2024/007) Find the value of k such that $(\frac{1}{x+k} + \frac{k}{x-k} + \frac{2k}{k^2-x^2})(| x-k | -k) = 0$ has exactly one non -ve root

Reference: https://www.physicsforums.com/threads/discovering-the-solution-problem-of-the-week-261-apr-23rd-2017.1037211/

Note that we cannot have or . Multiply both sides by , we get or As cannot be we have , this gives 2 values of , where or . Hence there is no solution to the problem.

Reference: https://www.physicsforums.com/threads/discovering-the-solution-problem-of-the-week-261-apr-23rd-2017.1037211/

Note that we cannot have $x=k$ or $x = -k$ 

So multiplying both sides by $x^2-k^2$

We get 

$((x-k) + k(x+k) + 2k) )(| x-k |  -k) = 0$

Or $(x+k)(k+1))(| x-k |  -k)=0$

As x cannot be -k  $(| x-k |  -k)=0$

So we get 2 values of x that is 2k or 0

Hence no solution