(−√9+√16+√/25)=}(−3+4+5)=+3,, (+9-16+/25)=(+3-4+5)=+2, x={(+√9+√16—√√√25)=(+3+4—5)=+1, And .., the four roots of the equation are 1, 2, 3, and 6. 1. Given x1-55x2-30x+-504-0, to find the four roots or values of x. Ans. 3, 7, 4, and -6. 2. Given x+2x3-7x2-Sx——12, to find the four roots or values of x. Ans. 1, 2, 3, and -2. 3. Given 1-8x3+14x2+4x=8, x2-17x2-20x-6—0, x1—3x -4x=3, x1— *19x+123x2-302x + 2000, to find the four roots or values of x. x* 1 2+/7,-2-7. Ans. 15, 5: Ans. 1-5.) 3. Ans. (1.02804, Ans.3, 3. + 3, { 4.00000 7.39543 4. Given x-27x+162x+356x-12000, x-12x2+12x —3—0, x1—36x2+72x-36-0, to find the four roots or values of z in each. Ans. 13.15306, 2.05608, -3.0000, .60601, -3.907572 14.7908. 2.85808, .443274 Ans. 0.8729836, 1.2679494, 4.7320506, -6.8729836 5. Given 21 5x3 +13x2· 17x+12 = 0, to find the 4 roots of the equation, which are all imaginary I have x2 2x+3=0, or 22. 3x + 4 - - 0, Ans. *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 whole expression so formed by the difference between the unknown quantity (x) 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 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 -15263x-50, the first root is found by approximation to be 1.02804. Hence x-1.02804) 15x2+63x-50(x2-13.97196x+48.63627=0. And the two roots of the quadratic equation, x2-13.97196x -48 63627, found in the usual way, are 6.57653 and 7.39543. So that the three roots of the given cubic equation x3-15x+ 63x-50, are 1.02804, .6.57653, and 7.39543; their sum being =15, the coefficient of the second term of the equation, as it ought to be when they are right. Of the resolution of Equations by approximation, or the method of successive substitutions. 102. Equations of the fifth power, and those of higher dimensions, cannot be resolved by any general rule, or algebraic formula, that has yet been discovered; except in some particular cases, where certain relations subsist between the coefficients of their several terms; or when the roots are rational, and, for that reason, can be easily found by means of a few trials. See p. 113. In these cases, therefore, recourse must be had to some of the usual methods of approximation, among which that commonly employed is the following, which is universally applicable to all kinds of numeral equations, whatever may be the number of their dimensions; and which, by continuing the process to a sufficient length, will give the value of the root sought to any required degree of exactness. See Index. RULE I. Find, by trials, a number nearly equal to the root sought, which call r; and let z be made to denote the difference between this assumed root and the true root z. Then, instead of x, in the given equation, substitute its equal r+z, or r―z, according as r is too little or too great, and there will arise a new equation, involving only z and known quantities. Reject all the terms of this equation in which z is of two or more dimensions, and the approximate value of z may then be determined by means of a simple equation. And if the value thus found be added to r, when it is too little, or subtracted from it when it is too great, it will give a near value of x, or of the root required. But as this approximation will seldom be sufficiently exact, the operation must be repeated, by substituting the number last found, for r, in the abridged equation exhibiting the value of z; when a second correction of z will be obtained, which being added to, or subtracted from that number, will give a nearer value of x than the former. And by again substituting this last number for r, in the above mentioned equation, and repeating the same process as often as may be thought necessary, a value of z, and consequently of the root sought, may be found to any degree of accuracy. Note. The decimal part of the root, as found both by this and the next rule, will in general about double itself at each eperation; and therefore it would be useless, as well as troublesome, to use a much greater number