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.
8 answers
hard problem
The answer is 439
wrong ans
Dude Please Help
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.
(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.
775 is the right answer
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.
Find the sum of squares of all real roots of the polynomial f(x)=x^5−7x^3+2x^2−30x+6.
Athul,can u explain why "implies ab-9=m^2 " can become " using ab=m^2 -9"?
4 answers