ficients become 21 and 63, which give 63-21=-3; consequently, 3 is a positive root. - 180÷30=-6, In like manner, by putting = 6, we get 30 and 180 for the coefficients of t and v, which give consequently 6 is the remaining positive root. ficients of t and v, and that the coefficient of v, divided by that of t, gives - 3 for the quotient, it follows that the equation is equivalent to (x-3)3 = 0. 5. To find the roots of n nan ra t n(n 1 2 n(n−1)(n −2), sa 1. 2 1. 2 Hence, by dividing a for the quotient; etc., for the coefficients of t and v. the coefficient of v by that of t, we get consequently, the equation is equivalent to (x − a)” = 0, without regard to the nature of n. Remark. It is hence clear that, whether n is positive or negative, rational or irrational, real or imaginary, we shall Lan-as+ etc., for the correct expansion of n(n − 1)(n-2) 3 (a)", according to the descending powers of a and the ascending powers of a. 6. To find the roots of 24 6x2 8x -3 = 0. Since the coefficients of t and v are 4x 12. 8 and - 12x2 - 24x-12, which have x2 + 2x + 1 = (x + 1)2 for a divisor, it follows that the given equation has three equal roots, having 1 for their common value. And dividing the equation by (x + 1)3 = ∞3 + 3x2 + 3x + 1, we get x 3 for the quotient, consequently we shall have a 3 for the remaining root of the equation. 7. To find the number of real roots of 24 15=0. - 8x2 Since the coefficients of t and v are 4x3 8.x + 8 and +24x60, by rejecting their common factor 4, they may clearly be reduced to a 2+2 and -2+6x-15; and putting x for x, these become 6x15, which will enable us to find the number of negative roots. - - - x3 + 2x + 2 and - 2x2 From 23-2x+2 and 2x2+6x-15, it is easy to perceive that the equation has one positive root, which lies between 1 and 2, whose first figures are easily found to be 1.9; and since a 2x + 2 and 2x2+6x 15 do not (both) change their signs for values of x that are greater than 1.9, it is clear that the equation has no other positive root. In like manner, from - x3 + 2x + 2 and 2x2-6x-15, it is easy to perceive that the equation has one negative root, which lies between 2 and -3; and that the equation has no other (real) negative root. Hence, the given equation has two real and two imaginary roots. 20 8. To find the number of real roots of 3x3 =0. The coefficients of t and v being 9-14x + 2 and - 72 +4x60, it is easy to show that the equation has only one real root, which lies between 2 and 3, the other roots being imaginary, since the coefficients do not change their signs when a is greater than 2. 9. To find the first figures of the roots of - 12x+8 =0," Here the coefficients of t and v are 3x2 12 and 24.x + 24, which will enable us to find the positive roots; and putting for a, they become 3.12 and 24x + 24, which will serve to find the negative roots. Putting 3 for a in 3x2 - 12 and 24x + 24, they become 13 x 5 and 96, and give 96 137 ; which, being greater than 13 3, it follows that a is greater than 3. Again, putting 4 for x, the preceding coefficients become 36 and 120, and give 120÷36=3, which, being less than 4, it follows that x is less than 4. 3' Hence, it results that 3 is the principal part of the negative root; and since 32-12 and 24 + 24 do not both change their signs when x is greater than 3, it is clear that the given equation has no other negative root. and To find the first figures of the positive roots, from 3x2-12 -24x24, we remark that when x= 0, they become 12 and 24, which have contrary signs, and it is clear that the least positive root must be found from 3-12 and 2424, when they have the same signs as - 12 and 24; and since x = 1 reduces 24x24 to 0, it is clear that the least positive root must be less than 1. Hence, after a few trials, we find that the least positive root lies between 0.6 and 0.7; consequently, its principal part is 0.6. Because the remaining root must be found from 3a2 - 12 and 24x24, after they have changed their signs, we soon find that 3 is its first figure. - 10. To find the first figures of the roots of 3- 2x 5=0. Here we have 3x2 2 and 4x 15 for the coefficients of t and v, from which to find the positive roots; and 3a2 2, 415 are the coefficients from which to find the negative roots. It is easy to show, from 3a2-2 and 4x-15, that the equation has no negative root; and from 32-2 and 4x 15, that it has only one positive root, whose first figure is 2. 