number, may now be used as new values of r and r' for obtain ing a still nearer approximation, and so on. It is generally most convenient to assume two numbers which differ only by unity in the last figure on the right, or one of the values of already used, together with the approximate root, may be employed for the two assumed numbers. This method of approximation is applicable to many equations which can not be solved by either of the preceding methods. EXAMPLES. 1. Find one root of the equation x3+x2+x-100=0. When 4 and 5 are substituted for x in this equation, the results are 16 and +55. Hence Therefore 55+16:5-4::16:.22. We now assume the two values 4.2 and 4.3, and, substituting them for x in the given equation, we obtain the results -4.072 and +2.297. Hence 4.072+2.297: 4.3-4.2:: 2.297: .036. Therefore x=4.264 nearly. Assuming again the two values 4.264 and 4.265, and substituting them for x, we obtain the results -.027552 and +.036535. Hence Therefore .064087.001:: .027552: .0004299. 2. Find one root of the equation x3+2x2-23x-70—0. 3. Find one root of the equation x-3x2-75x-10000=0. 4. Find one root of the equation Ans. x=10.2610. x5+3x+2x3-3x2-2x-2=0. Ans. x=1.059109. 471. The different Roots of Unity.—The equation "—a would appear to have but one root, that is, x=Va; but, by Art. 436, it must have ʼn roots; that is, the nth root of a must have n different values. Unity must therefore have two square roots, three cube roots, four fourth roots, five fifth roots, six sixth roots, and so on. Ex. 1. Find the two roots of the equation a2=1. Extracting the square root, we find x=+1 or -1. Ex. 2. Find the three roots of the equation a3=1. Since one root of this equation is x=1, the proposed equation must be divisible by x-1; and dividing, we obtain x2+x+1=0. Now the roots of this equation are x=±±±√3. Hence the required roots are +1, §(−1+√−3), and 1(—1—√—3), which are the cube roots of unity; and these results may be easily verified. Ex. 3. Find the four roots of the equation x1=1. The square root of this equation is x2=+1, or =−1. Hence the required roots are +1, −1, +√−1, −√—1. Ex. 4. Find the five roots of the equation x=1. Since one root of this equation is a=1, the proposed equation must be divisible by x-1; and dividing, we obtain which, being substituted in equation (1), gives v2+v-1=0. This equation, solved by the usual method, gives v=−1+1√5, or v=—1—1√5. Whence x=[+√v2 −4], and x=}[v−√v2 −4], from which, by substituting the value of v, we obtain Ex. 5. Find the six roots of the equation x6=1. These are found by taking the square roots of the cube roots. Hence we have +1, −1, 1±1√—3, −1±1√3. Ex. 6. Find the four roots of the equation x=-1, or x2+1=0. The first member may be made a complete square by adding 2x2; that is, x2+2x2+1=2x2, These four values, together with the four values found in Ex. 3, are the eight roots of the equation x8 = 1. |