equation one degree lower, and thence find a third root, and so on, till the equation be reduced to a quadratic; then the two roots of this being found, by the method of completing the square, they will make up the remainder of the roots. Thus, in the foregoing equation, having found one root to be 102804, connect it by minus with x for a divisor, and take the given equation with the known term transposed for a dividend: thus, *—1*02804)x3-15x2+63x-50(x2-13′97196x+48·63627=0. Then the two roots of this quadratic equation, or x2-1397196x=-48·63627, by completing the square, are 6'57653 and 7'39543, which are also the other two roots of the given cubic equation. So that all the three roots of that equation, viz. x3-15~2+63x=5°, And the sum of all the roots is found to be 15, being equal to the coefficient of the second term of the equation, which the sum of the roots always ought to be, when they are right. NOTE 3. It is also a particular advantage of the foregoing rule, that it is not necessary to prepare the equation, as for other rules, by reducing it to the usual final form and state of equations. Because the rule may be ap-. plied at once to an unreduced equation, though it be ever so much embarrassed by surd and compound quantities. As in the following example: 3. Let it be required to find the root of the equation 14420) + √196x*—+24=114, or the value of x in it. By By a few trials but little above 7. then that x=8. it is soon found, that the value of is Suppose therefore first, that x=7, and As 2'469 : 1 :: 0710 : 0'2 nearly. 7°0 7°2 nearly. Suppose again ➡72, and, because it turns out too great, suppose also x=71. *515: 123 : 'I: 024 the correction. 7.100 Therefore 7124 nearly, the root required. NOTE un NOTE 4. The same rule also, among other more difficult forms of equations, succeeds very well in what are called exponential equations, or those, which have an known quantity for the exponent of the power; as in the following example. 4. To find the value of x in the exponential equation *100. For the more easy resolution of this kind of equations, it is convenient to take the logarithms of them, and then compute the terms by means of a table of logarithms. Thus, the logarithms of the two sides of the present equation are, xX log. of x2, the log. of 100. Then by a few trials it is soon perceived, that the value of x is somewhere between the two numbers 3 and 4, and indeed nearly in the middle between them, but rather nearer the latter than the former. By taking therefore first x35, and then x=36, and working with the logarithms, the operation will be as follows: As 0984511 :: '002689 : 000273. Which correction, taken from 3.60000 On trial, this is found to be very little too small. = Take therefore again x 3'59727, and next x3'59728, and repeat the operation as follows: 0'0000099 difference of the errors. Then, As 0000099 00001 :: 0000047 : 000000474747 Which correction, added to = 3*59728000000 Gives nearly the value of x 3'59728474747 5. To find the value of x in the equation x3+1cx2+ 52600. Ans. 1100673. 6. To find the value of x in the equation x3-2x=5• Ans. 200455 7. To find the value of x in the equation x3+2x2 ---23x=70. Ans. x 51349. 8. To find the value of x in the equation x3-17x2+ 54x350. Ans. 14'95407. 9. To find the value of x in the equation x-3x275*10000. Ans. x 10'2615. 10. To find the value of x in the equation 2x2-16x3 +40x230x=—1. Ans. x1284724. II. To find the value of x in the equation x+2x2 + 3x3+4x+5x=54321. Ans. x8414455. 12. To find the value of x in the equation **= 123456789. Ans. x86400268. END OF ALGEBRA. |