.1 X440.02656

3103.55781

.0141; whence x 8.4.0131 = 3.4141,

nearly. Taking 8.414 and 8.415 for the numbers, we have 42170.867905346409824 — 2:5 = 42195.935727527009375

30.54158378121

54337.654666553259375 54337.65466655325

: *.001 :: 16.65466655325 : 000545

and a=r-z-8.415-.000545-8.414455, very near.

5. Given

(7x+4x2)+(20x2-10x)=28, to find an approxAns. 4.510661.

imate value of x.

By trial x is found to be between 4 and 5. Let these two numbe taken for the first value; then

=

1

1st Sup. 8 16.73 =

z=4.51, nearly. Assuming x 4.51 and 4.52, and repeating the operation, we get x=4.5106, the root, nearly.

6. Given

(144x2—(x2—20)2}+^{196x2—(x2+24)2}

to find a near value of x.

= 114,

Here the root is found, by a few trials, to be rather more than 7; let, therefore, 7 and 7.1 be assumed for the numbers; then 1st Sup. 7056 =

4761

✔ 2295 = 47.906

144x2 =

=

7259.04 2d Sup.

4957.5681

2302.4719=47.977

9880.36
5536.8481

4343.511965.905

(x2+20)2

113.289 7
.593

114. 113.882

: .1

2; whence 27.12, nearly. By assuming

7.12 and 7.13 for the numbers, we get z = 7.12209461, the answer, nearly.

1. Given x2+x2+x=100, to find an approximate value of x. Here it will be found, by a few trials, that the value of x lies between 4 and 5.

Hence, by taking these as the two assumed numbers, the operation will stand thus:

Of Reciprocal Equations.

104. Although no general method has hitherto been discov ered for the resolution of equations higher than those of the fourth power, there are, notwithstanding, some particular equations, of all orders under the 10th, which, on account of the relations subsisting between their coefficients, may be solved by the rules that have been already given for the first four orders.

....

1

x

2m

This is particularly the case with what have been usually called reciprocal equations; which are such that the coefficients of their terms, taken from the beginning of the equation, are the same as those of the corresponding terms, taken from the end, with the same signs; or which remain of the same form when the reciprocal of the unknown quantity, or is substituted for ; except that the terms are then reversed. Thus the equations m px2-1+9x2m2+....+qx2+px+1=0, and x2+1+px2+qx2-1_ •+qx2+px+1=0,may always be transformed into others of a degree which is denoted by half the exponent of the highest power of the unknown quantity, if it be an even number, or by half that exponent when it is diminished by 1, if it be an odd number, the method of resolving them, as far as to the 9th order, inclusively, being as follows. To this we may add, that the nature of these equations consists, as abovementioned, in their not being changed by substituting for z; from which it follows, that if a be any one of the roots, its reciprocal will also be a root; and as +1, or -1, is always a root of the equation, when the number of its dimensions is odd, it may be readily shown from these circumstances, that every equation of the 2mth or 2mth+1th order, can be reduced to another of the mth order.

CASE I. When the index of the highest term is an even number. RULE I. If the equation be of the fourth power, as x2+px3+ qx2+px+1=0, find the two values of z in the equation, zpz+ q-2-0, and let them be denoted by r and r'; then the roots of the two quadratic equations, 22-rx+1 = 0, and 2-r'x+1-0, will be the four roots of the proposed equation.

2. If it be of the sixth power, as x+px5+qx2+rx3+qx2+px +1=0, find the three values of z in the equation, z3+pz2+(q—3)z +r-2p0, and let them be denoted by r. r', r'; then the roots of the three quadratics 2-rx+1—0, x2—r'x+1=0, and x2—r'x+1=0, will be the six roots of the proposed equation.

3. And if it be of the eighth power, as x+p+qx®+rx3+sxa +rx3+qx2+px+1=0, find the four values of z in the equation, 2*+pz3 + (9—4) z2+(r—3p)z+s—2(2-1)=0, and let them be denoted by r, r', j'',

Then the roots of the four quadratic equations.

x2—rx+1=0, x2—r′x+1—0, x2—r'x+1=0, and x2—r””x+1 =0, will be the eight roots of the proposed equation.

CASE II. When the index of the highest term is an odd number. RULE I. If the equation be of the third power, as x3+px2+px +1=0, where one of its roots is evidently -1, find the value of z in the simple equation z+p-1-0, and let it be denoted by r; then the roots of the quadratic equation x2-rx+1=0, will be the other two roots of the proposed equation.

