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.
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.
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.
Athul,can u explain why "implies ab-9=m^2 " can become " using ab=m^2 -9"?
To find the maximum possible value of ab, we need to determine the range of values for a and b that satisfy the given conditions.
Let's start by finding the roots of the polynomial P(x). We can use the Vieta's formulas to relate the roots to the coefficients of the polynomial.
According to Vieta's formulas, the sum of the roots is given by:
r + s + t = a
Also, the product of the roots taken two at a time is given by:
rs + st + rt = a^2b^3
Furthermore, the product of all three roots is given by:
rst = -9a^2b^2
Now, our target is to find the value of (r+s)(s+t)(r+t). Let's expand it using the values we obtained.
(r+s)(s+t)(r+t) = (r^2s + rst + s^2t + rt^2 + srt + rst) = r^2s + s^2t + t^2r + 3rst
We can substitute the given values from the Vieta's formulas into this expression to simplify it:
(r+s)(s+t)(r+t) = (a - t)(a - r)(a - s) = -a^3 + (r + s + t)a^2 - (rs + st + rt)a + rst
Substituting the values we found earlier, the expression becomes:
(r+s)(s+t)(r+t) = -a^3 + a^3 - a^2(rt + rs + st) + (-9a^2b^2) = -a^2(rt + rs + st) - 9a^2b^2
Since we are given that (r+s)(s+t)(r+t) = k^2, we have:
k^2 = -a^2(rt + rs + st) - 9a^2b^2
Since a and b are positive integers, rt + rs + st must be a negative term. Therefore, the maximum possible value of (r+s)(s+t)(r+t) is 0 (which occurs when rt + rs + st = 0).
Now, we can set the expression for (r+s)(s+t)(r+t) to 0, and solve for a and b:
-a^2(0) - 9a^2b^2 = 0
9a^2b^2 = 0
From this equation, we can see that ab must be equal to 0 in order to satisfy the condition. Since a and b are positive integers, the maximum possible value of ab is 0.
Therefore, the maximum possible value of ab is 0.