Let A+B be the given binomial surd, in which both terms. are possible; the quantities under the radical signs whole numbers ; A greater than B; and n an odd number.

let Q be so assumed that (42 - B) × Q may be a perfect nth power,

as p", then x2 - y2 = p.

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

hence √(A + B)3 × Q + √(4 − B)2 × Q = 2x2 + 2y3,

which is always a whole number, when the root is a binomial of the supposed form; take therefore s and t the nearest integer values of

✓ (A + B)* × Q and √ (A – B)* × Q,

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 + 2y = 8+ t exactly, when the root of the proposed quantity can be obtained. We have therefore these two equations

therefore, if the

form x+y, it is

A+ B is

2

root of the binomial √ (A + B) × √ be of the √ 8 + t + 2p + √ 8 + t -2P; and the nth root of

2

s

√s + t + 2p + √s+t−2p

22/Q

335. In the same manner, the nth root of A B is

√s+t+2p-√8 s+t-2p
22/Q

;

in which expression, when A is less than B, p is negative.

336. If the index of the root to be extracted be an even number, the square root of the proposed quantity may be found by Art. 329, when it can be expressed by a binomial of the same description; and if half the index be an even number, the square root may again be taken, and so on, until the root remaining to be extracted is expressed by an odd number, and then the method of the preceding Art., or of Art. 332, may be applied.

Ex. 1. Required the cube root of 11 + 5 √7.

Here 45√7, B = 11, 42 - B2 = 54; therefore Q = 4, and p3 216, or p = 6.

Also √(A+B)2 × Q = √(296 +110 √7) × 4

3

= 2268.44

= 13 +f;

Similarly (AB)3× Q = 3-f;

s

or 8 = 13, and t = 3; therefore, by substitution, a =√7, and y = 1; hence x + y = √7+1; and the quantity to be tried for the root is

√7+1

which is found to succeed.

Ex. 2.

2√7+3√3.

Required the cube root of 2√7

Here A = 2√7, B = 3√3, 42 – B2 = 1; hence Q = 1, and

p = 1.

√7+√3

=

2

to succeed.

the quantity to be tried for the root, which is found

337. In the operation, it is required to find a number Q, such, that (42 - B2) × Q may be a perfect nth power; this will always be the case, if Q be taken equal to (42 - B2)-1; but to find a less number which will answer this condition, let 42 - B2 be divisible by a, a, &c....a times; b, b, &c....ß times; c, c, &c....y times, &c. in succession; that is, let A2-B2 = ab3 c, &c. Also let Q=a*b*c*. &c. then

which is a perfect nth power, if x, y, z, &c. be so assumed that a + x, ß + y, y + x, &c. are respectively equal to n, or some multiple of n. Thus, to find a number which multiplied by 180 will produce a perfect cube, divide 180 as often as possible by 2, 3, 5, &c. and it appears that 2.2.3.3.5 180; if, therefore, it be multiplied by 2.3.5.5, it becomes 23.33.53, or (2.3.5)3, which is a perfect cube.

=

338. If A and B be divided by their greatest common measure, either integer or quadratic surd, in all cases where the nth root can be obtained by this method, Q will either be unity, or some power of 2, less than 2". See Dr. Waring's Med. Alg. Chap. v.

339. The square root of a multinomial, of which one term is rational, and the rest quadratic surds, may sometimes be found by assuming

√a + √b + √c + √d = √x + √y + √z,

and proceeding to find x, y, and z, as in Art 329.

Ex. Required the square root of 21 +6√5 +6 √7+2√35.

Let √x+ √y+√z=√21+6 √√5+6√7+2√35,

then x+y+z+2√xy + 2 √xx+2√yz=21 +6√5+6 √7 +2 √35;

Now 2 √xy × 2 √xz = 4x √yz, or 6 √5 × 6 √7 = 4x . √35,

.. x=9, and √x = 3.

Also 2 √xy × 2√ys = 4y √xz, or 6 √5 × 2 √35 = 12y √7,

Hence √x+√y+√x=3+ √5+ √7, the root required.

INDETERMINATE COEFFICIENTS.

340. If A + Bx + Cx2+ &c. = a + bx + cx2 + &c. be an identical equation, that is, if it hold for all values whatever of x, then the coefficients of like powers of x are equal to each other, that is, A = a, B = b, C = c, &c.

an equation which admits of one value of x only (Art. 193), unless B-60, or B b, and therefore also A

=

Again, if A + B x + Cx2

A

a = 0, or A = a.

= a + bx + cx2, then

a + (B − b) x + (C − c) x2 = 0,

a quadratic equation with respect to x which admits of no more than two distinct values of x (Art. 204), unless C - - c = 0, or C = c, and B b = 0, or Bb, and therefore also A - a 0, or A = α.

=

Similarly, if any number of terms be taken, or

(A − a) + (B − b) x + (C − c) x2 + &c.

there are certain values of x, and none other, which will satisfy the equation as long as it remains an equation with respect to x.

But, by the supposition, the equation must be true for any value whatever which we may please to give to x, and consequently for any

number of values of x; and this, therefore, can only be attained by that which is apparently an equation with respect to a ceasing to be such, that is, by the coefficients of the powers of a being separately equal to 0 ; that is, we must have

COR. If there be found any power of x on one side of the proposed equation, and no corresponding one on the other, then the whole coefficient of that power is of itself equal to 0. Thus, if A + B x + Cx2 + &c. for all values whatever of x, then A 0, B = 0, C = 0, &c.

=

66

0,

341. It may be objected to the proof in the preceding Article, that it is rather assumed than proved that every equation has only a certain number of values of x which will satisfy it; and that it does not include those cases in which the assumed series is an 'infinite series." No such objections can be made to the following proof* : Let A + B + C x2 + ... = a + bx + c x2 + that is, hold for any value whatever of r; then

A

...

~ a + (B ~ b) x + (С ~ c) x2 +

:

be an identical equation,

and, if A ~a is not equal to 0, let it be equal to some quantity p; then we have

=

(B ~ b) x + (C ~ c) x2 + &c. p.

And since A, a, are invariable quantities, their difference p must be invariable; but p= = (B ~ b) x + (C~ c) x2 + ... a quantity which may have various values by the variation of x; that is, we have the same quantity (p) proved to be both fixed and variable, which is absurd. Therefore there is no quantity (p) which can express the difference A ~ a, or, in other words,

* In the proof which is usually given a is assumed equal to 0, and afterwards the equal quantities are divided by x, whereas it is not proved that we may divide any quantity by x when a stands for 0, in the same manner as when it stands for a finite magnitude; and that such a proceeding will in certain cases lead to erroneous results is well known.