11Nishant Sir, Olympiads do not allow usage of Calculus.....
Determine all pairs of positive integers (a,b) such that all roots of
x^3+8x^2+2^ax+2^b = 0 are real and at least one root is an
integer.
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5 Answers
62this seems simple to me...
I dont know if i am thinking in the right direction.
all we need is all +ve values of a for which
f(X) = x3+8x2+2ax should be <-2 for some value of x.
the derivative is
3x2+16x+2a
roots of which are {-16±√256-12.2a}/6
8/3{-1±√1-3.2a-3}
1-3.2a-3>0
1>3.2a-3
2a-3<1/3
a=1 is the only way by which the derivative will have 2 real roots.
in that case, the roots are given by
8/3{-1±1/2}
= -4, -4/3
These are the points where the derivative is zero...
now f(-4) should be +ve and f(-4/3) should be -ve
We can also say that all roots will be -ve
9x = -2λ for integer solutions
first use this and then try satisfying condition 1
9a=4
b=3
theres a nice method to solve
see gvn is p(x)
now P(x) =( x+2λ)f(x)
can someone proceed ?