2. If it be of the fifth power, as 25+px2+qx3+qx2+px+1=0, where one of its roots is also -1, find the two values of z in the equation 22(p-1)z+9-p-1-0, and then let them be denoted by r, and r'; then the roots of the two quadratics -rx+1=0, and x2-r'x+1 =0 will be the other four roots of the proposed equation.

3. If it be of the seventh power, as x2+px2+qx5+rx2+rx3+ qx2+px+1=0, where one of the roots is -1, as before; find the the three values of z in the equation z3+(p-1)2+(q—p—2)z+ r-p―q+1 0, and let them be denoted by r, r', r"; then the roots of the 3 quadratics 22-rx+1=0, x2-r'x+1 = 0, and -r'x+1=0, will be the other 6 roots of the proposed equation. 4. And if it be of the 9th

=

power, as

x2+px2+qx2+rx®+sx3+sx2+rx3+qx2+px+1=0,

where one of the roots is 1, as in the former cases, find the four values of z in the biquadratic equation

2*+ (p—1)23 + (q—p—3)22 + (r—q—2p+2)z+s—r—q+p+1 =0, and let them be denoted by r, r', r'', r''' ;

then the roots of the four quadratics

x2—rx+1=0, x2—r′x+1—0, x2—r''x+1=0, and x2-r""x+1 =0, will be the other eight roots of the proposed equation.

Note. If an equation of this kind be of an odd number of dimensions, or if the middle term of one of an even number of dimensions be wanting, the same rules will hold when the signs of the terms, taken from the beginning, are + and - alternately. Thus x3-px2+px-1=0, and 2*—px3-px-1—0, are reciprocal equations, like those above given, except that + 1 is now one of the roots, instead of -1.

1. Given x+4x3-19x2+4x+1=0, to find the four roots of the equation. Here p-4 and q=-19; hence, by Case 1,

2+pz 2-q becomes z+42-219-21,

Where the two values of z are +3 and
Consequently x-3x=-1, and x2+7x-1,
From the first of which equations 5,

And from the second x-(-5)=(±√45);

Therefore the four roots of the proposed equation are

1+1√5, §—1√5, +5, and ——3/5,

2. Given -11x+17x+17x11x+1=0, to find the five roots of the equation. Here p— - 11, and 917; hence, by case 2, z2+(p—1)z=1+p—q becomes 12x-27, where the two values of z are 9 and 3; x2-9x=-1, and x2-3x——1,| from the first of which equations x77, and from the second 25. Therefore the five roots of the proposed equation are 1,77, 2—177, +1√5, and 3—√√5.

105. If the coefficients of either of the orders of equations, mentioned in the rule, be in part literal, and so constituted as to render the different terms homogeneous, its roots may be determined by means of the simple substitution of a new unknown quantity, as if they were entirely numeral.

:

Thus, let there be taken, as an equation of this kind, x1+4ax3 -19a2x2+4a3x+a0, where the indices of the given and unknown quantities, when added together, are the same as those of the first and last terms. Then, by putting xaz, we shall have a'z'+4a1z3-19a1z2+4az+a* =0, or, dividing by a, za +4z3— —19z+4x+1=0; which equation, like the first of those given above, will have for its roots +√5, §—{√5, −1+3No5, and -5; and the four roots of the proposed equation are a(§ + 1/5), a(3—1√√√5), a(— + √5), and a(—7—3√√5). An homogeneous equation, containing only two letters, a and x, is of the general form

· x" + paxTM + qa2x2+ra3x3+····+kaTM-1x+la”—0, where p, q, r, k, l, are the numeral coefficients.

1. Given x-15x+38x2-15x+1=0, to find the four roots of the equation. Ans. 5, and 6±35. Here p-15, q=38, we have 22-15z--36, by case 2; z=12 or -3. Hence we have x2-12x=-1, and x2 + 3x——1; ... in the first we have x= 6±35, and in the second case we have x=5.

2. Given 1-4ax3+5a2x2-4a3x-a'=0, to find the four roots of the equation. Ans. a(5), and a(±√—3). Let p=-4, q=5; then 2-4-2-5--3; ..z3 or 1, and we have 2-3y--1, and 25. Again, x2 — x — — 1 ; .. x = {√3; consequently 3; consequently we have a(5), and a(}±±√3).

=

3. Given 25-212*+37x3 +37x2-21x+1=0, to find the five roots of the equation." Ans. 1, 5, and 1357. Here p-21, and q=37; then by case 2, 22—22z:

.. z=11±8=19 or 3, and .. y2—19x:

x=

-1; then

-57

357. Again x-3x-1, we find x=±±√5.

106. Equations of this kind, in which the given and the un