EXERCISES ON THE REDUCTION OF EQUATIONS. 1. Transform the equation 23 + 4x2 + 2x 2328 = 0, into one whose roots shall be less by 10; and this into another whose roots shall be still less by 2. Then transform the result into one whose roots shall be greater by 12. 2. Reduce the roots of 203 + 8x 34648584 = 0* successively by 300, 20, and 6. 3. Reduce the roots of the equation 23 18609625 0, successively by the numbers 200, 60, and 5; and then verify the process by increasing those of the result by 60, 200, and 5. PROBLEM III. 2n + Kxa To transform a given equation into another the roots of which shall be the same as those of the given one, but having all their signs reversed. Change the signs of the alternate terms, beginning with either the first or second; then this will be the equation required. For first, let the degree of the equation be even, as Amen + Bx2n-1 + Cx2n^2 + Kx? + L + M= 0, then writing - w instead of w, we shall have A(-x)2n+B (-x)21–1 +C (-2) 21–2+.... +K(-2)+L(-x)+M=0; or, Axen B.x2n-1 + Cx2n–2 Lx + M = 0. And, secondly, let it be of an odd degree, as Ax2n+1 + B.x2n + Cir2n-1 + Dx2n^2 + K23 + Luca + Mx + N= 0, in which, writing x for 2 we get A(-x) 3+1+B(-x)2 + C(-2)2--1+.... +K(-x)+L(-x)+M(-)+N=0; 2n* . * or, Ax2n+1 + Bx21 Cx2n-1 + K23 + Lx? M2 + N = 0; or, Ax2n+1 Bx2n + Cx2n=1 Læca + Mx — N= 0. Thus if the roots of the equation 24 + 0x3 - 25x2 + 60.3 36 = o be 3, 2, 1, — 6; then those of 24 F 0.x3 25x2 603 - 36 = 0 are -3, -2, -1, and + 6. + Kx3 EXERCISES. Change signs of the roots of the equations given in problems I. and II. the unctions in problem I. being equated to 0. PROBLEM IV. To transform an equation into another whose roots shall be the reciprocals of the roots of the given one. Reverse the order of the co-efficients : these will be the co-efficients of the new equation sought. For, if in Ax" + Bxn-1 + ...+ Lx2 + Mx + N= 0, we write y = 1 then substituting we get y' or x = * When any terms are deficient, their places must be kept and filled with 0; that being in such case the value of the coefficient of that term, as in the Synthetic Division. Find the equations whose roots are the reciprocals of those of the equations given in problems I. and II. equating the functions in the former to 0. Scholium. Any other proposed relations between the roots of a given equation and those of another to be determined, may be effected in an analogous manner, viz. : by first expressing the assigned relation between x and y, resolving for æ in terms of y, and substituting for x its value in the given equation. After the simplification of the expression to the utmost degree, we shall obtain the equation sought. For instance, to form an equation whose roots shall be m times those of the y given equation, put y = mx, or c = : then the equation m * When the coefficients of the direct and reciprocal equations are alike, that is, N = A, M=B, L = C, ... it is evident, that upon knowing the value of half the roots, the other half will be known from their being the reciprocals of these ; or, in other words, if ri, rg, . be roots, 1 1 then also will be roots also. This circumstance, as it lessens the work of solving an equation of such form, is important to be remarked. 1. Let the equation be of an even degree, having the above relation; then it may be written Ax2n + Bxm-1 + Cx2n-? + ...... Ca? + Bx + A=0, or again, 1 2n = 0; ar + =0 (1) nin Now 2n + A{r }+B {2-1 + }+c +c{2+*+ --++-;} + +={ e +}--n{x-=+=} ****+ } -"n="{ {++)} ---- (n = 1){ − c+==+ } = xn-4t -2) { ari }-(n = a-3 1 1 (n-1)(z It n 1.2 and so on. By these successive reductions we can convert (1) into the form Au" + B'un-1 + C'una? + = 0, ( where u = x 1 Then resolving (2) we find n values of u, and each of these substituted in x + = u gives two reciprocal values of x, and thereby furnish the 2n roots of the equation. The solution of Ex. 5, p. 194, is an exemplification of this circumstance. 