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CHAPTER XXII. SOLUTION OF NUMERICAL EQUATIONS OF HIGHER DEGREES.

461. Equations of the third and fourth degrees can sometimes be solved by direct methods; but these methods are complicated, and are of limited application. No general solution of an equation higher than the fourth degree has yet been discovered. To obtain the roots of numerical equations of degrees higher than the second, we must generally employ tentative methods, or methods which involve approximation.

462. Commensurable Roots of an Equation.--Any equation having fractional coefficients can be transformed into another which has all its coefficients integers, and the coefficient of its first term unity, Art. 441, and such an equation can not have a root which is a rational fraction, Art. 440; that is, every commensurable root of this equation must be an integer. Every integral root of this equation is a divisor of the last term, Art. 439. Hence, to find the commensurable roots of an equation, we need only make trial of the integral divisors of the last term.

463. Method of finding the Roots. In order to discover a convenient method of finding the roots, we will form the equation whose roots are 2, 3, 4, and 5. This equation, Art. 436, may be expressed thus,

(3-2)(x-3)(-4)(x–5)=0. If we perform the multiplication here indicated, we shall obtain

24 - 14x3 +71.x2—154x+120=0. We know that this equation is divisible by x-5, and we will perform the operation by an abridged method. Since the coefficients of the quotient depend simply upon the coefficients of the divisor and dividend, and not upon the literal parts of the terms, we may obtain the coefficients of the quotient by operating upon the coefficients of the divisor and dividend by the

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usual method. To the coefficients thus found the proper letters may afterward be annexed. The operation may then be exhibited as follows:

A B C D v
1-14+71–154+1201-5, divisor,
1- 5

1-9+26—24, quotient.
- 9+71
- 9+45

+26-154
+26-130

24+120

24 +120. Supplying the powers of x, we obtain for a quotient

23-9x2 +26x-24=0. In applying this method of division, care should be taken to arrange the terms in the order of the powers of x; and if the series of powers of x in the dividend is incomplete, we must supply the place of the deficient term by a cipher.

The preceding operation may be still further abridged by performing the successive subtractions mentally, and simply writing the results. Represent the root 5 by r, and the coefficients of the given equation by A, B, C, D, .... V.

We first multiply —y by A, and subtract the product from B; the remainder, -9, we multiply by — 1, and subtract the product from C; the remainder, +26, we multiply by –r, and subtract the product from D; the remainder, —24, we multiply by —r, and, subtracting from V, nothing remains. If we take the root r with a positive sign, we may substitute in the above process addition for subtraction; and if we set down only the successive remainders, the work will be as follows:

A B C D у
1-14+71–154+120(5

1- 9+26- 24, and the rule will be

Multiply A by r, and add the product to B; set down the sum, multiply it by r, and add the product to C; set down the sum, mul

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tiply it by r, and add the product to D, and so on. The final product should be equal to the last term V, taken with a contrary sign.

The coefficients above obtained are the coefficients of a cubic equation whose roots are 2, 3, and 4. The polynomial may therefore be divided by x-4, and the operation will be as follows:

1-9 +26-24(4

1-5+ 6. These, again, are the coefficients of a quadratic equation whose roots are 2 and 3. Dividing again by x-3, we have

1-5+6(3

1-2, which are the coefficients of the binomial factor -2.

These three operations of division may be exhibited together as follows:

1-14+71-154+1205, first divisor.
1- 9+26- 24 4, second divisor.
1- 5+ 6

3, third divisor.
1-2

464. How to find all the Integral Roots. The method here ex. plained will enable us to find all the integral roots of an equation. For this purpose, we make trial of different numbers in succession, all of which must be divisors of the last term of the equation. If any division leaves a remainder, we reject this divisor; if the division leaves no remainder, the divisor employed is a root of the equation. Thus, by a few trials, all the integral roots may be easily found.

The labor will often be diminished by first finding positive and negative limits of the roots, for no number need be tried which does not fall within these limits. Ex. 2. Find the seven roots of the equation

2? + 20 — 1425 – 14x4 +4933 +4932 - 36x-36=0. We take the coefficients separately, as in the last example, and try in succession all the divisors of 36, both positive and negative, rejecting such as leave a remainder. The operation is as follows:

.

1+1-14-14+49 +49-36-36 1, first divisor.
1+2-12-26+23+72+36 2, second divisor.
1+4-4-34-45-18

3, third divisor.
1+7+17+17+ 6

-1, fourth divisor. 1+6+11+ 6

-1, fifth divisor. 1+5+ 6

-2, sixth divisor. 1+3

-3, seventh divisor. Hence the seven roots are

1, 2, 3, -1,-1, -2, -3. Ex. 3. Find the six roots of the equation

26 +525 -8124 -85x3 +964x2+780x—1584=0.
1+ 5-81 – 85+964+ 780–1584 1.
1+ 6-75-160+804+1584

4.
1+10-35-300-396

6.
1+16+61 + 66

2.
1+14+33
1+11

11. The six roots, therefore, are

1, 4, 6, -2, -3, -11. Ex. 4. Find the five roots of the equation

26 + 6* - 1023-112cc2-207-110=0.

1+6-10-112-207-110-1.
1+5-15- 97-110 - 2.
1+3-21-55

-5.
1-2-11
Three of the roots, therefore, are

-1, -2, -5. The two remainining roots may be found by the ordinary method of quadratic equations. Supplying the letters to the last coefficients, we have

22 -2.-11=0. Hence

x=-1+V12. Ex. 5. Find the four roots of the equation Oct - 1203 +472 - 72x+36=0.

Ans. 1, 2, 3, and 6. Ex. 6. Find the four roots of the equation

Xc4 +233 — 702 —8x+12=0.

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Ex. 7. Find the four roots of the equation

c* -5522 - 30x+504=0. Ex. 8. Find the four roots of the equation

24 - 25x2 + 60x-36=0. Ex. 9. Find the four roots of the equation

24 - 2x3 +19x-42=0. Ex. 10. Find the five roots of the equation

cco +5x4 +2°— 16x2 — 20x—16=0.

465. Incommensurable Roots.-If a high numerical equation is found to contain no commensurable roots, or, if after removing the commensurable roots, the depressed equation is still of a higher degree than the second, we must proceed by approximation to find the incommensurable roots. Different methods may be employed for this purpose; but the following method, which is substantially the same as published by Horner in 1819, is generally to be preferred.

Find, by Sturm's Theorem, or by trial, Art. 446, the integral part of a root, and transform the given equation into another whose roots shall be less than those of the preceding by the number just found, Art. 444. Find, by Art. 446, the first fig. ure of the root of this equation, which will be the first decimal figure of the root of the original equation. Transform the last equation into another whose roots shall be less than those of the preceding by the figure last found. Find, as before, the first figure of the root of this equation, which will be the second decimal figure of the root of the original equation. By proceeding in this manner from one transformation to another, we may discover the successive figures of the root, and may carry the approximation to any degree of accuracy required. Ex. 1. Find an approximate root of the equation

23 + 3x2 +5x=178. We have found, page 330, that this equation has but one real root, and that it lies between 4 and 5. The first figure of the root therefore is 4. Transform this equation into another whose roots shall be less than those of the proposed equation by 4, which is done by substituting y+4 for x. We thus obtain

y + 15y2 + 77y=46.

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