EXERCISES XI. Determine the number and location of the real roots of the following equations. Also, find each real positive root correct to three decimal places : 71. To transform an equation into another whose roots are the reciprocals of the roots of the given equation. or x = then each value of y is the reciprocal of a value of x. ... any" + an-1yn−1 + An-2yn−2 + That is, the coefficients of the transformed equation are the coefficients of the given equation written in reverse order. Ex. The equation whose roots are the reciprocals of the roots of the equation 2 - 3 x2 + 7 x + 5 = 0 is 5 y3+7y2-3y+2=0. 72. A Reciprocal Equation is one whose roots, in pairs, are reciprocals one of the other. That is, if r be a root of a reciprocal equation, 73. Now, if r be a root of the equation 1 is also a root. Equations (1) and (2) will in general be different; but if equation (1) will be a root of it also. be a reciprocal equation, 1 That is, each root of (2) is a root of (1), and vice versa, and the two equations are equivalent. Hence, corresponding coefficients must be equal or proportional, and That is, in a reciprocal equation, the coefficients of terms equally distant from the beginning and end are equal, or equal and opposite. Thus, 2x2+3x2 + 3x + 2 = 0, and ar1 bx3 + bx — a = 0, are reciprocal equations. - Observe that an equation of even degree has one middle term. When the equidistant coefficients are opposite in sign, this term must be wanting, since it cannot be equal and opposite to itself. Reciprocal Equation of Odd Degree. be a reciprocal equation of odd degree. Grouping the equidistant terms, we have аo(x2±1) + α1x(x2−2 ± 1) + A2x2 (xn−4 ± 1) + ··· = 0, wherein the upper signs go together, and also the lower. When the upper signs are taken, the equation is divisible by x + 1, and — 1 is a root. When the lower signs are taken, the equation is divisible by x − 1, and 1 is a root. - 1 by x- 1, the first and last These terms, multiplied by a, When +1 is divided by x + 1, or x" terms of the quotient have the same sign. are the first and last terms of the depressed equation. In like manner, it follows that other equidistant terms in the depressed equation have the same sign. Hence, a reciprocal equation of an odd degree has a root-1 when equidistant coefficients have the same sign, and a root +1 when these coefficients have opposite signs. The depressed equation is a reciprocal equation of even degree, in which the coefficients of equidistant terms have the same sign. Reciprocal Equations of Even Degree. 75. Let at + A1x2-1 + A ̧ì¬2 + · — ɑπx2 — α1x — α = 0 be a reciprocal equation of even degree, in which equidistant coefficients have opposite signs. Grouping terms, we have 1, and 1 are roots. аo(x-1)+ α1x (x2−2 − 1) + A2x2 (x”−4 − 1) + ··· = 0. Since n is even, x" — 1 is divisible by x2 Therefore, a reciprocal equation of even degree, in which the coefficients of equidistant terms are opposite, has the roots +1. The depressed equation is a reciprocal equation of even degree in which the coefficients of equidistant terms have the same sign. 76. Standard Form of Reciprocal Equations. It follows from the preceding articles that any reciprocal equation can be reduced to one of even degree in which the coefficients of equidistant terms have the same sign, if it be not already in this form. This is therefore taken as the standard form of reciprocal equations. Ex. Reduce the equation 2+2 xa − 3 xo3 − 3 x2+2x+1=0 to the standard form. By Art. 74, this equation has the root — 1. Dividing by +1, we obtain the depressed equation x2 + x3 −4x2+x+1=0. 77. A reciprocal equation of the standard form can be reduced to an equation of half its degree. be a standard reciprocal equation, wherein a is the middle term. Dividing by x", and grouping terms having equal coefficients, we obtain 1 Evidently, " + and so on. is of the nth degree in z. Therefore, making the above substitutions, we obtain an equation of the nth degree in z. or Ex. Solve the equation -2x+2x-2x3+2x2-2x+1=0. Dividing by a and grouping terms, 2x2+ +2(x+ 2= ( ~ + 1) − 2 ( ~ + 1 ) + 2 (x + 1) − 2 Substituting z for x+ we have x 23-3z-2(x-2)+2x-2=0, 23-222 z+2=0. The roots of this equation are found to be 1, 1, 2. Sincea, or "+a, and the derived function na"-1 do not have a common factor, the n roots of a binomial equation are distinct. From (1) and (2), we obtain x = √/a, or x = -a. Since each equation has n distinct roots, we conclude that any positive or negative number has n distinct algebraical nth roots. 79. As in Ch. XXII., Art. 1, Ex. 1, the n roots of the equations · a = 0, x" + a = 0, can be obtained by multiplying the n roots of the equations 2"-10, x" + 1 = 0, by the arithmetical nth root of a. But the latter equations. are evidently reciprocal equations and can be solved by the methods of Arts. 74-77. Ex. Solve the equation 25+1 = 0. From (2), by the method of Art. 77, we obtain (1) (2) 3. 3x-2x3-34 x2-2x+3=0. 4. 6x4x3- 14 x2 −x + 6 = 0. 5. x56x + 13 x3 — 13x2 + 6x − 1 = 0. 42 +20 gio +5024 – 55 r2 – 20 tr – 4 = 0. 8. 12x6 + 16x5 – 25 x1 – 33 x3 — 25 x2 + 16x + 12 = 0. 80. Hitherto the only applications of the principles developed have been to the solution of numerical equations. Nowhere |