ComplexPhi wrote: Let $a_1,a_2,\cdots ,a_n$ be real numbers, not all zero. Prove that the equation: \[\sqrt{1+a_1x}+\sqrt{1+a_2x}+\cdots +\sqrt{1+a_nx}=n\] has at most one real nonzero root. LHS is concave and so equation has at most two real roots, one of them being $0$. Hence the result.

can you pls elaborate the solution?

What part of the proof you dont understand ? 1) LHS is concave ? 2) So equation has at most two real roots ? 3) $0$ is a real root ? 4 ) hence equation has at most one real nonzero root ?

Alternatively, $f(x) = \sum_{k=1}^n \sqrt{1+a_k x} - n = \sum_{k=1}^n (\sqrt{1+a_k x} - 1) = x\sum_{k=1}^n \dfrac {a_k}{\sqrt{1+a_k x} + 1}$. The derivative of $f(x)/x$ is $-\sum_{k=1}^n \dfrac {a_k^2}{2\sqrt{1+a_k x}(\sqrt{1+a_k x} + 1)^2} < 0$, thus $f(x)/x$ is decreasing on its domain, and so vanishes at most at one point $x\neq 0$.