Find the sum of all positive integers k for which 5x2−2kx+1&lt;0 has exactly one integral solution.My attempt is as follows:(x−d2k−4k2−2010)(x−d2k+4k2−2010)&lt;0(x−dk−k2−55)(x−dk+k2−55)&lt;0x∈(dk−k2−55,dk+k2−55)As it is given that it has got only one integral solution, so there must be exactly one integer between dk−k2−55 and dk+k2−55.Let x1=dk−k2−55 and x2=dk+k2−55, then [x2]−[x1]=1 where [] is a greater integer function.But from here, how to proceed? Please help me in this.

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Find the sum of all positive integers k for which 5x2−2kx+1&amp;lt;0 has exactly one integral solution.My attempt is as follows:(x−d2k−4k2−2010)(x−d2k+4k2−2010)&amp;lt;0(x−dk−k2−55)(x−dk+k2−55)&amp;lt;0x∈(dk−k2−55,dk+k2−55)As it is given that it has got only one integral solution, so there must be exactly one integer between dk−k2−55 and dk+k2−55.Let x1=dk−k2−55 and x2=dk+k2−55, then [x2]−[x1]=1 where [] is a greater integer function.But from here, how to proceed? Please help me in this.

Frain4i62 · Accepted Answer

Your idea is good; you want to find all positive integers k for which there is precisely on integer between the roots of5x2−2kx+1=0.Then the distance between the roots can be at most 2, where the distance between the roots is precisely15(−2k)2−4⋅1⋅5=25k2−5,as you already found. This is at most 2 if and only if k2−5≤5, or equivalently k≤5. This leaves only 5 values of k to check.

Find the sum of all positive integers k

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