In try to figure out the exercise:
Let $f(x)=\sum_{k=1}^{n}c_ke^{\lambda_kx}$where $\lambda_i \not=\lambda_j,i\not=j$,and $c_1^2+c_2^2+\dots+c_n^2\not=0$, then the number of $f(x)$'s roots is strictly less than $n$.
My approach(this way can't deal with $f(x)$ has repeated root):
assume $x_1,x_2,\dots,x_n$ are $f(x)$'s roots,and $x_i\not=x_j$ if $i\not=j$. then I get a linear equations about $c_1,c_2,\dots,c_n$: $e^{\lambda_1x_1}c_1+e^{\lambda_2x_1}c_2+\dots+e^{\lambda_nx_1}c_n=0$ $e^{\lambda_1x_2}c_1+e^{\lambda_2x_2}c_2+\dots+e^{\lambda_nx_2}c_n=0$ $\dots\dots\dots\dots\dots\dots$ $e^{\lambda_1x_n}c_1+e^{\lambda_2x_n}c_2+\dots+e^{\lambda_nx_n}c_n=0$ I want to show that the solution to this linear equations are $0$,it will be a contradiction.but i can't figure out its determinant of coefficient: $\begin{vmatrix} e^{\lambda_1x_1}& e^{\lambda_2x_1} &\dots &e^{\lambda_nx_1}\\ e^{\lambda_1x_2}& e^{\lambda_2x_2} &\dots&e^{\lambda_nx_2} \\ \dots&\dots &\dots&\dots \\ e^{\lambda_1x_n}&e^{\lambda_2x_n} &\dots &e^{\lambda_nx_n} \end{vmatrix}\not=0$