Ex. 7. To find the value of in the equation r3 + 2x2 -23.r 70. Ans. r513457.

Ex. 8. To find the value of r in the equation r3 — 17x2 +54x350. Ans. x = 14·95407. Ex. 9. To find the value of r in the equation 4-3x2 75x=10000. Ans. 10.2609. Ex. 10. To find the value of r in the equation 2xa — 16x3 +40x2-30x = −1. Ans. 1.284724.

Ex. 11. To find the value of x in the equation x2 + 2x1 † 3x3 + 4x2 + 5x = 54321. Ans. x 8414455. Ex. 12. To find the value of x in the equation_x* = 123456789. Ex. 13. Given 2x4 -7x3 + 11x2 - 3x =

Ans. x

8.6400268.

11, to find x. Ex. 14. To find the value of r in the equation

(3.r2 — 2.√/ x + 1 )3 − (x2 − 4x √ x + 3 √/ x )3.

= 56.

Ans. 18-360877.

To resolve Cubic Equations by Cardan'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 Cardan'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. z3 azb, that is, wanting the second term, or the term of the 2d power z2. Therefore, after any cubic equation has been reduced down to its final usual form, x3 +px2 + qx = r, freed from the coefficient of its first term, it will then be necessary to take away the 2d term pr2; which is to be done in this manner: Takep, or of the coefficient of the second term, and annex it, with the contrary sign, to another unknown letter z, thus z-p; then substitute this for x, the unknown letter in the original equation x3 +px2 + qx = r, and there will result this reduced equation 23 azb, of the form proper for applying the following, or Cardan's rule. Or take ca, and db, by which the reduced equation takes this form, z3 ★ 3c ≈≈ 2d.

Then

Then substitute the values of c and d in this

form, z=3d+ √(d2 + c3) +3⁄4d− √ (d2 + c3),

or z=d+

C

3d + √ (d2 + c3)−3⁄4√d + √ (ď2+c3),

and the value of the root z, of the reduced equation 3* azb, will be obtained. Lastly, take x=z-p, which will give the value of x, the required root of the original equation x3 + px2 + qx =r, first proposed.

One root of this equation being thus obtained, then depressing the original equation one degree lower, after the manner described P. 250 and 251, the other two roots of that equation will be obtained by means of the resulting quadratic equation.

Note. When the coefficient a, or c, is negative, and c3 is greater than d2, 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 x3-6x2 + 10x=8. First, to take away the 2d term, its coefficient being - 6, its 3d part is 2; put therefore x = x + 2; then

-

Theref. 3d+√(d2 +c3) = 3/2+√ (4−287) = 3/2 + √12070

=

and 3d − √ (ď2+c3) = 3/2 − √√/ (4 −337) = 3/2 — √/ 2200 =

then the sum of these two is the value of x = 2.

27

Hence x = x + 2 = 4, one root of x in the eq. x3 — 6x2 +

10x=8.

To find the two other roots, perform the division, &c, as in p. 251, thus:

x-4) x3-6x2 + 10x-8 (x2 - 2x + 2 = 0

x3-4.x2

- 2x2+10x
-2x2 + 8x

2x-8

2x-8

Hence

Hence x2-2x=

= ±√1; x = other sought.

2, or x2-2x+1=1, and r- 1 1 + √−1 or 1-1, the two

--

Ex. 2. To find the roots of x3-9x2 + 28x = 30. Ans. 3, or 3+1, or 3-√-1. x =

Ex. 3. To find the roots of x3 —7x2 + 14x = 20. Ans. 5, or 1+/-3, or 1-√ −3. x = = =

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 interest for that time. That is, if there be put

= the rate of interest of 1 pound per annum, pany principal sum lent,

t = the time it is lent for, and

=

a = the amount or sum of principal and interest; then is prt the interest of the sum p, for the time t, and conseq. p+prt or p× (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:

1st, ap + prt, the amount,

For Example. Let it be required to find, in what time any principal sum will double itself, at any rate of simple

interest.

=

In this case, we must use the first theorem, a = p + prt, in which the amount a must be made 2p, or double the principal, that is, p + prt = 2p, or prt = p, or rt = 1;

1

and hence ==

Here,

Here, being the interest of 17. for 1 year, it follows, that the doubling at simple interest, is equal to the quotient of any sum divided by its interest for 1 year. So, if the rate of interest be 5 per cent. then 100 520, is the time of doubling at that rate.

Or the 4th theorem gives at oncé

t =

a-p

2p-p 2-1 1

= the same as before.

COMPOUND INTEREST.

BESIDES the quantities concerned in Simple Interest, namely,

p the principal sum,

a

the rate or interest of 11. for 1 year,

the whole amount of the principal and interest, t = the time,

there is another quantity employed in Compound Interest, viz. the ratio of the rate of interest, which is the amount of 17. for 1 time of payment, and which here let be denoted by R, viz.

Řir, the amount of 11. for 1 timě.

Then the particular amounts for the several times may be thus computed, viz. As 17. is to its amount for any time, so is any proposed principal sum, to its amount for the same time; that is, as

:

R: p
: PR, the 1st year's amount,
R pR PR2, the 2d year's amount,
R; PR2: pR3, the 3d year's amount,
and so on.

Therefore, in general, pRta is the amount for the t year, or t time of payment. Whence the following general theorems are deduced:

"

From which, any one of the quantities may be found, when the rest are given.

As to the whole interest, it is found by barely subtracting the principalp from the amount a.

Example. Suppose it be required to find, in how many years any principal sum will double itself, at any proposed rate of compound interest.

In this case the 4th theorem must be employed, making = 2; and then it is

t =

log. a-log.plog. 2p-log.p log. 2

log. R.

=

=

log. R. log. R

So, if the rate of interest be 5 per cent. per annum; then R=1+05 1·05; and hence

t =

log. 2 log..1.05

that is, any sum doubles itself in 14 years nearly, at the rate of 5 per cent. per annum compound interest.

Hence, and from the like question in Simple Interest, above given, are deduced the times in which any sum doubles itself, at several rates of interest, both simple and -compound; viz.