of figures than these, in the several substitutions for the values of r. 1. Given x+x+x=90, to find the value of x, by approximation. Here the root is soon found, by a few trials, to lie between 4 and 5; but nearer to 4 than to 5. Let therefore 4-r, and x-r+z, Then x3=r3+3r2z+3rz2+z3 x=r+z 90. And, by rejecting the terms 23, 3rz2 and z2, or, to avoid trouble, omitting them in the operation, as being small in comparison with z, we shall have r3+r2+r+3r2z+2rz+z=90; 90772 -r 90-64-16-4 Whence z= 3r2+2r+1 6 48+8+1 57 10, and, consequently, a 4.1, nearly. Again, if 4.1 be substituted in the place of r, in the last equation, we shall have. And consequently -4.1+.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, an approximate value of x may be obtained, to six or seven places of decimals. And by proceeding in the same manner, the root may be found still more correctly. 1. Given 20x=100, to find the value of x by approximation. Ans. x 4.1421356237. Given x3-9x=10, to find the value of x by approximation. Ans. 3.4494897428. +9x+4x=80, to find the value of x by approxiAns.x-2.47213596. Here, by a few trials, the root is found to be between 2 and 3, and very nearly equal to 2.5; let, therefore, ar+z=2,5+z; then 3. Given mation. 80, and z = 2:3 9x2 4x 3r2z + &c. 9r2 + 18r z + &c. = 4r + 4 z + 9p2 + 4r+z(3r2+18r+4r) 80392—4r -1.875 .0276. Whence 3r2+18r+4r 67.75 x=2.5—.0276—2.472, the root nearly. And if this value be substituted in the above expression for z, we get x=2.4721359. 4. Given 38x+210x2+538x+2890, to find the value of x by approximation. Ans. x=30.5356537529. Here x is found, by trial, to be very nearly 30; let, therefore, xr+z=30+z; then whence ―30.56, nearly. Again, taking 30.5 for the root, and substituting it instead of r in the expression for z, we get z = 30.5356. 5. Given +6x-10x3-112x2-207x-110 = 0, to find the value of z by approximation. : Ans. 4.46410161. By a few trials, x is found to be between 4 and 5, and nearly equal to 4.5. Let, therefore, xr+z=4.5+z; then 75 5rtz + &c. 24r3z + &c. 107-3 30r2z + &c. 112,2 207x= 207r 207z 756-10-112r2-207r+ z(5r1+24r3—30r2-227-207)=110; therefore 110—75—6r1+10r3+112r2+207r 51+243-302-227r-207 -.0354; whence = 4.5—.035—4.465; and, taking this for the root, and substituting again in the value for z, we get x=4.464101, the root nearly. 103. 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 certain cases, superior to the former, as it can be applied at once to any unreduced equation, consisting of surds, or compound quanties, as readily as if it had been brought to its usual form. RULE II. Find, by trial, 'two numbers nearly equal to the root sought, 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 ques tion, and either of the former, to the correction of the assumed 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. 1 Given 3+3x2+3x=130, to find a near approximate value of x. Ans. x 4.078753. 2. Given 23+10x2+5x=2600, to find a near approximate value Ans. 11.00673. of x. It is soon found by trial that x is rather more than 11; let, therefore, 11 and 11.1 be taken for the assumed number; then fore, 1.2 and 1.3 be assumed for the numbers; then 1st Sup. 2d Sup. 1331 1367.631 : = .0067; whence .4199 59.231 11.0067 11.0067, Ans. .1 X.0801 : 3. Given 2-16x+40x230x+1=0, to find a near value of x. Dividing by 2, we have 21 8x+20x-15x=-5. By a few trials it is found that x is a little greater than 1; let, there .4199 1.2 .5305 : .1 :: .0801 =.01509, whence .5305 x=1.3-.0151-1.2849. Now take 1.284 and 1.285 for the assumed numbers, and repeat the operation. Then x will come out 1.284724. 4. Given 2+2x-3x+4x2+5x=54321, to find a near value of x. By a few trials x is found to be nearly 8; take, therefore, 8.4 and 8.5 for the assumed numbers; then 1st Sup. 41821.19424 25 44370.53125 2d Sup. 9957.4272 =2a410440.1250 1778.112 282.24 42. 53880.97344 3x3 = 1842.375 4x2 = 289. 5x = 42.5 results 56984.53125 8.5 Therefore 53880.97344 8.4 3103.55781 : .1 56984.53125 54321 440.02656 |