Posted by **Mathslover Please help** on Monday, May 6, 2013 at 12:05am.

Suppose a and b are positive integers satisfying 1≤a≤31, 1≤b≤31 such that the polynomial P(x)=x^3−ax^2+a^2b^3x+9a^2b^2 has roots r, s, and t.

Given that there exists a positive integer k such that (r+s)(s+t)(r+t)=k^2, compute the maximum possible value of ab.

- Math (algebra) -
**hans**, Monday, May 6, 2013 at 12:14pm
hard problem

- Math (algebra) -
**John**, Monday, May 6, 2013 at 7:14pm
The answer is 439

- Math (algebra) -
**Anonymous**, Tuesday, May 7, 2013 at 12:01am
wrong ans

- Math (algebra) -
**Mathslover Please help**, Tuesday, May 7, 2013 at 12:11am
Dude Please Help

- Math (algebra) -
**Athul**, Tuesday, May 7, 2013 at 11:12am
775

(r+s)(s+t)(r+t)= (r+s+t)(rs+sr+st)-rst

=a(a^2.b^3)- (-9ab)=k^2

implies

a^2.b^2(ab-9)= k^2

implies ab-9=m^2 where m is an integer as k^2 is a perfect square

given max value of a and b can be 31 and without loss of generality we take a<b, then we find b-a =6 using ab=m^2 -9

therefore max of b=31 then a = 25 therefore ab= 775 which is 9 less than 28^2.

- Math (algebra) -
**amgad ahmed**, Tuesday, May 7, 2013 at 11:21am
775 is the right answer

- Math (algebra) -
**Mathslover Please help**, Wednesday, May 8, 2013 at 1:50am
thanxxxx Sir . Can yiu please do this question also

Find the sum of squares of all real roots of the polynomial f(x)=x^5−7x^3+2x^2−30x+6.

- Math (algebra) -
**hans**, Wednesday, May 8, 2013 at 8:07am
Athul,can u explain why "implies ab-9=m^2 " can become " using ab=m^2 -9"?

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