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Find all values of parameter $a$, when sum of solutions of following equation is $100$. $$ \sin(\sqrt{ax-x^2})=0 $$

I tried to get rid of that $sin$ and there was quadratic equation with two parameters ($a$ and $k$($\pi$$k$)). After that I did pretty much nothing productive. Could you please help me with this equation?
Thanks in advance

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If $a\leq0$ then $ax-x^2\geq0$ only when $a\leq x\leq 0$; such $x$-values cannot sum up to $100$. Therefore we may assume $a>0$. Put $x:={a\over2}+y$. Then $$\sqrt{(a-x)x}=k\pi, \quad k\geq0,$$ translates into $$y^2={1\over4}\bigl(a^2-(2k\pi)^2\bigr)\ .\tag{1}$$ Assume that $$2n\pi<a<2(n+1)\pi, \qquad n\geq0\ .$$ Then $(1)$ admits $2(n+1)$ real solutions $\pm y_k\ne0$ by letting $0\leq k\leq n$. These give rise to $2(n+1)$ corresponding $x$-values $$x_k^+={a\over2}+y_k,\qquad x_k^-={a\over2}-y_k\ .$$ Their sum is $(n+1)a$, which then enforces $a={100\over n+1}$. We now have to check whether this value is within the bounds assumed for $a$, i.e., whether $$2n\pi<{100\over n+1}<2(n+1)\pi\ .\tag{2}$$ On the one hand $(2)$ enforces $(n+1)^2>{50\over \pi}\doteq15.9$, hence $n\geq3$. On the other hand $n(n+1)<{50\over\pi}$ is true for $n=3$, but is already violated for $n=4$. It follows that $n=3$ is the only admissible $n$, so that we obtain the only solution $a={100\over n+1}=25$ of the original problem.

I leave the case $a=2n\pi$, $n\geq1$ to you.

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hint: notice $x$ belongs $(0,a)$ where $0$ and $a$ both are roots , and if $x=\frac{a}{2}+k$ is a root so is $x=\frac{a}{2}-k$ . now try to look into how many roots does this equation have. assuming $a$ is positive for now , you can use the same logic for negative $a$ as for positive $a$

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  • $\begingroup$ Why does $x\in(0,a)$? What about $x=-\frac{a}{2}$? $\endgroup$ – almagest Jun 24 '16 at 11:29
  • $\begingroup$ Sqrt value becomes negative $\endgroup$ – avz2611 Jun 24 '16 at 11:30
  • $\begingroup$ Only if $a>0$. We are not told that is the case. $\endgroup$ – almagest Jun 24 '16 at 11:32
  • $\begingroup$ Your right i should edit that $\endgroup$ – avz2611 Jun 24 '16 at 11:33
  • $\begingroup$ but basic logic of the solution still remains the same $\endgroup$ – avz2611 Jun 24 '16 at 11:35

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