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 Another Quintic Problem (Posted on 2006-05-31)
The equation x5 –23x3 –66x2 +px +q =0 has three non-zero roots, and two of them are integers, with multiplicity 2.

Find p and q and determine all the roots of the given equation.

 Submitted by K Sengupta Rating: 3.6667 (3 votes) Solution: (Hide) p= - 68; q = -24 and the roots of the equation under reference are -1, -1, -2, -2 and 6. EXPLANATION: FIRST METHOD ( On the basis of the solution submitted by Old Original Oskar! and comments by Frederico Kereki) The sum of the five roots is 0. Accordingly, the roots are a, a, b, b, and c=-2(a+b). The sum of the pairwise products of the roots is: a2+b2+4ab+2(a+b)c = -3(a2+b2)-4ab= -23; so that: 3a2+4ba+(3b2-23)= 0. Solving this equation for a gives a=(-4b+/-sqrt(276-20b2))/6, which produces integer solutions only whenever, b=+/-1 (then a=-2) or b=+/-2 (then a=-1) Since a and b are symmetric, the feasible solutions are : (a,b) = (-2, -1); giving, c = -2(a+b) = 6. Consequently, the required roots are: -2, -2, -1, -1 and 6. This checks out since: (x+2)2(x+1)2(x-6) = x5 - 23x3 -66x2- 68x - 24; so that p = -68 and q = -24 in terms of the problem under reference. --------------------------------------------------------------------------------------------------------------------------------- SECOND METHOD Let the two distinct roots each occurring with multiplicity 2 be denoted by a and b. Then, by conditions of the problem: x^5 - 23*(x^3) - 66*(x^2)+ px + q = (x-a)^2 *(x-b)^2*( x +(q/((ab)^2))) Comparing the coefficients corresponding to the various exponents of x in both sides of the aforementioned relationship, we obtain: (i) q/(ab)^2 = 2(a+b) (ii) (-2q(a+b)/((ab)^2)+ b^2 + 4ab + a^2 = -23 (iii) -2ab(a+b) +(b^2+4ab+a^2)*(q/((ab)^2)) = -66 (iv) (ab)^2 - 2ab(a+b)*(q/((ab)^2)) = 1 By (i); q = 2*(ab)^2 *(a+b) Substituting the above value of q in (ii) and simplifying, we obtain: ab = (3*(a+b)^2 - 23)/2 -----------------(#) Substituting the above value of ab in (iii) and simplifying, we obtain: 5*(a+b)^3 - 23*(a+b) = -66 It can easily be established with a little trial and error that the above equation is satisfied whenever a+b = -3 and substituting this in (#), we obtain ab = 2 Solving for (a+b= -3, ab=2) we obtain (a,b)=(-2,-1),or(-1.-2) Without any loss of generality, we take (a,b)=(-2,-1) giving: q = 2*((ab)^2)*(a+b) = -24 and p = (ab)^2 - 2ab(a+b)*(q/((ab)^2))= -68 Hence, p=-68; q= -24; a= -2; b=-1 giving q/((ab)^2))= -6, so that: f(x) = x^5 -23*(x^3) -66*(x^2)-68x - 24 or, f(x) = ((x+1)^2)* ((x+2)^2)*(x-6) Hence, f(x) = 0 gives x=-2,-2,-1,-1 and 6 so that: p =- 68 ; q =- 24 and the roots of the equation under reference are -2,-2,-1,-1 and 6.

 Subject Author Date Acknowledgements K Sengupta 2006-06-15 03:33:42 re(2): Who needs p and q? (SPOILER) Old Original Oskar! 2006-05-31 21:49:13 re: Who needs p and q? (SPOILER) Federico Kereki 2006-05-31 14:51:12 re: Who needs p and q? (SPOILER) K Sengupta 2006-05-31 14:40:45 Who needs p and q? (SPOILER) Old Original Oskar! 2006-05-31 13:11:01

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