# Find all values of parameter a, when sum of solutions of following equation is 100

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$$k)). After that I did pretty much nothing productive. Could you please help me with this equation? Thanks in advance ## 2 Answers If a\leq0 then ax-x^2\geq0 only when a\leq x\leq 0; such x-values cannot sum up to 100. Therefore we may assume a>0. Put x:={a\over2}+y. Then$$\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. 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$a$• Why does$x\in(0,a)$? What about$x=-\frac{a}{2}$? – almagest Jun 24 '16 at 11:29 • Sqrt value becomes negative – avz2611 Jun 24 '16 at 11:30 • Only if$a>0\$. We are not told that is the case. – almagest Jun 24 '16 at 11:32
• Your right i should edit that – avz2611 Jun 24 '16 at 11:33
• but basic logic of the solution still remains the same – avz2611 Jun 24 '16 at 11:35