The lower bound!

Algebra Level 5

The expression x 4 + 2 p x 3 + x 2 + 2 p x + 1 = 0 x^4+2px^3+x^2+2px+1=0 has at least two distinct negative roots \textbf{at least two distinct negative roots} for p > a b p>\frac{a}{b} where a , b a, b are co-prime. Find a b a^b .


The answer is 81.

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1 solution

Pranjal Jain
Nov 1, 2014

Divide throughout by x 2 x^{2} and substitute x + 1 x x+\frac{1}{x} = α \alpha

This will give a quadratic in α \alpha . α 2 + 2 p α 1 = 0 \alpha^{2}+2p\alpha-1=0 Since atleast two -ve values of x must satisfy, one root of this quadratic must be less than 2.

Range of x + 1 x x+\frac{1}{x} is (-∞,-2] \cup [2,∞)

This will give a condition that p> 3 4 \frac{3}{4} So answer will be 3 4 = 81 \boxed{3^{4}=81}

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