Polynomial but quadratic this time
Consider polynomials of the form with integer coefficients which have two distinct zeros in the interval . Find the least positive integer value of for which such a polynomial exists.
The answer is 5.
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1 solution
Let f(x)= ax²-bx+c(coefficientof x is changed so that we can solve for a positive value of b). Product of roots is c/a<1=>c<a, but the minimum value of c can be 1(not 0 since that would make 0 a root which doesn't lie in the interval as stated). Now, f(1)>0 since the roots are less than 1, which implies a+c>b => a+c>=b+1 => a>=b+1-c=b (since min value of c was 1) => a=b for min values. Since the quadratic has real, distinct roots, b²-4ac>0 => a²-4a>0 => a>4. Thus min value of a,b,c are 5,5,1.