MHB What Values of k Make the Equation Have One Non-Negative Root?

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    2017
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The discussion focuses on determining the values of k for which the equation involving x has exactly one non-negative root. Participants are encouraged to solve the equation and share their findings. A solution has already been provided by a user named kaliprasad, which is acknowledged in the thread. The equation combines rational expressions and absolute values, adding complexity to the problem. The thread emphasizes the importance of following the guidelines for problem-solving submissions.
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Hi MHB, sorry for missing one week of high school's POTW, I guess I can make it up by posting two POTWs this week.(Blush)

Here is this week's another POTW:

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Find all values of $k$ such that the equation

$\left(\dfrac{1}{x+k}+\dfrac{k}{x-k}-\dfrac{2k}{k^2-x^2}\right)(|x-k|-k)=0$

has exactly one non-negative root.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to kaliprasad for his correct solution, which you can find below::)

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

Multiply both sides by $ (x^2-k^2)$, we get

$((x-k) + k(x+k) + 2k) (\left| x -k \right| -k) = 0$

or $(x+k) (k+1) (\left| x -k \right| -k) = 0$

As $x$ cannot be $-k$ we have

$(\left | x -k \right| -k) = 0$

$|x-k | = k$, this gives 2 values of $x$, where $x = 0$ or $2k$.

Hence there is no solution to the problem.
 
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