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In transforming an equation having fractional, into another with entire coefficients, in terms of another unknown quantity, it is important to have the transformed equation in the lowest possible terms. The least common multiple of the denominators will not necessarily be the least value of a that will give the required equation. If, in each case, the denominators be resolved into their prime factors, it will be easy to decide upon the powers of these factors to be taken as the factors of a.

The following illustration will render further explanation unnec

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into another of the same form with the smallest possible entire

coefficients.

Writing y for x and multiplying the second, third and fourth terms, by a, a3, a3, respectively, we have

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3
35 ay2+

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The denominators, resolved into their prime factors, are

7.5, 7.5.2, 7.5.23;

and assuming a 7.5 2, the equation may be written.

y3

=

7.5.2,

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7.5 which reduces to

y'-6y'+26y-850.

In this example, the least common multiple of the denominators is 7.5.23; and had this value been taken for a, instead of 7.5.2, the coefficients of the transformed equation would have been much larger than they are, as found above.

When a root of the transformed equation is known, the corresponding root of the original equation will be given by the relation

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COMMENSURABLE ROOTS.

430. A number is commensurable with unity when it can be expressed by an exact number of units or parts of a unit; a number which can not be so expressed is incommensurable with unity.

431. Every equation having unity for the coefficient of the first term, and for all the other coefficients, whole numbers, can have only whole numbers for its commensurable roots.

This being one of the most important principles in the theory of equations, its enunciation should be clearly understood. Such equations may have other roots than whole numbers; but its roots can not be among the definite and irreducible fractions, such as ?, 11, etc. Its other roots must be among the incommensurable

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quantities, such as √/2, (3)3, etc.; i. e., surds, indeterminate deci

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irreducible fraction, to be a root of the equation,

m

m-1

x+AxTM-1+Вx.... Tx+U = 0,

A, B, etc., being whole numbers.

Substituting this supposed value of x, we have

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Transpose all the terms but the first, and multiply by 6-1, and we have

m

α = −(AaTM~1+Bam-b.... TabTM-2+UbTM~1).

b

Now, as a and b are prime to each other, b can not divide a,

or any number of times that a may be taken as a factor; for

α

b

a being irreducible, Xa is also irreducible, as the multiplier a will

b

not be measured by the divisor b; therefore can not be expressed

b

in whole numbers. Continuing the same mode of reasoning,

am

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can not express a whole number, but every term in the other member of the equation expresses a whole number.

a

Hence, the supposition that the irreducible fraction is a root of b

the equation, leads to this absurdity, that a series of whole numbers is equal to an irreducible fraction.

Therefore, we conclude that any equation corresponding to these conditions can not have a definite commensurable fraction among its roots.

432. It has been shown (429) that an equation having fractional coefficients, that of the first term being unity, can be changed into another of the same form, with entire coefficients. The expression entire must there be understood in its algebraic sense; that is, the new coefficients being entire merely in algebraic form, may be irrational or imaginary. In the preceding article it is proved that if these coefficients are whole numbers, all the commensurable roots of the equation are also whole numbers; moreover, these roots must be found among the divisors of the absolute term; (428). If the divisors of the absolute term are few and obvious, those answering to the roots may be found by trial substitutions; but in most cases the labor will be abridged by the rule suggested by the following investigation:

Suppose a to be a commensurable root of the equation,

m- 1

x+Ax1+.... +Rx3 + Sx2+ Tx+U = 0.

Writing a for x, transposing all the terms except the last to the second member, and dividing through by a, we have

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-Ra'-Sa-T.

U

is an entire number; trans

α

U

Tto the first member of the last equation, make + T =

a

N,, and divide both members of the resulting equation by a; it then becomes

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The second member of this equation is a whole number; the first

member is therefore entire; and if -S be transposed to this mem

ber, and N+S be denoted by N2, we shall again have, after di

a

viding through by a, the equation,

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of which the second, and therefore the first member, is an entire number.

By continuing this process of transposition and division, we shall finally arrive at the equations,

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Every whole number which is a root of the proposed equation will satisfy all of the above conditions, from the first down to that expressed by the equation,

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by which the root will be recognized.

All of the commensurable roots of an equation of the assumed form may then be found by the following

RULE.-I. Write all of the exact divisors of the absolute term in a line, and beneath them write their respective quotients.

II. Add to these quotients, severally, the coefficient of the next to the last term, with its proper sign.

III. Divide such of these sums by the divisors to which they correspond as will give exact quotients, neglecting others.

IV. Add to these quotients the coefficient of the third term from the last, with its proper sign, and divide again as before, and so on, until the coefficient of the second term has been added to the preceding quotients, and these last in turn are divided by their respective divisors. Those divisors which correspond to the final quotients, minus 1, are roots of the equation.

NOTE.-Absent powers of the unknown quantity must be introduced with 10 for their coefficient.

EXAMPLES.

1. Required the commensurable roots (if any) of the equation,

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Quotients,

1, 2, 5, -10, 10, 5, 2,

1.

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Thus, there are two final quotients equal to -1; and the corres

ponding divisors are 1 and -5.

commensurable roots, 1 and -5.

Hence, the given equation has two

If we divide the given equation by (x−1)(x+5), or x2+4x—5, the quotient will be x2-2. Hence,

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and the four roots are 1, 5, +√2, −√2.

2. Required the commensurable roots of the equation, x2-6x' +11x-6=0.

Ans. 1, 2, 3.

3. Required the commensurable roots of the equation, x*+4x3— x2-16x-12 = 0. Ans. 2, -1, -2, −3.

4. Required the commensurable roots of the equation, x*—6x216x+21= 0. Ans. 3 and 1.

NOTE. Supplying the absent term, the equation will be x10x3-6x2 -16x+21= 0. In the operation, go through the form of adding 0.

5. It is required to find all the roots of the equation x-6x+ 5x+2x-100. Ans. —1, +5, 1+√—1, 1—√—1.

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