17. If the coefficient of the highest power of x in a rational integral equation with integral coefficients be unity, the equation cannot have a rational fractional root. For, by the preceding article, the denominator of any fractional root must be a factor of this coefficient, and hence be 1. Surd Roots. 18. If the coefficients in a rational integral equation be rational, quadratic surds enter in conjugate pairs. That is, if a + √b be a root of ƒ (x) = 0, then a −√b is also a root. Divide f(x) by (x- a − √b) (x − a + √b), and let Q stand for the quotient, and Rx + S for the remainder, if any. Then, f(x) = (x − a − √b) (x − a + √b) Q + Rx + S. f(a + √b) = R(a + √b) + S = Ra + S + R√b. Ra+S+R√b = 0. Whence, by Ch. XIX., Art. 31, R√b=0, (1) and Ra + S = 0. From (1), since b‡0, R = 0; and then, from (2), S = 0. f(x) = (xa√√b) (x − a + √b) Q, and ab is also a root. (2) The quadratic factor (x-a-b)(x−a+√√b), = (x—a)2 —b2, corresponding to a pair of conjugate surd roots, is rational in a and b. Imaginary Roots. 19. If the coefficients of a rational integral equation be real, imaginary or complex roots enter in conjugate pairs. That is, if a+bi be a root of ƒ (x)=0, then a−bi is also a root. The proof is similar to that given in the preceding article. The quadratic factor (x — a − bi)(x − a + bi), = (x − a)2 + b2. corresponding to a pair of conjugate complex roots, is real. Ex. One root of the equation 1 - 5x3+3x2+19 x − 30 = 0 is 2+1; solve the equation. The quadratic factor corresponding to the roots 2+√1 and 2 √1 is (x-2-√1)(x − 2 + √ − 1), = (x − 2)2 + 1, −= x2 − 4 x + 5. Determining the other quadratic factor, we have 24-5x+3x2+19 x 30 (x2-4x+5)(x2-x-6). 260 are found to be -2, 3. Therefore the required roots are 2, 3, 2±√-1. - EXERCISES IV. Find the factors of the first members of, and hence solve, the following equations. Also solve by Art. 11: 1. x3 + 3x2 + 2x+6=0; one root: -2. 19x+20=0; one root: √3. 0; one root: 1 + √2. 23x2+4x+ 190; one root: 8 x3 + 21 x2 26x+14=0; one root: 3 7. 36 x 24x3- 143 x2-146 x 500; one root: 8. x5 16 x3- 4 x2 - 17 x 4 = 0; two roots: i, 2 + √5. To transform an Equation into Another whose Roots are any Multiples of the Roots of the Given Equation. into another equation whose roots are k times the roots of the given equation. and Let y be the unknown number of the new equation. Then ALGEBRA. [CH. XL That is, to form an equation whose roots are k times the roots of a given equation: Multiply the coefficient of the second term by k; that of the third term by k2; that of the fourth term by k3; and so on. Ex. Form the equation whose roots are 3 times the roots of the equation 2-2-x+6= 0. or We have y3-3.2y-32 y + 33·60, y3-6 y2-94 + 162 = 0. 21. Ex. 1. Form the equation whose roots are ten times the roots of the equation 2 - 3 3 + 8 x2 + 2 x − 19 = 0. The transformed equation is or, y-10 x 3y+102 x 8 y2+ 10 x 2 y 10' x 19= 0, y30y+800 y2+ 2000 y 190,000. To obtain the transformed equation, when the multiplier is 10: Annex one cipher to the coefficient of the second term; two ciphers to the coefficient of the third term; three ciphers to the coefficient of the fourth term; and so on. If any term be wanting, it must be supplied mentally with a zero coefficient. Ex. 2. Form the equation whose roots are ten times the roots of the equation -5x+6=0. We have y3 - 500 y 6000 = 0. 22. An important application of this transformation is to obtain an equation in which the coefficient of the highest power of the unknown number is 1 and the other coefficients are integral. Ex. Transform the equation 12x-5x-3x+1=0 into an equation without fractional coefficients, in which the coefficient of the third power of the unknown number shall be 1. Dividing by 12, x3- √31⁄2 x2 - 4x + 11⁄2 = 0. - Forming an equation whose roots are k times the roots of the given equation, we have y3 — ég ky2 — 24 k2y + j1⁄2 k3 = 0. We find by inspection that 12 is the least value of k which will make all the coefficients integral. Substituting 12 for k, we obtain y3 – 5 y2 – 42 y + 144 = 0. To transform an Equation into Another whose Roots are Those of the Given Equation with Signs Changed. 23. Let f(x) = (x − r1)(x — V2) ··· (x — rn) = 0 ... ... (1) (2) - rn. + An-2x2 + An-1x + An= = 0, + ( − 1)"−1an-19 + ( − 1)′′a, = 0. That is, to obtain the transformed equation: Change the sign of every alternate term, beginning with the second; or the signs of the terms containing odd powers of x. If any term be wanting it must be mentally supplied with a zero coefficient. Ex. Transform the equation +5x-3x-2=0 into another whose roots are those of the given equation with signs changed. 24. If the signs of the terms in the equation. 24-3x-4x2+2x-3 be arranged in order, we have + - -+ A Permanence of Sign is a succession of two like signs, as A Variation of Sign is a succession of two unlike signs, as +- or In the above equation there is one permanence of sign and three variations. 25. If f(x) be multiplied by x-r, the number of variations of sign in the product will be at least one greater than the number of variations of sign in f(x). Grouping in parentheses all successive terms which have like signs, we have +(+x8+2x2) - ( + 5 x2 + x2 +4 x1) + ( +7 x3) − ( + 3 x2) + ( + 4x+1) +x-1 +(+x3 + 2 x8 ) − ( + 5 x2 + x + 4 x3 ) + ( +7 x1) − ( + 3 x3) + ( + 4 x2 + x) ( x8) − ( +2 x7 −5 x6 − x·3) + ( +4 x1) − ( + 7 x3) + ( + 3 x2 − 4 x) − 1 +(+x+×8) − ( +7 x7 −4 x6 +3 x3) + ( + 11 xa ) − ( + 10 x3) + ( +7 x2 — 3 x) — 1 - - Observe first that in the given expression, as grouped, the number of variations of sign is one less than the number of groups; in fact, that in counting variations we need give attention only to the signs preceding the parentheses. Also, that the grouping in the partial products and in the result corresponds to that in the given expression, although thereby variations of sign enter in some of the groups. We see that the number of variations of sign in the signs before the parentheses in the product is one more than in the given expression. That this should be so, can be inferred from the following considerations. The distribution of signs in the first partial product is evidently the same as in the given expression. In the second partial product there is no term in 29. Therefore in the result the sign of the first term in the first parentheses, and therefore the sign before the parentheses, is the same as in the given expression. The first term in the second parentheses is obtained by multiplying the last term in the first parentheses in the given expression, + (+ 2 x2), by - 1, and is (+2x), of the same sign as the like term in the first partial product. Therefore, in the result the sign of the first term in the second parentheses, and therefore the sign before the parentheses, is the same as in the given expression. Similar considerations hold for the other groups. Therefore no variation of sign in the signs of the groups in the given expression is lost. But to the last term in the second partial product there is no corresponding term in the first partial product. Since this term is obtained by multiplying the last term in the last parentheses by 1, its sign will be opposite to that before the last parentheses. Hence at this point one additional variation of sign in the signs of the groups is obtained. |