Ex. 8. How many real roots has the equation x-12x2+12x-3=0? Ex. 9. How many real roots has the equation x-8x+14x2+4x-8=0? Ans. Four. Ans. Four. PROPOSITION XIV. (324.) To discover a method of elimination for equations of any degree. The principle of the greatest common divisor affords one of the most general methods for the elimination of unknown quantities from a system of equations. Suppose we have two equations involving x and y reduced so the form of A=0, If we proceed to find the greatest common divisor of A and B, we shall have, according to Art. 249, A=QB+R. But since A and B are each equal to zero, it follows that R must equal zero. Hence we see that, if we divide one of the polynomials by the other, as in the method of finding the greatest common divisor, each successive remainder may be put equal to zero. If we arrange the polynomials before division with reference to the letter x, we shall at last obtain a remainder which does not contain x; which remainder, being put equal to zero, is the equation from which x has been eliminated. Ex. 1. Eliminate x from the equations Divide the first polynomial by the second, as follows: x2+y' -13x+y-5 −(y−5)x+ y2-13 —(y—5)x— y2+10y-25 2y-10y+12= remainder, This remainder we have already proved must be equal to zero; that is, 2y2-10y+12=0, an equation from which x has been eliminated. Ex. 2. Eliminate x from the equations If we have three equations containing three unknown quantities, we must first eliminate one of the unknown quantities by combining either of the equations with each of the others. We thus obtain two new equations involving but two unknown quantities, from which we may obtain a final equation involving but one unknown quantity. Ex. 5. Eliminate x and y from the equations xyz-c=0, xz+xy+yz-b=0, Ans. z-az+bz-c=0. Ex. 6. Eliminate x and y from the equations Ans. z-72z+1930%-22824x2+z+100476=0. SECTION XX. SOLUTION OF NUMERICAL EQUATIONS. (325.) We will first consider the method of finding the integral roots of an equation, and will begin with forming the equation whose roots are 2, 3, 4, and 5. This equation must be composed of the factors. (x−2) (x−3) (x−4) (x−5)=0. If we perform the multiplication (which is most expeditiously done by the method of detached coefficients shown in Art. 64), we obtain the equation. x2-14x3+71x2-154x+120=0. We know that this equation is divisible by x-5. Let us perform the division by the method of detached coefficients shown in Art. 80. A B C D V α Supplying the powers of x, we obtain for a quotient x3-9x2+26x-24=0. This operation may be still further abridged, as follows: Represent the root 5 by a, and the coefficients of the given equation by A, B, C, D, . . . . . V. We first multiply -a by A, and subtract the duct from I B; the remainder, -9, we multiply by -a, and ubtract the product from C; the remainder, +26, we multiply again by -a, and subtract from D; the remainder, -24, we multiply by —a, and, subtracting from V, nothing remains. If we take the root a with a positive sign, we may substitute addition for subtraction in the above statement; and if we set down only the successive remainders, the work will be as follows: A B C D Va and the rule will be, 1-14+71-154+12015 1- 9+26- 24, Multiply A by a, and add the product to B; set down the sum, multiply it by a, and add the product to C; set down the sum, multiply it by a, and add the product to D, and so on. The final product should be equal to the last term V, taker with a contrary sign. The coefficients above obtained are the coefficients of a cubic equation whose roots are 2, 3, 4. The equation may therefore be divided by x-4, and the operation will be as follows: 1-9+26-24|4 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 x-2. These three operations of division may be exhibited together as follows: 1-14+71-154+120/5, first divisor. (326.) The method here explained 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. bund. Ex. 2. Find the seven roots of the equation x+x-14x-14x1+49x3+49x2-36x-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+2-12-26+23+72+36 1+4-4-34-45-18 1+7+17+17+6 1+6+11+ 6 1+5+ 6 1, first divisor. 2, second divisor. 1, fourth divisor. -3, seventh divisor -1, −1, −2, -3. Ex. 3. Find the six roots of the equation xo+5x3—81x1—85x3+964x2+780x−1584=0. 1+ 5-81- 85+964+ 780-1584 1+ 6-75-160+804+1584 1+10-35-300-396 1+16+61+ 66 1+14+33 1+11 The six roots, therefore, are 1, 4, 6, -2, -3, -11. Ex. 4. Find the five roots of the equation 1. 4. 6. 2. 3. -11. |