11. To find the number of real roots of "x4 2x2+6x+ 100." - 4x2 + 18x + 40 Since Ꮞ 4x + 6 and cients of t and v for the positive roots, and and-418x + 40 those for the negative to show that the equation has no real roots. are the coeffi 4x3 + 4x + 6 roots, it is easy 12. To find the first figures of the roots of "203 +11æ3· 102x+181 = 0." - Here the coefficients of t and v for the positive roots are 3x2+22x-102 and 11x2 204x+543; and those for the negative roots are 3x2 22x 102 and 1122 + 204x + 543. Because the coefficients for the negative roots must clearly both be positive, it is easy to show that -17 is the principal part of the only negative root which the equation can have. Also, because the coefficients for the positive roots become 102 and 543, when 0 is put for x, it is clear that one positive root must be found from the coefficients, when the first is negative and the second positive, and that the second positive root must be found from the coefficients after they have changed their signs. Hence, we easily find that the equation has two positive roots which lie between 3 and 4, whose principal parts are 3.213 and 3.229. 13. To find the first figures of the real roots of “x — 5æ1 +223 — x2-1=0." 2x and - 10x+3x3 4x2 -- 6, Here 6x20x3-3x2 being the coefficients of t and v for the positive roots, and 6x+20x3 + 3x2 + 2x and 10x · 3x3- 4x2- 6 those for the negative roots, it is evident that the equation can have only one positive and one negative root, since the coefficients of v can not change their signs when x is positive. Because the substitution of 2 and 3 for x in the coefficients for the negative roots gives 16,206, and — 885, - 933 for the corresponding coefficients of t and v, it clearly follows that 2 is the first figure of the negative root. And in a similar way we get 2.2 for the first figures of the positive root. 14. To find the number of real roots of "10-10x3. -204 +-110." Because 10x3 — 80x7 4+1 and - 20x8 6x+9x 110 are the coefficients of t and v for the positive roots, and 10x+80x+4x2 + 1 and 20x 6x 9x110 those for the negative roots ( being positive), it is easy to perceive that the coefficients of v can not change their signs, and of course the equation can not have more than one positive and negative root; and it is easy to show that 3 and -3 are the first figures of these roots. 15. To find the first figures of the roots of "Sa3 1=0." Because 24x2 6 and - 12x 3 are the coefficients of t and for the positive roots, it is easy to perceive that the equation can have only one positive root, whose first figures are 0.9. And because 24x2 - 6 and 12x 3 are the coefficients for the negative roots, and that 12x-3 changes its 1 4 sign when a is greater than it is easy to perceive that one of the roots must be found by making a less than 1; 4 ; conse quently, we easily find that - 0.1 is the principal part of the root. Similarly, since 24x2-6 changes its sign when x is greater than 1, 2' we must evidently find the remaining negative root, by making a greater than ; indeed, its principal part is found to be - 0.7. 16. To find the roots of "x7 — 9x+6x2 + 15x3 12x2 7x+6=0," Here we have 7x-45x + 24x3 + 45x2 − 24x — 7, and 18x5 +18x+60x3- 60x2 - 42x+42, for the coefficients of t and v; and since the coefficient of v is reducible to (x − 1)3. (x + 1) × (− 18x2 + 42), it is easy to perceive that the coefficient of t is also reducible to (x-1)2. (x + 1). (7×3 + 7x2 - 31x-7). From the coefficients of t and v, it follows that the equation has (x-1)3. (x + 1)2 for a factor, and of course it has three roots whose common value is 1, and two roots whose common value is — 1. To find the unequal roots of the equation, it is clear that we may take 7x3 +7x2 31x7 and 18x2+42 for the coefficients of t and v for the positive roots; and putting-e for x, we may take - 7+7x2+31x-7 and 18.2+42 for the coefficients for the negative roots. After a few trials, we find that by putting 3 for x in the coefficients for the negative roots, they become - 40 and – 120, which give — 120 ÷ -120-40 = 3; consequently, - 3 - is a root of the equation. Similarly, by putting 2 for x in the coefficients for the positive roots, they become 15 and -30, which give −30÷ 152; consequently, 2 is a root of the equation. Hence, the proposed equation is reduced to x-9 + 6x+15x3-12-7x+6= (x - 1)3. (x + 1)2. (x-2). (x+3)= 0. |