highest power of the unknown quantity in the given equation.

4. Divide that term of the progression, thus found, which stands against the term O in the first progression, by the ratio or common difference.

5. To the quotient last found, prefix the sign + or according as the progression is increasing or decreasing, and this number being substituted for the unknown quantity, will be found to be one of the roots of the equation.

Note. When there is more than one progression, the roots must be taken out of each.

EXAMPLES.

1. Let x3-x2-10x+6=0, be the equation proPosed.

Then, by substituting successively the terms of the progression 2, 1, 0, -1, instead of x, the work will stand as follows:

And-3, the term standing against o, being substituted for x, gives-27—9+30+6=0; and therefore 3 is a root of the equation.

2. Let 2x35x2+4x-10=0, be the equation proposed.

Then, by substituting successively the terms of the progression, 2, 1, 0, -1, -2, instead of x, the work will stand as follows:

Here 5, the term standing against o, being divided by 2, the common difference, gives 24; and this being substituted for x, gives 311-314+10-10=0; and, therefore, 21 is a root of the equation.

3. Let x4x3.

proposed*.

29x2. 9x+180-0 be the equation

Then, by substituting as before, the work will stand as

follows:

So that here are four progressions, and the numbers 3, 4, -3, and -5, being each substituted for x, make the whole equation vanish, and are therefore the roots required.

4. Given a 33x246x72-0, to find the values of , by the method of divisors. Ans +9, -2, and -4. 5. Given x4. ·4x3-19 x2+46x+120=0, to find the values of x, by the method of divisors.

Ans. +5, +4, --3, -2.

* Several other rules for discovering the roots of equations may be found in Newton's and Maclaurin's Algebra.

PROBLEM V.

To find the roots of cubic equations, according to the method of CARDAN.

RULE*.

1. Take away the second term of the equation, by problem second, and it will be reduced to this form: x3±ax=±b.

2. Substitute the values of a and b, with their proper signs, in the following expression, and it will give the Thus:

root required.

Toot

required.

The rule, from whence this method is derived, is x= ŵ36+√(4b2+1a3)+33/3b−√($62+13); and the investiga

tion of it is as follows:

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Let the equation, whose root is required, be x3+ axb.
And assume y+2=x, and 3yz=- -a.

Then, by substituting these values in the given equation, we shall have y3+3y2z+3yz2+z3+a×(y+z)=y3+z9+3yz×(y+2)+a X(y+2)=y3+23—a×(y+z)+a×(y+z)=y3+z3=b.

And if, from the square of this last equation, there be taken 4 times the cube of the equation yz=-ļa, we shall have y6—2y3z3 +26=62+4a3, or y3—23=√(62 +4a3).

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But the sum of this equation and (y+23b) is 2y3b+/ (62+14a3) and their difference is 223=b~√(b2-|-1a3); whence · √3/+b+✓(:462+21,a3), and z—

y is found

Note*. When a is negative, and a3 is greater than 162, the solution, by this rule, cannot be generally obtained.

EXAMPLES.

1. Let y3+3y2+9y=13, be the equation proposed; it is required to find the value of y.

1. In order to destroy the second term, let y=x-1; then y3—x3-3x2+3x—1 3y2+3x2-6x+3

And from hence it appears, that y+z, or its equal x, is = 3/36+√(462+21-a3)+ŵ§b—√(462+1a3), which is the theo

rem.

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This method of solving cubic equations is usually ascribed to Cardan; but the invention is not his. The authors of it were Scipio Ferreus, and Nicholas Tartalea, who discovered it about the same time, independently of each other, as is proved by M. de Montucla, in his Histoire des Mathématiques.

*This is called the irreducible case; and, notwithstanding several of the most eminent mathematicians in Europe have attempted the solution of it, no general rule has yet been discovered.

The usual method is by a table of sines, or by throwing the expression into an infinite series, and finding the sum of a certain number of terms, according to the degree of exactness required.

2. For a put 6, and for b 20, and we shall have

x=2, and consequently y=1=root required.

3. Given x3-6x=— -9, to find the value of x.

———1—2———3; that is, x=—-3= root required.

EXAMPLES FOR PRACTICE.

1. Given x3-6x2+10x=8, to find x.

Ans. x=4.

2. Given y3+30y=117, to determine y.
3. Given y3-36y=91, to determine y.
4. Given y3-3y=18, to determine y.
5. Given y3+24y=250, required y.
6. Given y6-3y4——2y2-8=0, to find y.

Ans. y=3.

Ans. y=7.

Ans. y=3.

Ans. y=5.05.

Ans. y=2.