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Algebra

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4 Mar 2008 19:27:04 IST

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Quadratic Prob

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Find all positive integers k, such that 5x²-2kx+1<0 for exactly one integer.

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sreeraman nagasubramaniyan

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4 Mar 2008 21:46:55 IST

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5x²-2kx+1

since k is positive integer ..-2kx is going to positive for all negative x and hence 5x²-2kx+1cannot be less than 0 for negative numbers

now another condition is that <0 shud hold for only one integral x

now if it s true for 2 lets say ..then it is going to be true for 1 also..

so that makes it 2 integers

so i think only way is condition is to be satisfied for 1 but not for 2

5-2k+1<0

k>3

20-4k+1>0

k<=5

so i think the two values are 4 and 5

Hari Shankar

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4 Mar 2008 22:12:50 IST

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f(x) = ax²+bx+c = a(x-p)(x-q), where p and q are the roots.

If a>0, and p and q are complex, then f(x)>0 for all x. Hence, here p and q are to be real.

Hence, D>0. This means k²>=5.

Now, when a>0, f(x)<0 for p<x<q.

If it should be positive for only one integer, p-q<2

p-q =D/a<2. This gives k²<30.

The natural numbers that satisfy this condition are k=3,4 and 5

k=3 however doesnt satisfy the condition which leaves only 4 and 5.

This is a rigorous solution, but I think sboosy's is a nice and intuitively appealing solution and definitely worth at least two looks.

Sai Ganesh Bandiatmakur

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5 Mar 2008 21:48:19 IST

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why p-q<2?

Hari Shankar

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5 Mar 2008 21:59:07 IST

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So that not more than one integer lies between p and q.

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