2. Let the equation be of an odd degree, as the (2n + 1)th: then, since it is the same thing as A{220+1 + 1} + B{x2-1 + 1}++ C{121–3+ 1}x? + ...... = 0, in which, as all the powers of x within the brackets are odd, every bracketted term is divisible by x + 1. Hence x +1=0 is one of the component factors of the equation; and the given equation being divided by x + 1, gives a depressed equation of the 2nth degree; and hence also this is soluble by means of an equation of the nth degree, and a quadratic as in the last case. It would be just the same if the latter half of coefficients were written instead of +, since it would only change the sign between each two bracketted terms, and every term would then be divisible by x – 1= 0, and x = I would be the corresponding single root. The problem suggested at p. 194 may be taken as an illustration. Equations in which this relation exists are called reciprocal recurrents. ܪ m2 m Ammo + B2-1 + .... + L.x2 + Mx + N= 0, becomes My Ay" + mByn-1 + m Cyn-? + m"2. Lya + ma-1 My + m".N=0. Or again, to form an equation whose roots shall be 1 m m -1 equation, we have y or x = my, and the transformation leads to Am"y" + Bm"-ya-1 + ...... Lmạy: + Mmy + N = 0. These transformations are often useful in the solution of equations, and the arrangement of the work for effecting them is sufficiently obvious, without any examples. PROBLEM V. TO FORM THE EQUATION WHOSE ROOTS ARE ANY GIVEN NUMBERS. 1. Change the signs of all the given roots. 2. Put down 1, having annexed to it by its proper sign one of these changed roots ; and multiply this binomial by another of the changed roots, beginning the products one place to the right, and add up the columns into one single horizontal row. These will be the co-efficients with their proper signs of the quadratic equation whose roots are those two roots already used. 3. Multiply this horizontal row by another of the changed roots, placing as before the first product in the second column. The several sums will be the co-efficients with their proper signs of a cubic equation, whose roots are the three roots already used. 4. Proceed thus through all the roots : then the equation will be completed. For this is evidently only working by detached co-efficients, and availing ourselves of the contrivance of allowing the multiplicand to stand as the product of the multiplicand by 1, and also of omitting the actual exhibition of the multiplier beneath the multiplicand. The second term of the multiplier may be put in the margin, as in the example annexed to the rule. This will appear quite clearly upon working out one example at length. For this is the only application of the method of detached coefficients to the multiplication of a - a= 0, x — b= 0, together : and as (x—a) (2-6)..... = 0, is fulfilled by x = 0, x = b, therefore a, b, are roots of the equation, by the definition of a root. Thus, suppose it were required to form the equation whose roots are - 1, 1, 3, 4, and 5, would be as follows :- -1 - 1 .... a 1-0) - 1 1-3 - 3 + 0 + 3 1 - 3 1+ 3 L-4 4 + 12 + 4 12 1-7 + 11 + 7 12 [+ 5 5 - 35 + 55 + 35 · 60 1 - 2 - 24 + 62 + 23 - 60 Hence the equation is 2.5 224 — 2423 + 62x2 + 23% – 60 = 0. EXAMPLES. T 1. Form the equation whose roots are 1, 5, — 4, — 3, 1, and compare it with the example worked above by means of Problem III. 2. Form the equation whose roots are the reciprocals of each of these, viz. of }, }, - \, - , and - j; and also of — }, – , 4, }, and }, by means of Problem IV. 3. Form the equation whose roots are severally the first nine digits, taken alternately + and —, viz. + 1, - 2, + 3, .... ; and then form that whose routs are the reciprocals of these. 4. The three roots of a cubic equation are in arithmetical progression, they are all integers, and their sum is equal to their continued product. What is that cubic equation? 5. If any number of pairs of roots be of the form a + bv - 1 and a – 6v-1, a, + b, N-1, and a, – b, N-1, show that the co-efficients of the equation will be real; or, in other words, that the imaginary symbols will disappear from the result. THEOREM III. If an equation given in terms of x be transformed into one in terms of x - a, then the several coefficients will be f.