Nice Prob

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.

5 Answers

62
Lokesh Verma ·

this 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

11
Devil ·

Nishant Sir, Olympiads do not allow usage of Calculus.....

9
Celestine preetham ·

x = -2λ for integer solutions
first use this and then try satisfying condition 1

9
Celestine preetham ·

small mistake nishant sir

2a-6<1/3

9
Celestine preetham ·

a=4
b=3

theres a nice method to solve

see gvn is p(x)

now P(x) =( x+2λ)f(x)

can someone proceed ?

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