=s +t, exactly, when the root of the proposed quantity can be obtained. We have therefore these two equations, x2—y2 =n, and x2+y2=s+t;.. by addition, 2x2=n+s+t, and x={√(2n+s+); and by subtraction, 2y=1s+1t—n, and y = √(s+t-2n).

Consequently, if the root of the binomial /[(A+B)×✔ Q] be of the form x+y, it is (2n+s+t)+{√(s+t-2n); and the cth root of A+B is√(2n+s+1) +√(s+t—2n)

Ex. 1. Required the cube root of 10+/108.

In this case, ✓108 is > 10 ; .. A=✓✓108, B=10, A2—B2 =108-100=8, and 8Q=n3. Now, since 8 is a cube number, Q may be taken equal to 1; then 8Q=8=n3 ; .. n=2. Also, V/[(A+B)3]=7+ƒ; 3⁄4/[(A—B)3]=1-f, where f is some √12+2 fraction less than unity; .. s=7, t=1 ; and x+y=· 2

=√3+1.

If therefore the cube of 10+108 can be expressed in the proposed form, it is 3+1; which on trial is found to succeed.

Ex. 2. Find the cube root of 26+15/3.

Ex. 3. Find the cube root of 9/3—11√/2.
Ex. 4. Find the cube root of 4/5+8.

Ans. 2+√3.

Ans. √3-2.

Ans.

√5+1

3/2

371. In the operation, it is required to find a number Q, such, that (AB)XQ may be a perfect cth power; this will be the case, if Q be taken equal to (AB); but to find a less number which will answer this condition, let A2- B2 be divisible by a, a, . . . (m); b, b,... (n) ; d, d, . . . (r) ; &c. in succession, that is, let A2 - B2=ambndr &c.; also, let Q=&*