(a) f., (a) fu-,(a) ), , fi (a) film) 1.2 f(a), ..n' 1.2 (n-1) 1.2 (n-2)' 1.2' wherein a takes the place of x in the given equation and its successive n derivatives. This is established at once by completing the transformation in terms of a, and resolving the numerical coefficients into factors in terms of n. The only difficulty in giving the successive steps of the work in this place, is the extent of page which it would require ; but as this does not hold in the student's practice, he should exercise himself in the actual reduction of four or five of these coefficients *. -2 n + Bla + 2)--- = Ba-1 + (n = 1 ) * This theorem admits of an elegant demonstration, by means of the binomial theorem, as follows: Put x = a + z: then expanding the function, A.xin + Brna + so that like powers of z stand in the same vertical columns, we have nAan-1 nln - 1) AanAla + )" nan -- )(n— 2) Aan—3 ] 23 +... 1.2.3 (n-1)(n-2) Ba"-3 (n-1) (n—2)(n-3) Bar-4 23 + ... 1.2 1.2.3 C(a + zja—2 = Can- + (n—2) Can—3 (n—2)(n-3) Can- (n—2)(n-3)(n - 4) Can-5 z+ z? + 1 z3 + ... 1.2 1.2.3 and so on to the end. In this we see at once that by adding vertically, and attending to the forms taken by the derivatives, the several coefficients of z are as stated in the text; and that we have f2(a) fr-1(a) f(a + x) = f(a) + fila) fr (a) 2-1 + 1.2 fr (a) x (x -- a)"; (܂ ...... ... n f(x) = f(a)+1; (a)(x – a) + f(a) (r – a)? + ... + 1.2 or, again, reversing the order of the terms, it is 1.2 ... 1 = THEOREM IV. If the equation f (x) = 0 have p equal roots, then fi () = 0 will have р · 1 of them, f2(X) = 0 will have p 2 of them, and so on; till fp-1(x) = 0 will have p (P — 1) = 1 of them, and fp (x) = 0 will have P-p= 0 of them. For, since f (2) = 0 contains p roots each equal to r, the function f(x) is divisible р times successively by x r without remainder (theor. 1). Hence the p last coefficients, viz. N,, M2, Lz ..... in the operation of Problem II. will But these coefficients are respectively fr-1 () 1.2.3' 1.2...(p-1); and as the denominators are all finite, it follows that the several numerators are equal to 0. That is, fi (r) = 0, $2 (r) = 0,.. 0; and therefore the value of r, which fulfils the equation f (x) = 0, fulfils also the p - 1 equations fi(x) = 0, $2(x) = 0, .... f-1() = 0*. be zero. fp (r) 2 пA fula) fr_ila) f (x)= fula 1.2 which is the same result as above stated.' One important use of this problem in the older methods of solution of equations, was to enable us to remove any specified term from the equation by rendering its coefficient equal to zero. It only required us to solve an equation of an inferior degree, viz. of the first degree to remove the second term, of the second degree to remove the third term, and so on. For evidently, if we find such a value of a in these several coefficients as would render them respectively zero, our object would be aceomplished. That is, to solve the equation fola)=0, which is of the (144p)th degree, for the unknown quantity a. B Thus, fn-1(a)=(nAa+B)(n-1)(n—2) ...... 2.1=1, or nAa+B=0; or a=fr-2(a)={n(n-1)Aa +(n-1)Ba+C}(n-2)......2.1=0; or, n(n-1)Aa?+(n-1)Ba+C=0. Hence there are two values of a (either real or imaginary), which will remove the third term. In the same way by solving fr-3(a)=1, fn-(a)=0, and so on, we may remove any term whatever. It may be remarked, that to remove the last term we must solve f(a)=0, or the given equation. This use is, however, superseded by improved methods of solution ; but it is still applicable to many others of great importance, some of which will be made apparent in this work. * The following method may perhaps be found more intelligible to some students than that in the text. Since f(x)= 0 contains p roots equal to r, the function f (x) is divisible by x — r successively p times ; and as r. is a root of the depressed equations till the last, we may write x for it in each of them successively. But this transforms the several depressed equations into the several derivatives; and thus we have Axn + Bxn-1 + Corne2 + + Lx2 + Mx + N=0= f (x) nAXn-1 + (n − 1) Bxa2 + (n − 2) Cwm-3 + + 2Lx + M =1= filx) nan — 1) Axr-+ (n − 1) (n − 2) Bxn-3 + =1= f(x) P + 2L =0 = f(x) (1 – 1) (n - 2 + 1) Axn-pt. f (x) = 0 has p roots equal to r |