Problem: For which $a$ will the sum of solutions be equal to $100$, in $\sin(\sqrt{ax-x^2})=0$.
The attempt at a solution: For $\sin(x)=0$, $x$ must be equal to $0$, so we get $$\sqrt{ax-x^2}=0\\ax-x^2=0\\x(a-x)=0\\x=0\\a-x=0$$last 2 are the solutions as I understand but I can't seem to finalize the solution, please help. Thank you in advance.
On
I think there is no such $a$. Note that
$$\sin(\sqrt{ax-x^2}) = 0 \iff \sqrt{ax-x^2}=k \pi$$
for some $k \in \mathbb{N}$. Squaring both sides of the second equation and solving the second degree equation shows
$$x_{1,k} = \frac{a}{2} +\frac{\sqrt{a^2-4k^2\pi^2}}{2} \text{ and }x_{2,k} = \frac{a}{2} -\frac{\sqrt{a^2-4k^2\pi^2}}{2}$$
In particular we have $x_{1,k}+x_{2,k} = a$. These solutions exists as long as $$a^2 - 4\pi^2k^2 \geq 0 \iff \frac{|a|}{2\pi} \geq |k|.$$ In particular if $N$ denotes the integer part of $\frac{|a|}{2\pi}$, then the sum of all solutions is give by $S=a2N$. Asking for $S=100$ leads to the equation $$a(2\lfloor |a|/2\pi\rfloor+1) = 100,$$ where $\lfloor \cdot \rfloor$ denotes the floor function. It follows that there is no solution.
You are correct so far, to answer the question, all you need to do is set sum of the solutions equal to $100$, $$x+a-x=100\\a=100$$ so a must be equal to $100$ in order for sum of solutions to be $100$.