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Therefore x = 7.124 nearly the root required. Note 5. The same rule also among other more difficult forms of equation, succeeds very well in what are called exponential ones, or those which have an unknown quantity in the exponent of the power; as in the following example : Ex. 4. To find the value of x in the exponental equation

= 100. For more easily resolving such kinds of equations, it is convenient to take the logarithms of them, and then compute the terms by means of a table of logarithms. Thus, The logarithms of the two sides of the present equation are * X log of z - 2 the log of 100. Then, by a few trials, it is soon perceived that the value of x is somewhere between the two numbers 3 and 4, and indeed nearly in the middle between them, but rather nearer the latter than the former. Taking therefore first x = 3.5, and then 3.6, and working with the logarithms, the operation will be as follows: First supp. x = 3.5.

Second Supp. x= 3.6. Log. of 3.5 = 0.544068 Log. of 3•6 = 0.556303 then 3.5 x log. 3 5 = 1 904238 then 3:6 x log 3.6 = 2.002689

the true number 2.000000 the true number 2 000000

error, too little, -095762

002689

error, too great +.002689

098451 sum of the errors. Then,

As .098451 : :1: : .002689: 0.00273 the correction

taken from 3.60000

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On trial, this is found to be a very small matter too little. Take therefore again, x 3.59727, and next = 3.59728, and repeat the operation as follows:

First, Supp, I

= 3.59727. Second, Supp. x = 3.59728 Log. of 3 59727 is 0.555973 Log. of 3 59728 is 0.555974 3.59727 x log.

3 59728 x lng of 3.59727 = 1.9999854 of 3.59728 = 1.9999953 the true number 2 0000000 the true number 2•0000000

error, too little, -0.0000146 error, too little, -0.0000047

-0.0000047

0.0000099 diff. of the errors. Then, As .0000099 : 00001 : : 0000047 : 0000: 47 4747* the cor. added to

3.59728000000

gives nearly the value of x = 3.5972847 4747

Ex. 5. To find the value of x in the equation x3 + 10.02 45% = 260.

Ans. I = 4•1179857.

Ex. 6. To find the value of x in the equation x3 -27 = 50

Ans. 3.8648854.

The Author has here followed the general rule in finding as many additional figures as were known before : viz. The 6 figures 3.59:28, but as the logarithms here used are to places only, we cannot depend on more than 6 figures in the answer ; we bave no reason, therefore, to suppose any of the figures in :00000474747, to be correct.

The log. of 3.59728 to 15 places is 0.55597 424 31 34677

the log of 3:59727 to 15 places is 0.55597 30358 47267 which logarithms multiplied by their respective numbers give the following products :

1.99999 50253 435127

1.99998 51226 62298 Therefore, the errors are

and and the difference of errors

both true to the last figure,

49746 56488 148773 37702 99026 81214.

Now since only. 6 additional figures are to be obtained, we may omit the last three figures in these errors; and state thus : as diff. of errors 9902681 : diff. of sup. 1:: error 4974656: the correction 502354, which imited to 3 59728 gives us the true value of x =359728502354.

17 r2

Ex. 7. To find 'the value of x in the equation 3 + 2x2 -23 = 70.

Ans. r = 5 i3457. Ex. 8. To find the value of x in the equation is + 543 = 350.

Ans. x = 14.95407. Ex. 9. To find the value of x in the equation et - 333 750 = 10000.

Ans x = 10.2609. Ex. 10. To find the value of x in the equation 2r• -16 + 40 r' 30x -1.

Ans. x = 1.284724. Ex. 11. To find the value of x in the equation x5 + 23% + 3r3 + 4x2 + 5x = 54321.

Ans. x = 8.414455, Ex 12. To find the value of x in the equation : 123456789

Aas. x = 8 6400268. Ex. 13. Given 2x 7353 + 113? 3x = ll, to find 3.

Ex. 14. To find the value of x in the equation (3x2 – 2 V 3+*- (* – 43 V 3+ 3 VI)* = 56.

Ans. x = 18-360877.

To resolve Cubic Equations by Carden's Rule.

