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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.

 See The Solution Submitted by K Sengupta Rating: 3.6667 (3 votes)

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 Who needs p and q? (SPOILER) | Comment 1 of 5
The sum of the five roots is 0, so 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 3a²+4ba+(3b²-23)= 0.

Solving this equation for a gives a=(-4b+/-√(276-20b²))/6, which produces integer solutions only for b=+/-1 (then a=2) or b=+/-2 (then a=1); since a and b are symmetric, the solution is a=2 and b=1.

So, the roots are 2, 2, 1, 1, and -6. The polynomial is (x-2)².(x-1)².(x+6)= x^5-23x^3-66x^2-68x+24.

Edited on May 31, 2006, 1:11 pm

Edited on May 31, 2006, 1:12 pm

Edited on May 31, 2006, 1:13 pm
 Posted by Old Original Oskar! on 2006-05-31 13:11:01

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