as well as troublesome, to use a much greater number of figures than these in the several substitutions for the values of r. (i) . EXAMPLES. 1. Given x3 tox? itx=90, to find the value of x by approximation. Here the root, as found by a few trials, is nearly equal to 4. Let therefore 4=r, and r+z=X. 23=73 +372z+3r22 +23 Then r2 =p2 +2rz+22 =90. x =rtz. And by rejecting the terms 23, 3r22 and 23, as small in comparison with 2, we shall have 73+72+r+3r22+2r2+z=90; 6 Whence za .10 3r3+2+1 48+8+1 57 And consequently x=4.1, nearly. (i) It may here be observed, that if any of the roots of an équation be whole numbers, they may be determined by substituting 1, 2, 3, 4, &c successively, both in plus and in minus, for the unknown quantity, till a result is obtained equal to that in the question ; when those that are found to succeed, will be the roots required. Or, since the last term of any equation is always equal to the continued product of all its roots, the number of these trials may be generally diminished, by finding all the divisors of that term, and then substituting them both in plus and minus, as before, for the unknown quantity, when those that give the proper result will be the rational roots sought : but if none of them are found to sucoeed, it may be concluded that the equation cannot b resolved by this method ; the roots, in that case, being either irrational or imaginary as Again, if 4.1 be substituted in the place of r, in the last equation, we shall have 90-93 m2 - 90- 68.921-16.81-4.1 -.00283 3r2 +2r+1 50.43 +8.2+1 And consequently =4.17..00283=4.10283, for a second approximation. And, if the first four figures, 4.102, of this number be again substituted for r, in the same equation, a still nearer value of the root will be obtained ; and so on, far as may be thought necessary. 2. Given x2 +20x=100, to find the value of x by approximation. Ans. x=4.1421356 3. Given x3 +9x3+4x=80, to find the value of x by approximatioa. Ans. x=2.4721359 4. Given x4.38.3 +210x2 +538x+289=0, to find the value of x by approximation. Ans. Is 30.53565375 5. Given x5 +64-1023_112x2 - 2074 110 = 0, to find the value of x by approximation. Ans. 4.46410161 The roots of equations, of all orders, can also be determined, to any degree of exactness, by means of the following easy rule of double position; which, though it has not been generally employed for this purpose, will be found, in some respects, superior to the former, as it can be applied, at once, to any unreduced equation, consisting of surds, or compound quantities, as readily as if it had been brought to its usual form. RULE. Find, by trial, two numbers as near the true root as possible, and substitute' them in the given equation instead of the unknown quantity, noting the results that are obtained from each. Then, as the difference of these results is to the difference of the two assumed numbers, so is the difference between the true result, given by the question, and either of the former, to the correction of the number belonging to the result used; which correction being added to that number when it is too little, or subtracted from it when it is too great, will give the root required, nearly. And if the number thus determined, and the nearest of the two former, or any other that appears to be more accurate, be now taken as the assumed roots, and the operation be repeated as before, a new value of the unknown quantity will be obtained still more correct than the first ; and so on, proceeding in this manner, as far as may be judged necessary. (k) (& The above 'rule' for Double Position, which is much more simple and commodious than the one commonly employed for this purpose, is the same as that which was first given at p. 311 of the octavo edition of my Arithmetic, published in 1810. To this we may farther add, that when one of the roots of an equation has been found. either by this method or the former, the rest may be determined as follows: Bring all the terms to the left hand side of the equation, and divide the wbole expression, so formed, by the difference between the unknown quantity (3C) and the root first found; and the resulting equation will then be depressed a degree lower than the given one. Find a root of this new equation, by approximation, as in the first instance, and the number so obtained will be a second root of the original equation. Then, by means of this root, and the unknown quantity, depress the second equation a degree lower, and thence find a third root; and so on, till the equation is reduced to a quadratic; when the two roots of this, together with the former, will be the roots of the equation required. Thus, in the equation 23-15x2 + 63x =50, the first root is found, by approximation, to be 1.02804 Hence -102304(03 - 15x2 + 63.x -- 50(303 - 13 97196x+48.63627=0 And the two roots of the quadratic equation, x? 13.9719675 48.63627, found in the usual way, are 6.57653 and 7.39543. EXAMPLES. 1. Given 3+2?t=100, to find an approximate value of x. Here it is soon found, by a few trials, that the value of x lies between 4 and 5. Hence, by taking these as the two assumed numbers, the operation will stand as follows: First Sup. Second Sup. 4 5 25 125 71 : 84 Results 155 155 5 100 84 Therefore 1. 16: .225 And consequently x=4+.225=4225, nearly. Again, if 4.2 and 4.3 be taken as the two assumed numbers, the operation will stand thus : First Sup. Second Sup. 4.2 4.3 17.64 x2 :18.49 74.088 79.507. X 3 . So that the three roots of the given cubic 'equation 13 - 1572 +63x50, are 1.03804, 6.5765), and 7.39543; their sum Dzing 13, the coefficient of the second term of the equationas it ought to be when they are right. And consequently r=4.3–.036=4.264, nearly. Again let 4.264 and 4.265 be the two assumed numbers ; then First Sup. Second Sup. 4.265 . And consequently 2. Given (1#2 – 15)2 +xvx=90, to find an approximate value of x. Here, by a few trials, it will be soon found, that the value of x lies between 10 and 11; which let, therefore, be the two assumed numbers, agreeably to the directions given in the rule. Then Second Sup. XxX .36.482 56.622 121.122 56.622 Results 121, 122 Hence 64.5 31.122 ; .482 And consequently a=11_482=10.518 Again, let 10.6 and 10.6 be the two assumed numbers, |