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Commensurable Roots of any Equation.

is not always the one which leads to the most simple result. But when A has two or more equal factors, it is often the case that the substitution

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leads to an equation of the desired form, A' being the product of the prime factors of A, and each factor need scarcely ever be repeated more than once.

315. EXAMPLES.

1. Find the commensurable roots of the equation

64 x4

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328 x3 + 574 x2 393 x - 90 = 0.

Solution. We have, in this case,

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hence we may take A' equal to some power of 2; and it is easily seen that the third power will do, so that we may make

x= ty.

Hence the given equation becomes

y441y3574 y2-3144 y+57600.

The commensurable roots of which are found, as in art. 311, to be

y = 4, 6, 15, and 16;

so that the roots of the given equation are

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2. Find the commensurable roots of the equation

8334x2-79x+30= 0.

Ans. 1, §, and —6.

Commensurable Roots of any Equation.

3. Find the commensurable roots of the equation

24 x3 — 26 x2 + 9 x − 1 = 0.

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5. Find the commensurable roots of the equation

8x438349 x2-22x+3=0.

Ans.,, 1, and 3.

6. Find all the roots of the equation

6x37x2+39x+63 0

which has a commensurable root.

Ans., and

✔-251.

7. Find the commensurable roots of the equation 9x6+30 x5 +22 x4 +10 x3 + 17x2-20x+4=0.

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Value of Continued Fractions.

CHAPTER IX.

CONTINUED FRACTIONS.

316. A continued fraction is one whose numerator is unity, and its denominator an integer increased by a fraction, whose numerator is likewise unity, and which may be a continued fraction.

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317. Problem.

To find the value of a continued

fraction which is composed of a finite number of fractions.

Solution. Let the given fraction be

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and this method can easily be applied in any other case.

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An approximate value of this fraction is obviously obtained by omitting all its terms beyond any assumed fraction, and obtaining the value of the resulting fraction, as in the previous article.

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and each of these values is easily shown to be more accurate than the preceding; for the second value is what the first becomes by substituting, for the denominator a, the

1

b

more accurate denominator a+ ; the third is what the second becomes by substituting, for the denominator b, the

1
C

more accurate denominator b + ; and so on.

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