THOUGH the foregoing general method, by the application of Double Position, be the readiest way, in real practice, of finding the roots in numbers of cubic equations, as well as of all the higher equations universally, we may here add the particular method commonly called Carden's Rule, for resolving cubic equations, in case any person should choose occasionally to employ that method.

The form that a cubic equation must necessarily have to, be resolved by this rule, is this, viz. zs + az = b, that is, wanting the second term, or the term of the 2d power zo. Therefore, after any cubic equation has been reduced down to its final usual form, x3 + px' + 92 = r, freed from the co-efficient of its first term, it will then be necessary to take away the 2d term poza; which is to be done in this manner: Take it, or of the co-efficient of the second term, and annex it, with the contrary sign, to another unknown letter z, thus z-ih; then substilute this for xj the unknown letter in the original equation is + px? + qx = r, and there will result this reduced equation zs + az = b, of the form proper for applying the following, or Carden's rule. Or take

fb, by which the reduced equation takes this form z3 + 3c z 2d.

Ja, and d

Then

с

Then substitute the values of cand d in this

form z = y + d +63) +- (d + 3),

or z = 3d tv (d2 + (8) 3d tv laz + c3), and the value of the root z, of the reduced equation 23 + az = b, will be obtained. Lastly, take x = z

*M, which will give the value of x, the required root of the original equation x3 + px2 + 9x = r, first

proposed. One root of this equation being thus obtained, then depressing the original equation one degree lower, after the munner described p. 260 and 261, the other two roots of that equation will be obtained by means of the resulting quadratic equation.

Note. When the co-efficient a, or c, is negative, and c3 is greater than d’, this is called the irreducible Case, because then the solution cannot be generally obtained by this rule.

Ex. To find the roots of the equation x3 -63% + 10x 8.

First, to take away the 2d term, its co-efficient being 62 its 3d part is 2 ; put therefore x = 2 + 2; then

x3 = 23 + 6z2 + 12z + 8 -6.72

24Z 24 + 10x =

+ 10Z + 20

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or z3

100

theref. the sum z3

2z + 4 8

2z Here then a= 2,6 = 4,0 =

şid=2. Theref. 3d+v(de + c3)=32 +v(4-09)=32+v

$2 + OV 3 = 1 57735 and 3d-V (d2 +63)=12-V (4-7)=32-v= 32

v 3 0.42265 then the sum of these two is the value of z = 2. Hence x = 2 + 2 = 4, one root of x in the eq. x3

10r = 8. To find the two other roots, perform the division, &c. as in p. 261, thus: x – 4) 33 6r? + 10x - 8( 32 23 + 2

4x2

10

63" +

3.

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Hence x2 2.0 -2, or x? - 2x + 1 1, and I-1 I IV

1;= =1+ ✓ I or =1-N-1, the two other sought. Ex. 2. To find the roots of x3 9x2 + 287 = 30.

Ans. I = 3, or 3 + V - 1, or = 3 V-1. Ex. 3. To find the roots of x3 7x3 + 14x 20. Ans. x = 5, or Itv - 3, or=1

3.

OF SIMPLE INTEREST.

As the interest of any sum, for any time, is directly proportional to the principal sum, and to the time; therefore the interest of 1 pound, for 1 year, being multiplied by any given principal sum, and by the time of its forbearance, in years and parts, will give its interesi for that time.

That is, if there be put

the rate of interest of 1 pound per annum,

any principal sum lent,
t= the time it is lent for, and

the amount or sum of principal and interest; then

the interest of the sum |, for the time t, and conseq. ptprt or p X(1 + rt)=a, the amount for that time.

From this expression, other theorems can easily be deduced, for finding any of the quantities above mentioned; which theorems collected together, will be as below:

Ist, a =p + pirt, the amount,

a

is prt

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For Example. Let it be required to find, in what time any principal sum will double itself, at ang rate of simple interest. In this case, we must use the first theorem, a =

= pt pri, in which the amount a must be made = 2n, or double the. principal, that is, n + fort 2p, or pri=p, or re=1; and

1 hence =>

Here,

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